problem solving of cubic equation

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Happiness Hub Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• Happiness Hub
• This Or That Game
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Solve a Cubic Equation

Last Updated: August 15, 2023 Fact Checked

This article was co-authored by David Jia and by wikiHow staff writer, Christopher M. Osborne, PhD . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,150,926 times.

Solving Cubic Equations without a Constant

• You can also factor it by grouping .

Finding Integer Solutions with Factor Lists

• Synthetic division is a complex topic that’s beyond the scope of describing fully here. However, here's a sample of how to find one of the solutions to your cubic equation with synthetic division: -1 | 2 9 13 6 __| -2-7-6 __| 2 7 6 0

Using a Discriminant Approach

• A cubic equation always has at least one real solution, because the graph will always cross the x-axis at least once.

• You can solve the example by checking the answer when n is equal to 1 , 2 , and 3 . The answers you get from these tests are the possible answers to the cubic equation — any that give an answer of 0 when plugged into the equation are correct.

Practice Problems and Answers

Expert Q&A

You Might Also Like

• ↑ https://brilliant.org/wiki/cubic-equations/
• ↑ https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-cubicequations-2009-1.pdf
• ↑ https://sciencing.com/solve-cubic-equations-8136094.html
• ↑ https://www.purplemath.com/modules/quadform.htm
• ↑ https://math.vanderbilt.edu/schectex/courses/cubic/
• ↑ https://www.geeksforgeeks.org/solving-cubic-equations/
• ↑ https://mathworld.wolfram.com/CubicFormula.html
• ↑ https://courses.cs.washington.edu/courses/cse590b/13au/lecture_notes/cubic.pdf

About This Article

To solve a cubic equation, start by determining if your equation has a constant. If it doesn't, factor an x out and use the quadratic formula to solve the remaining quadratic equation. If it does have a constant, you won't be able to use the quadratic formula. Instead, find all of the factors of a and d in the equation and then divide the factors of a by the factors of d. Then, plug each answer into the equation to see which one equals 0. Whichever integer equals 0 is your answer. Read on to learn how to solve a cubic equation using a discriminant approach! Did this summary help you? Yes No

• Send fan mail to authors

Reader Success Stories

Sep 14, 2017

Did this article help you?

Feb 5, 2017

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

Cubic Equation

A cubic equation is an equation of the form:

A cubic equation has 3 roots , either all real OR one real, two complex .

• 1.1.1 TLDR?
• 1.2.1 TLDR?
• 2.1 If you're just asking for the formula for a monic cubic...
• 2.2 If you're asking for the formula for a depressed monic cubic...

Solving Cubic Equations

If you're too lazy to follow, look at subsection "TLDR" for each section.

Converting to a Depressed Equation

Solving the Depressed Equation

The Cubic formula

The cubic formula can be obtained by using the above method. These are the steps:

One last piece of advice: Don't try to memorize this. Memorize the process (shortcut: just look at TLDR for each section). Here is another way to do it.

If you're just asking for the formula for a monic cubic...

If you're asking for the formula for a depressed monic cubic...

Something appears to not have loaded correctly.

Click to refresh .

Solving Cubic Equations

In these lessons, we will learn how to solve cubic equations using the Remainder Theorem and the Factor Theorem.

Related Pages Cubic Functions Factor Theorem Remainder Theorem More Algebra Lessons More Algebra Worksheets More Algebra Games

What is the Remainder Theorem?

If a polynomial, f(x), is divided by x - k, the remainder is equal to f(k).

What is the Factor Theorem?

x - k is a factor of the polynomial f(x) if and only if f(k) = 0

How to solve cubic equations using the Factor Theorem?

In these lessons, we will consider how to solve cubic equations of the form px 3 + qx 2 + rx + s = 0 where p, q, r and s are constants by using the Factor Theorem and Synthetic Division.

The following diagram shows an example of solving cubic equations. Scroll down the page for more examples and solutions on how to solve cubic equations.

Example: Find the roots of f(x) = 2x 3 + 3x 2 – 11x – 6 = 0, given that it has at least one integer root.

Step 1: Use the factor theorem to test the possible values by trial and error.

f(1) = 2 + 3 – 11 – 6 ≠ 0 f(–1) = –2 + 3 + 11 – 6 ≠ 0 f(2) = 16 + 12 – 22 – 6 = 0 We find that the integer root is 2.

Step 2: Find the other roots either by inspection or by synthetic division .

2x 3 + 3x 2 – 11x – 6 = (x – 2)(ax 2 + bx + c) = (x – 2)(2x 2 + bx + 3) = (x – 2)(2x 2 + 7x + 3) = (x – 2)(2x + 1)(x +3)

Example: Solve the cubic equation x 3 – 7x 2 + 4x + 12 = 0

Solution: Let f(x) = x 3 – 7x 2 + 4x + 12

We find that f(–1) = –1 – 7 – 4 + 12 = 0

So, (x + 1) is a factor of f(x)

x 3 – 7x 2 + 4x + 12 = (x + 1)(x 2 – 8x + 12) = (x + 1)(x – 2)(x – 6) So, the roots are –1, 2, 6

What are The Remainder Theorem and the Factor Theorem? How to use the Theorems to find the linear factorization of a polynomial? Example: Factor F(x) = 2x 3 − 3x 2 − 5x + 6

How to use the Factor Theorem to factor polynomials? Examples:

• Factor P(x) = 3x 3 − x 2 − 10x + 8
• Factor P(x) = 2x 3 − 9x 2 + x + 12

How to use the Factor Theorem to solve a cubic equation? If f(x) is a polynomial and f(p) = 0 then x - p is a factor of f(x) Example: Solve the equation 2x 3 −5x 2 − 10 = 23x

How to solve Cubic Equations using the Factor theorem and Long Division? Example: Find the roots of the cubic equation 2x 3 − 6x 2 + 7x − 1 = 0

How to solve Cubic Equations using the Factor theorem and Synthetic Division? Example: Show that x + 3 is a factor of x 3 − 19x − 30 = 0. Then find the remaining factors of f(x)

How to solve cubic equation problems? Example: 3x 3 −4x 2 − 17x = x 3 + 3x 2 − 10 Step 1: Set one side of equation equal to 0. Step 2: Collect like terms. Step 3: Factorize using the Factor Theorem and Long Division

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators
• Solving Cubic Equations – Methods & Examples

Solving Cubic Equations – Methods & Examples

This article will discuss how to solve the cubic equations using different methods such as the division method, Factor Theorem, and factoring by grouping.

But before getting into this topic, let’s discuss what a polynomial and cubic equation is.

A polynomial is an algebraic expression with one or more terms in which an addition or a subtraction sign separates a constant and a variable.

How to Solve Cubic Equations?

The traditional way of solving a cubic equation is to reduce it to a quadratic equation and then solve it either by factoring or quadratic formula.

Like a quadratic equation has two real roots , a cubic equation may have possibly three real roots. But unlike a quadratic equation, which may have no real solution, a cubic equation has at least one real root.

The other two roots might be real or imaginary.

Whenever you are given a cubic equation or any equation, you always have to arrange it in a standard form first.

For example, if you are given something like this, 3x 2 + x – 3 = 2/x, you will re-arrange into the standard form and write it like, 3x 3 + x 2 – 3x – 2 = 0. Then you can solve this by any suitable method.

Let’s see a few examples below for better understanding:

Determine the roots of the cubic equation 2x 3  + 3x 2  – 11x – 6 = 0

Since d = 6, then the possible factors are 1, 2, 3 and 6.

Now apply the Factor Theorem to check the possible values by trial and error.

f (1) = 2 + 3 – 11 – 6 ≠ 0 f (–1) = –2 + 3 + 11 – 6 ≠ 0 f (2) = 16 + 12 – 22 – 6 = 0

Hence, x = 2 is the first root.

We can get the other roots of the equation using synthetic division method. = (x – 2) (ax 2  + bx + c) = (x – 2) (2x 2  + bx + 3) = (x – 2) (2x 2  + 7x + 3) = (x – 2) (2x + 1) (x +3)

Therefore, the solutions are x = 2, x = -1/2 and x = -3.

Find the roots of the cubic equation x 3 − 6x 2 + 11x – 6 = 0

x 3 − 6x 2 + 11x – 6

(x – 1) is one of the factors.

By dividing x 3 − 6x 2 + 11x – 6 by (x – 1),

⟹ (x – 1) (x 2 – 5x + 6) = 0

⟹ (x – 1) (x – 2) (x – 3) = 0

This of the cubic equation solutions are x = 1, x = 2 and x = 3.

Solve x 3  – 2x 2  – x + 2

Factorize the equation.

x 3  – 2x 2  – x + 2 = x 2 (x – 2) – (x – 2)

= (x 2  – 1) (x – 2)

= (x + 1) (x – 1) (x – 2)

x = 1, -1 and 2.

Solve the cubic equation x 3  – 23x 2  + 142x – 120

First factorize the polynomial.

x 3  – 23x 2  + 142x – 120 = (x – 1) (x 2  – 22x + 120)

But x 2  – 22x + 120 = x 2  – 12x – 10x + 120

= x (x – 12) – 10(x – 12) = (x – 12) (x – 10)

Therefore, x 3  – 23x 2  + 142x – 120 = (x – 1) (x – 10) (x – 12)

Equate each factor to zero.

x – 10 = 10

The roots of the equation are x = 1, 10 and 12.

Solve the cubic equation x 3  – 6 x 2  + 11x – 6 = 0.

To solve this problem using division method, take any factor of the constant 6;

Divide the polynomial by x-2 to

(x 2  – 4x + 3) = 0.

Now solve the quadratic equation (x 2  – 4x + 3) = 0 to get x= 1 or x = 3

Therefore, the solutions are x = 2, x= 1 and x =3.

Solve the cubic equation x 3  – 7x 2  + 4x + 12 = 0

Let f(x) = x 3  – 7x 2  + 4x + 12

Since d = 12, the possible values are 1, 2, 3, 4, 6 and 12.

By trial and error, we find that f (–1) = –1 – 7 – 4 + 12 = 0

So, (x + 1) is a factor of the function.

x 3  – 7x 2  + 4x + 12 = (x + 1) (x 2  – 8x + 12) = (x + 1) (x – 2) (x – 6)

Therefore x = –1, 2, 6

Solve the following cubic equation:

x 3  + 3x 2  + x + 3 = 0.

x 3  + 3x 2  + x + 3 = (x 3  + 3x 2 ) + (x + 3) = x 2 (x + 3) + 1(x + 3) = (x + 3) (x 2  + 1)

Therefore, x = -1 ,1 -3.

Solve x 3 − 6x 2 + 11x − 6 = 0

x 3 − 6x 2 + 11x − 6 = 0 ⟹ (x − 1) (x − 2) (x − 3) = 0

Equating each factor to zero gives;

x = 1, x = 2 and x = 3

Solve x 3 − 4x 2 − 9x + 36 = 0

Factorize each set of two terms.

x 2 (x − 4) − 9(x − 4) = 0

Extract the common factor (x − 4) to give

(x 2 − 9) (x − 4) = 0

Now factorize the difference of two squares

(x + 3) (x − 3) (x − 4) = 0

By equating each factor to zero, we get;

x = −3, 3 or 4

Solve the equation 3x 3  −16x 2  + 23x − 6 = 0

Divide 3x 3  −16x 2  + 23x – 6 by x -2 to get 3x 2  – 1x – 9x + 3

= x (3x – 1) – 3(3x – 1)

= (x – 3) (3x – 1)

Therefore, 3x 3  −16x 2  + 23x − 6 = (x- 2) (x – 3) (3x – 1)

Equate each factor to zero to get,

x = 2, 3 and 1/3

Find the roots of 3x 3  – 3x 2  – 90x=0

factor it out 3x

3x 3  – 3x 2  – 90x ⟹3x (x 2  – x – 30)

Find a pair of factors whose product is −30 and sum is −1.

⟹- 6 * 5 =-30

⟹ −6 + 5 = -1

Rewrite the equation by replacing the term “bx” with the chosen factors.

⟹ 3x [(x 2  – 6x) + (5x – 30)]

Factor the equation;

⟹ 3x [(x (x – 6) + 5(x – 6)]

= 3x (x – 6) (x + 5)

x = 0, 6, -5

Solving cubic equations using graphical method

If you cannot solve the cubic equation by any of the above methods, you can solve it graphically. For that, you need to have an accurate sketch of the given cubic equation.

The point(s) where its graph crosses the x-axis, is a solution of the equation. The number of real solutions of the cubic equations is same as the number of times its graph crosses the x-axis.

Find the roots of x 3  + 5x 2  + 2x – 8 = 0 graphically.

Simply draw the graph of the following function by substituting random values of x:

f (x) = x 3  + 5x 2  + 2x – 8

You can see the graph cuts the x-axis at 3 points, therefore, there are 3 real solutions.

From the graph, the solutions are:

x = 1, x = -2 & x = -4.

Practice Questions

Previous Lesson  |  Main Page | Next Lesson

• Number Theory
• Data Structures
• Cornerstones

Solving The General Cubic Equation

The tschirnhause-vieta approach.

Having now covered the basics of trigonometry, let's see how we can put this together with the depressed terms method of solving quadratic equations to solve cubic equations whose roots are all real. As a concrete example, let us consider the following cubic equation: $$x^3 - 2x^2 - 5x + 6 = 0$$ Note that we have $3$ occurences of $x$ in the above. We hope to eliminate $2$ of them, so that we can apply the socks and shoes principle.

Francois Viete (whom we've discussed before) and German mathematician Ehrenfried Walther von Tschirnhaus realized that the method of depressed terms provides the opportunity to eliminate $1$ of these fairly easily.

Suppose that the function $f(x)$ outputs the value of the left side of our equation. Then consider the function $g(x) = f(x+h)$, whose graph is some shift of the graph of $y=f(x)$. As when we applied this method to solving quadratics, note that the graph of $y=g(x)$ is simply a horizontal shift of the graph of $y=f(x)$.

While the sign of $h$ determines whether this shift is to the left or right, should we find any roots to $g(x) = 0$ we can deduce corresponding roots to $f(x)=0$ by simply adding $h$ to each of the found roots to $g(x) = 0$.

Plugging in $(x+h)$ as input to $f$ and expanding the result as quadratic in $x$, we find $$\begin{array}{rcl} g(x) &=& f(x+h)\\ &=& (x+h)^3 - 2(x+h)^2 - 5(x+h) + 6\\ &=& x^3 + 3x^2 h + 3xh^2 + h^3 -2x^2 - 4xh - 2h^2 - 5x - 5h + 6\\ &=& x^3 + (3h-2)x^2 + (3h^2 - 4h - 5)x + (h^3 - 2h^2 - 5h + 6)\\ \end{array}$$ As before, we may find it useful to note that the constant term of $g(x)$ must always be $f(h)$, given the way it was defined.

Note the coefficient on the quadratic term, however. Just as we can depress the linear term in a quadratic to find its solutions, let us here depress the quadratic term for the resulting cubic to solve this problem. That is to say, let us find $h$ so that $(3h-2) = 0$, so that $g(x)$ has no quadratic term.

This of course is almost immediate, as $3h-2 = 0$ implies $h=\frac{2}{3}$

Now, finding the rest of $g(x)$ for this special $h$ value, we have $$\begin{array}{rcl} g(x) &=& x^3 + \left(3 (\frac{2}{3})^2 - 4 (\frac{2}{3}) - 5\right) x + \left((\frac{2}{3})^3 - 2(\frac{2}{3})^2 - 5(\frac{2}{3}) + 6\right)\\ &=& x^3 + \left(\frac{4}{3} - \frac{8}{3} - \frac{15}{3}\right) x + \left( \frac{8}{27} - \frac{24}{27} - \frac{90}{27} + \frac{162}{27}\right)\\ &=& x^3 - \frac{19}{3} x + \frac{56}{27}\\ \end{array}$$ As such, the roots of $f(x) = 0$ can be found by adding $h=\frac{2}{3}$ to each root of $x^3 - \frac{19}{3} x + \frac{56}{27} = 0$. The problem, however, is that this last equation still has more than one occurrence of $x$. This is where trigonometry can help!

Notice that if we solve for the constant in this equation, we get $$-\frac{56}{27} = x^3 - \frac{19}{3}x$$ Now recall the triple-angle formula for the cosine: $$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$$ Notice the similarities between the two. On their right sides, we have in both situations a cubic and a linear term -- in terms of $x$ for the first, and in terms of $\cos \theta$ for the second.

Our aim is to make these two look even closer in form, and then assume some nice relationship exists between them (i.e., one where we can solve for either in terms of the other), allowing us to make a clever substitution (more on that in a minute).

With this in mind, and noting that there is a coefficient on the $\cos^3 \theta$ term but not on the $x^3$ term, -- let's divide the equation in $\theta$ by $4$ to get $$\frac{\cos 3\theta}{4} = \cos^3 \theta - \frac{3}{4} \cos \theta$$ Now for the clever substitution! Suppose we choose $\theta$ and some real value $a$ so that $x = a\cos \theta$. There are lots of $(a,\theta)$ pairs possible here. To see this, note that when $0 \lt \theta \lt \frac{\pi}{2}$, increasing $\theta$ by a "smidge" will require decreasing $a$ by some amount to keep their product equal to $x$. Similar things happen when $\theta$ is in other ranges.

The many options we have for the particular pair of $\theta$ and $a$ values we will use gives us some flexibility to make some desired thing happen -- just as there were many $h$ we could have used to define $g(x)$, so we picked the one that would make the quadratic term disappear.

Let's consider what happens if $x = a\cos \theta$ and we use this to rewrite our equation currently in terms of $x$ to be in terms of $\cos \theta$: $$-\frac{56}{27} = (a \cos \theta)^3 - \frac{19}{3}(a \cos \theta)$$ Dividing both sides by $a^3$ (to return the coefficient on $\cos^3 \theta$ to $1$, we have $$-\frac{56}{27a^3} = \cos^3 \theta - \frac{19}{3a^2} \cos \theta$$ Recall again the earlier relationship stemming from the triple-angle formula: $$\frac{\cos 3\theta}{4} = \cos^3 \theta - \frac{3}{4} \cos \theta$$ If we can make the right side of our equation look exactly like the right side of the equation immediately above, then we could replace it with $\frac{1}{4} \cos 3\theta$. This would reduce the number of occurrences of our variable down to a single occurrence -- allowing us to solve the rest with the socks and shoes principle. Given this, let us choose the value $a$ to be the one that will make the coefficient on $\cos \theta$ equal to $\frac{3}{4}$.

We should pause for a moment here and consider what could have happened just now had we attacked a slightly different cubic equation. Suppose the $19$ above had been negative. This could easily happen! (Can you make a cubic where it does?) In such a case, the two values associated with the variable $a$ will then be complex. As will be seen in the paragraphs coming, such a situation would require we consider when the cosine of an angle is complex! Certainly this never happens when the input to the cosine is real -- but what if the cosine function (and others) could accept complex-valued inputs? Lest we get ahead of ourselves, let us leave those exciting possibilities linger for now. But don't worry, we'll pick them up again soon!

With two options for $a$ that both accomplish what we seek to do, let us pick the positive one. With $a = \frac{2}{3}\sqrt{19}$, we then have $$ \frac{-56}{27 ( \frac{2}{3} \sqrt{19} )^3} = \cos^3 \theta - \frac{3}{4} \cos \theta $$ Simplifying the left side and then swapping sides (just to make it look prettier), we have $$\cos^3 \theta - \frac{3}{4} \cos \theta = -\frac{7}{19^{3/2}}$$ Finally, we have thus set ourselves up to replace the two occurrences of $\theta$ with a single one -- using the relationship resulting from the triple-angle formula for cosine: $$\frac{\cos 3\theta}{4} = -\frac{7}{19^{3/2}}$$ From here, we use socks and shoes. Of course, we must be careful when navigating the one "wrinkle" that the cosine function is technically not invertible. That said, we know how to find (using the arccosine function, $\cos^{-1}$) all angles for which their cosine equals some given value. $$\begin{array}{rcl} \displaystyle{\frac{\cos 3\theta}{4}} &=& -\displaystyle{\frac{7}{19^{3/2}}}\\\\ \cos 3\theta &=& \displaystyle{-\frac{28}{19^{3/2}}}\\\\ 3\theta &=& \pm \displaystyle{\cos^{-1} \left(\frac{-28}{19^{3/2}}\right)+2\pi n \quad \textrm{where $n$ is an integer}}\\\\ \theta &=& \pm \displaystyle{\frac{1}{3} \cos^{-1} \left(\frac{-28}{19^{3/2}}\right)+\frac{2\pi n}{3} \quad \textrm{where $n$ is an integer}}\\\\ \end{array}$$ Our last steps are to work backwards from this solution for $\theta$, first to a solution $x$ to $g(x)=0$, and then to a solution $x$ to $f(x)=0$.

Recall that we let $x = a \cos \theta$, except now we know the value of $a$ we wanted -- namely, $a = \frac{2}{3}\sqrt{19}$. As such, the following $x$ solves $g(x)=0$: $$x = \frac{2}{3}\sqrt{19} \cos \left(\pm \frac{1}{3} \cos^{-1} \left(\frac{-28}{19^{3/2}}\right)+\frac{2\pi n}{3}\right) \quad \textrm{where $n$ is an integer}$$ All that remains is to add $h=\frac{2}{3}$ to these solutions to find the $x$ values that solve $f(x)=0$ $$x = \frac{2}{3}\sqrt{19} \cos \left(\pm \frac{1}{3} \cos^{-1} \left(\frac{-28}{19^{3/2}}\right)+\frac{2\pi n}{3}\right) + \frac{2}{3}\quad \textrm{where $n$ is an integer}$$ Of course, we probably want to simplify these if we can! Evaluating the above with a calculator for $n=0$, $n=1$, and $n=2$ and the "$+$" option (things start to repeat after that), we shockingly get the below as our solution! $$x = 1, x = -2, \textrm{ or } x = 3$$ Note we can easily verify this, as if true our original polynomial should factor into: $$x^2 - 2x^2 - 5x + 6 = (x-1)(x+2)(x-3)$$ Expanding the right side, we see that this is true. Wow! 😀

Cardano's Method

Just as we saw multiple ways to solve quadratic equations, there are also mutiple ways to solve cubic equations. Interestingly, some of these give the solutions in a radically different form. While the method above expressed it's solutions in terms of trigonometric functions and their inverses, the method below known as Cardano's method, after Italian mathematician Gerolamo Cardano, will reveal it's solutions in a more algebraic form -- one that involves nested roots of various values. Seeking to give attribution to where it is due, one should note that while Gerolamo Cardano published the method of solution seen below -- it was first discovered by Scipione del Ferro, another Italian mathematician, born in $1465$.

Consider the following cubic equation: $$x^3 + 3x^2 + 4x + 3 = 0$$ We begin in the same manner as the previous method -- we depress the quadratic term in the related cubic function $f(x) = x^3 + 3x^2 + 4x + 3$.

Recall this means defining a new function $g(x) = f(x+h)$, and then writing it as a cubic function of $x$. Thus, $$\begin{array}{rcl} g(x) &=& (x+h)^3 + 3(x+h)^2 + 4(x+h) + 3\\ &=& (x^3 + 3x^2 + 3xh^2 + h^3) + 3(x^2 + 2xh + h^2) + 4(x+h)+3\\ &=& x^3 + (3h+3)x^2 + (3h^2+6h+4)x + (h^3+3h^2+4h+3)\\ \end{array}$$ We wish the coefficient on $x^2$ to be zero, so we choose $h=-1$.

Much like the previous solution of a cube used the triple-angle formula for cosine as a trick to reduce the now two occurrences of $x$ down to a single occurrence of an unknown variable -- we seek the same goal, but use a different trick to get there.

Recalling the cube of a binomial: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, rearrange the terms to discover the following: $$\underbrace{(a+b)^3}_{\textrm{a cubic term}} - 3ab\underbrace{(a+b)}_{\textrm{a linear term}} - (a^3 + b^3) = 0$$ Here's the trick: Noting the similarity in form between our depressed cubic and the equation immediately above, let us equate the following: $$\begin{array}{rcl} m &=& -3ab\\ n &=& -(a^3+b^3) \end{array}$$ and let $x=a+b$, so that $x$ solves both $$x^3 + mx + n = 0$$ and $$x^3 - (3ab)x - (a^3 + b^3) = 0$$ Then, we simply solve for $a$ and $b$ in terms of $m$ and $n$ and add them together to find $x$, a solution to the depressed cubic whose roots we seek.

To solve this system, it will be easier to only see the same powers of $a$ and $b$, so we cube both sides of the first equation. This gives us the system: $$\begin{array}{rcl} m^3 &=& -27a^3 b^3\\ n &=& -(a^3+b^3) \end{array}$$ Now let us hide the detail of the cubing, so we can better see the structure of what is left, by letting $A = a^3$ and $B = b^3$: $$\begin{array}{rcl} m^3 &=& -27 A B\\ n &=& -(A + B) \end{array}$$ Now, it is clear that $$\begin{array}{rcl} A B &=& \displaystyle{-\frac{m^3}{27}}\\ A + B &=& -n \end{array}$$ As seen below, solving for $B$ in the first and substituting the result into the second reveals an equation easily convertible into a quadratic in $A$! $$A B = -\frac{m^3}{27} \quad \longrightarrow \quad B = -\frac{m^3}{27A}$$ Then, $$\begin{array}{rcl} A - \frac{m^3}{27A} &=& -n\\ 27A^2 - m^3 &=& -27nA \quad \quad {\scriptsize \textrm{(after multiplying both sides by $27A$ to eliminate the fractions)}}\\ 27 A^2 + 27n A - m^3 &=& 0 \end{array}$$ This quadratic in $A$ can be quickly solved with the quadratic formula: $$A = \frac{-27n \pm \sqrt{(27n)^2 - 4(27)(-m^3)}}{2\cdot 27}$$ Just as a bit of flourish, we can rewrite the above with a bit more symmetry (allowing us to remember it more easily, if that be desired). The algebra to convert the above form to the one given below is straight-forward and not too long (so the reader should try their hand at doing so!) $$A = -\frac{n}{2} \pm \sqrt{\left(\frac{n}{2}\right)^2 + \left(\frac{m}{3}\right)^3}$$ By symetry one of these will be the solution for $A$, and the other will be a solution for $B$. Given that $a = \sqrt[3]{A}$, $b = \sqrt[3]{B}$, and one solution sought will be given by $x = a + b$, we have $$x = \sqrt[3]{-\frac{n}{2} + \sqrt{\left(\frac{n}{2}\right)^2 + \left(\frac{m}{3}\right)^3}} + \sqrt[3]{-\frac{n}{2} - \sqrt{\left(\frac{n}{2}\right)^2 + \left(\frac{m}{3}\right)^3}}$$ Recalling that in our particular depressed cubic $g(x) = x^3 + mx + n$ both $m$ and $n$ equaled $1$, we now have one solution in hand to $g(x) = 0$: $$x = \sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{1}{27}}} + \sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} + \frac{1}{27}}}$$ Finally, adding $h=-1$ to the above solution to $g(x) = 0$ shifts it to a solution of $f(x) = 0$, our original cubic equation $x^3 + 3x^2 + 4x + 3 = 0$: $$x = \sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{1}{27}}} + \sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} + \frac{1}{27}}} - 1 \approx -1.682327804$$ One will note that an approximation to the exact solution has been provided above. While finding the other two roots will be relatively simple from here -- the arithmetic involved can be ...well, awful! As such, we will use approximations from this point forward, but will provide enough detail to allow the ambitious (masochistic?) reader to find the other roots exactly, if they so desire.

Armed with the knowledge that $r_1 \approx -1.682327804$ is a root to $f(x) = 0$, by the factor theorem we know $f(x)$ has a factor of $(x-r_1)$. Dividing $f(x)$ by this factor produces a quadratic whose roots (let us call them $r_2$ and $r_3$) provide the remaining two solutions to $f(x) = 0$. These of course, can be found via the quadratic formula.

In this case, doing that division yields the additional (complex) roots $r_2$ and $r_3$: $-0.658836 \pm 1.16154i$. As a double-check, plugging these roots into $f(x)$, we find $f(r_2) = f(r_3) \approx 3.06404 \times 10^{-6} + 4.0109 \times 10^{-6} i$, which is quite close to zero indeed!

Just as we can apply "completing the square", "Po-Shen Loh's method", or "the method of depressed terms" to a general quadratic equation $ax^2 + bx + c = 0$ to arrive at the quadratic formula, you could also apply Cardano's method (or others) to the general cubic equation $ax^3+bx^2+cx+d=0$ to find its three roots.

Should you be interested enough to do so, you would get the following:

While many have the quadratic formula memorized -- for obvious reasons, one probably doesn't want to try to commit the above to memory! Still, it is very cool that a general solution exists! Makes one wonder about a quartic (i.e., fourth-degree) equation, doesn't it?

• Math Formulas

Cubic Equation Formula

The cubic equation formula expresses the cubic equation in Mathematics. An equation with degree three is called a cubic equation. The nature of roots of all cubic equations is either one real root and two imaginary roots or three real roots. If the polynomials have degree three, they are known as cubic polynomials. 

What is Cubic Equation Formula?

To plot the curve of a cubic equation, we need cubic equation formula. This formula helps to find the roots of a cubic equation. If the degree of the polynomial is n, then there will be n number of roots. The roots of cubic equation are also called zeros.

The  cubic  equation formula is given by:

Depressing the Cubic Equation

Substitute  \(\begin{array}{l}\large x= y-\frac{b}{3a}\end{array} \) in the above cubic equation, then we get,

Simplifying further, we obtain the following depressed cubic equation –

It must have the term in x 3   or it would not be cubic ( and so a≠0 ), but any or all of b, c and d  can be zero. For instance:

The examples of cubic equations are:

• x 3  – 6x 2  + 11x – 6 = 0
• 4x 3  + 57 = 0
• x 3  + 9x = 0

Solved Examples on Cubic Equation Formula

(x-1)(x-2)(x-3)=0

This equation has three real roots, all different – the solutions are x = 1, x = 2 and x = 3.

Question 2: Solve the cubic equation x 3 – 23x 2 + 142x – 120.

Solution:  First factorize the polynomial to get;

x 3 – 23x 2 + 142x – 120 = (x – 1) (x 2 – 22x + 120)

But x 2 – 22x + 120 = x 2 – 12x – 10x + 120

= x (x – 12) – 10(x – 12)

= (x – 12) (x – 10)

Therefore, x3 – 23×2 + 142x – 120 = (x – 1) (x – 10) (x – 12)

Equate each factor to zero to get;

The roots of the equation are x = 1, 10 and 12.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Course: Algebra 2   >   Unit 10

Solving cube-root equations.

Want to join the conversation?

• Upvote Button navigates to signup page
• Downvote Button navigates to signup page
• Flag Button navigates to signup page

Video transcript

• School Guide
• Mathematics
• Number System and Arithmetic
• Trigonometry
• Probability
• Mensuration
• Maths Formulas
• Class 8 Maths Notes
• Class 9 Maths Notes
• Class 10 Maths Notes
• Class 11 Maths Notes
• Class 12 Maths Notes

Solving Cubic Equations

Cubic Equation is a mathematical equation in which a polynomial of degree 3 is equated to a constant or another polynomial of maximum degree 2. The standard representation of the cubic equation is ax 3 +bx 2 +cx+d = 0 where a, b, c, and d are real numbers. Some examples of cubic equation are x 3 – 4x 2 + 15x – 9 = 0, 2x 3 – 4x 2 = 0 etc.

Table of Content

Polynomial Definition

Degree of equation, cubic equation definition, how to solve cubic equations, solving cubic equation using factors, solving cubic equation using graphical method, problems based on solving cubic equations, practice problems on solving cubic equations.

For learning How to Solve Cubic Equations we must first learn about polynomials, the degree of the polynomial, and others. In this article, we will learn about, Polynomials, Polynomial Equations, Solving Cubic Equations Or how to solve cubic equations, and others in detail.

Polynomial is defined as follows,

A polynomial is an algebraic expression in which the power of a variable is a non-negative integer. The general form of a polynomial is a 0 x n + a 1 x n-1 + a 2 x n-2 +… + a n . Depending upon the maximum power of the variable, a polynomial can be classified as a monomial, binomial, trinomial, and so on.

What is an Equation?

An Equation is defined as follows,

An equation is a polynomial that is equated to a numerical value or any other polynomial. For Example, x + 2 is a polynomial but x + 2 = 5 is an equation. Similarly, 2x + 3 = x + 1 is also an equation whereas, 2x + 3 and x + 1 are polynomials individually.

The definition of the Degree of Equation is stated below:

Degree of a Equation is defined as the maximum power possessed by the variable in an Equation.

Based on the degree of the Equation, an Equation can be classified as follows:

Linear Equation

• Quadratic Equation

Cubic Equation

• Biquadratic Equation

The Equation in which the maximum power of the variable is 1 is called a Linear Equation.

• For example 3x +1 = 0

Quadratic Polynomial

The Equation in which the maximum power of the variable is 2 is a Quadratic Equation.

• For example 3x 2 +x+1 = 0

The Equation in which the maximum power of the variable is 3 is called a Cubic Equation.

• For example 5x 3 +3x 2 +x+1 = 0

Biquadratic Polynomial

The Equation in which the maximum power of the variable is 4 is called a Biquadratic Polynomial or Quartic Polynomial.

• For example 5x 4 +4x 3 +3x 2 +2x+1 = 0

Cubic Equation is an algebraic equation where the highest degree of the polynomial is 3. Some examples of cubic equations are 5x 3 +3x 2 +x+1 = 0, 2x 3 +8 = x ⇒ 2x 3 -x+8 = 0, etc.

 The general form of a cubic equation is,

ax 3 + bx 2 + cx + d = 0, a ≠ 0 Where, a, b, and c are the coefficients of variable and their exponenats and d is the constant, and a, b, c and d are real numbers.

A cubic equation is a equation with degree three. It has three solution and it can be solved easily by following the steps added below,

Step 1: Find one solution to the cubic equation by hit and try method. Suppose we have a cubic equation P(x) then find for any x = a, P(a) = 0 by taking, x = 0, ±1, ±2, ±3, … and so. Step 2: When we get, P(a) = 0, find the factor (x – a) of P(x) Step 3: Divide P(x) by (x – a) to get a quadratic equation say Q(x) using polynomial division. Step 4: Factarize the quadratic equation Q(x) to get the factors as (x – b), and (x – c). Step 5: (x – a), (x – b), and (x – c) are the factors of P(x) and solving each factors we gets the roots of equation as, a, b, and c.

Learn more about, Dividing Polynomial

A Cubic Equation can be solved by two methods

By reducing it into a quadratic equation and then solving it either by factoring or the quadratic formula  By Graphical Method

A Cubic Equation has three roots. These roots may be real or imaginary. Also, there can be there distinct roots or two same and one different root and all three same roots.

Point to be noted that for any equation, including Cubic Equations , the equation must always be arranged in its standard form first before solving the equation. 

For instance, if the given equation is 2x 2 -5 = x + 4/x, then we have to re-arrange this into its standard form, i.e., 2x 3 -x 2 -5x-4 = 0. Now, we can solve the equation using any appropriate method.

The solution of cubic equation using factor theorem is explained using the example added below,

Example:  Find the roots of equation f(x) = 3x 3 −16x 2 + 23x − 6 = 0.

Given expression: f(x) = 3x 3 −16x 2 + 23x − 6 = 0 First, factorize the polynomial to get roots Since the constant is -6 the possible factors are 1, 2, 3, 6 f(1) = 3 – 16 + 23 – 6 ≠ 0 f(2) = 24 – 64 + 46 – 6 = 0 f(3) = 81 – 144 + 69 – 6 = 0 f(6) = 648 – 576 + 138 – 6 ≠ 0 We know that, iaccording to Factor Theorem if f(a) = 0, then (x-a) is a factor of f(x) So, (x – 2) and (x – 3) are factors of f(x). Therefore, the product of (x – 2) and (x – 3) will also be factor of f(x). Now to find the remaining factors use the long division method and divide f(x) by product of (x – 2) and (x – 3) Hence, Divisor = (x – 2)(x – 3) = (x 2 – 5x + 6) and Dividend = 3x 3 −16x 2 + 23x − 6. Now divide as shown below, After division we obtain (3x- 1) as quotient and remainder is 0. Now as per Division Algorithm we know that Dividend = Divisor×Quotient+Remainder. ⇒ f(x) = (3x 3 −16x 2 + 23x − 6) =  (x 2 – 5x + 6)(3x-1) Since f(x) = 0 ⇒ (x 2 – 5x + 6)(3x-1) = 0 ⇒ x 2 – 5x + 6 = 0 or 3x-1 = 0 Now we will take 3x-1 = 0 ⇒ x = 1/3 as we already know two roots from x 2 – 5x + 6 which are 2 and 3 So, Roots of the given Cubic Equation are 1/3, 2, and 3.

A cubic equation is solved graphically when you cannot solve the given equation using other techniques. So, we need an accurate drawing of the given cubic equation. The equation’s roots are the point(s) at which the graph crosses the X-axis if the equation is in the terms of x and if the equation is in the terms of y then the roots of the equation are the points at which the graph cuts the Y-axis.

The number of real solutions to the cubic equation is equal to the number of times the graph of the cubic equation crosses the X-axis. 

Example: Find the roots of equation f(x) = x 3 − 4x 2 − 9x + 36 = 0, using the graphical method.

Given expression: f(x) = x 3 − 4x 2 − 9x + 36 = 0. Now, simply substitute random values for x in the graph for the given function: x -4 -3 -2 -1 0 1 2 3 4 5 f(x) -56‬ 0 19‬ 40 36 24‬ 10‬ 0 0 16 We can see that the graph has cut the X-axis at 3 points, therefore, there are 3 real solutions. From the graph, the solutions are: x = -3, x = 3, and x = 4. Hence, the roots of the given equation are -3, 3, and 4.

Linear Equation Solving Quadratic Equation Factoring Polynomials

Problem 1: Find the roots of f(x) = x 3 – 4x 2 -3x + 6 = 0.

Given expression:  f(x) = x 3 – 4x 2 -3x + 6 = 0. First, factorize the polynomial to get roots. Since the constant is +6 the possible factors are 1, 2, 3, 6. f(1) = 1 – 4 – 3 + 6 = 7 – 7 = 0 f(2) = 8 – 16 – 6 + 6 ≠ 0 f(3) = 27 – 36 – 9 + 6 ≠ 0 f(6) = 216 – 144 -18 + 6 = -48 ≠ 0 So,  according to Factor Theorem (x – 1) is a factor of the given equation. Now to find the remaining factors use the long division method. According to Division Algorithm we can write, So, f(x) = x 3 – 4x 2 -3x + 6 = (x – 1) (x 2 – 3x – 6) = 0 ⇒ (x – 1) = 0 or (x 2 – 3x – 6) = 0 We know that the roots of a quadratic equation ax 2 + bx + c = 0 are, x = [-b ± √(b 2 -4ac)]/2a Hence, for (x 2 – 3x – 6) = 0 x = [3 ± √(3 2 – 4(1)(-6)]/2(1) x = (3 ± √33)/2 Hence, the roots of the given cubic equation are 1, (3+√33)/2, and (3–√33)/2.

Problem 2: Find the roots of equation f(x) = 4x 3 – 10x 2 + 4x = 0.

Given expression:  f(x) = 4x 3 – 10x 2 + 4x = 0 ⇒ x (4x 2 – 10x + 4) = 0 ⇒ x (4x 2 – 8x – 2x + 4) = 0 ⇒ x(4x(x – 2) – 2(x – 2)) = 0 ⇒ x (4x – 2) (x – 2) = 0 ⇒ x = 0 or 4x – 2 = 0, x – 2 = 0 ⇒ x = 0 or x = 1/2 or x = 2 Hence, the roots of the given equation are 0, 1/2 and 2.

Problem 3: Find the roots of equation f(x) = x 3 + 3x 2 + x + 3 = 0.

Given expression:  f(x) = x 3 + 3x 2 + x + 3 = 0. ⇒ x 2 (x + 3) + 1(x + 3) = 0 ⇒ (x + 3) (x 2 + 1) = 0 ⇒ x + 3 = 0 or x 2 + 1 = 0 ⇒ x = -3, ±i So, the given equation has a real root, i.e., -3, and two imaginary roots, i.e., ±i.

Problem 4: Find the roots of equation f(x) = x 3 – 7x 2 – x + 7 = 0.

Given expressions, f(x) = x 3 – 3x 2 – 5x + 7 = 0 First, factorize the equation, f(x): x 3 – 3x 2 – 5x + 7= 0 It can be factored into (x-7)(x+1)(x-1) = 0 After factoring the polynomial, we can find the roots by equating each factor to zero. For example: x – 7 = 0, so x = 7 x + 1 = 0, so x = -1 x – 1 = 0, so x = 1 So the roots of the equation f(x): x 3 – 3x 2 – 5x + 7 = 0 are x = 7 x = -1 x = 1

Problem 5: Find the roots of equation f(x) = x 3 − 6x 2 + 11x − 6 = 0, using the graphical method.

Given expression: f(x) = x 3 − 6x 2 + 11x − 6 = 0. Now, simply substitute random values for x in the graph for the given function: x 1 2 3 4 5 f(x) 0 0 0 6 24 We can see that the graph has cut the X-axis at 3 points, therefore, there are 3 real solutions. From the graph, the solutions are: x = 1, x = 2, and x = 3. Hence, the roots of the given equation are 1, 2, and 3.

Various practise problems related to cubic equations are added below. Solve these problems to fully grasp the concept of How to Solve Cubic Equation?

P1. Solve the cubic equation, 3x 3 + 2x 2 – 11x + 7 = 0.

P2. Find the roots of cubic equation, 4x 3 – 12x 2 + 17 = 0.

P3. Solve the cubic equation, x 3 + 4x 2 – x + 3 = 0 using graphical method.

P4. Find the number which satisfy, -9x 3 + 11x 2 – 8x + 2 = 0.

FAQs on Solving Cubic Equations

1. what are cubic equations.

Cubic Equations are the algebraic equations in which the maximum power of a variable is 3

2. How do you Factor a Cubic Equation?

We can factor a cubic equation in two ways. First by taking a linear expression common from the given cubic equation, then we will have a linear and a quadratic expression as a product. This quadratic equation can be further factorized to get all the factors. The second method is to find a zero of the given cubic equation by putting random values. The value for which we get the value of the equation to be zero will be one of the zeros of the given cubic equation. Now using the factor theorem form a linear expression let’s say x-a and divide the given cubic equation by this expression which will give a quadratic equation as quotient. This obtained quadratic equation can be further factorized to get all the factors.

3. How do you Solve a Cubic Equation Graphically?

To solve a cubic equation graphically put random values for x in the given cubic equation and solve, you will get the values of y. Plot these obtained values on the graph. Find the coordinates at which the graph is intersecting the x-axis. These coordinates are the solution of the cubic equation.

4. Can all Cubic Equations be Solved Exactly?

Any equation that has odd power must have one real root. Hence, a cubic equation must have at least one real root, unlike a quadratic equation where both roots can be imaginary when the discriminant is less than zero.

5. Can a Cubic Equation have Multiple Solutions?

Yes, cubic equations can have multiple solutions as a cubic equation can have up to three distinct real roots.

6. What do you mean by the Degree of an Equation?

The maximum power possessed by the variable in an equation is called the degree of a polynomial.

7. What is the Difference between a Polynomial and an Equation?

Polynomial is simply an algebraic equation in which the power of the variable is non-negative integer. This polynomial when equated (=) with a numerical value or another polynomial then it is called an Equation.

8. What is the Factor Theorem for cubic equations?

Factor Theorem states that if r is a root (solution) of the cubic equation ax 3 + bx 2 + cx + d = 0, then x – r is a factor of the equation.

9. What if I can’t find exact solutions using formulas?

If finding exact solutions seems impossible, we can use the numerical methods like iterative methods (e.g., Newton’s method) to approximate the roots of the equation. Learn more about Newton Raphson’s Method .

Please Login to comment...

Similar reads.

• School Learning
• Maths-Class-12

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

PROBLEMS BASED ON CUBIC EQUATIONS

(1)  If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.              Solution

(2)  Construct a cubic equation with roots

(i) 1, 2 and 3      (ii) 1,1, and −2         (iii) 2, 1/2 and 1.

(3)  If α , β and γ are the roots of the cubic equation x 3  + 2x 2  + 3x + 4 = 0 , form a cubic equation whose roots are

(i) 2α , 2β , 2γ

(ii)  1/α, 1/β, 1/γ

(iii) −α, -β, -γ          Solution

(4)   Solve the equation 3x 3  −16x 2  + 23x − 6 = 0 if the product of two roots is 1.          Solution

(5)   Find the sum of squares of roots of the equation 2x 4  −8x 3  + 6x 2  − 3 = 0.          Solution

(6)   Solve the equation x 3  − 9x 2  + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3: 2.    Solution

(7)   If α, β  and γ are the roots of the polynomial equation ax 3  + bx 2  + cx + d = 0 , find the Value of  ∑  α/βγ in terms of the coefficients          Solution

(8)   If  α, β, γ  and δ are the roots of the polynomial equation 2x 4 + 5x 3  − 7x 2  + 8 = 0 , find a quadratic equation with integer coefficients whose roots are  α + β + γ +  δ and  αβγ δ.       Solution

(9)   If p and q are the roots of the equation lx 2  + nx + n = 0, show that  √(p/q) +  √(q/p) +  √(n/l)  =  0       Solution

(10)   If the equations x 2  + px + q = 0 and x 2  + p'x + q' = 0 have a common root, show that it must be equal to (pq'-p'q)/q q' or (q - q') / (p' - p)       Solution

(11)  Formalate into a mathematical problem to find a number such that when its cube root is added to it, the result is 6.       Solution

(12)  A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away       Solution

Kindly mail your feedback to   [email protected]

We always appreciate your feedback.

© All rights reserved. onlinemath4all.com

• Sat Math Practice
• SAT Math Worksheets
• PEMDAS Rule
• BODMAS rule
• GEMDAS Order of Operations
• Math Calculators
• Transformations of Functions
• Order of rotational symmetry
• Lines of symmetry
• Compound Angles
• Quantitative Aptitude Tricks
• Trigonometric ratio table
• Word Problems
• Times Table Shortcuts
• 10th CBSE solution
• PSAT Math Preparation
• Privacy Policy
• Laws of Exponents

Recent Articles

Digital sat math problems and solutions (part - 31).

Aug 20, 24 12:09 PM

SAT Math Resources (Videos, Concepts, Worksheets and More)

Aug 20, 24 01:35 AM

Geometry Problems with Solutions (Part - 5)

Aug 19, 24 10:56 PM

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Is there a systematic way of solving cubic equations?

According to my text book, to solve cubic equations, I need to

• By trial & error find what value $a$ will make the cubic $0$ . The factor will be $(x-a)$ .
• Then the other factor will be $Ax^2+Bx+C$ which I can solve by comparing coefficients.

But for the first part though, is there any other way to find $a$ besides trail & error? Is there some technique to guess the factor of an equation?

The cubic I am currently factoring happens to be $2x^3+3x+4=9x^2$ .

• polynomials

• $\begingroup$ Have you learned the "rational root theorem"? $\endgroup$ –  The Chaz 2.0 Commented Sep 4, 2011 at 3:09
• $\begingroup$ $2(-0.5)^3 - 9(0.5)^2 + 3(-0.5) + 4 =$ $-2(1/8) -9(1/4) - 3(1/2) + 4 = $ $-(1/4) -(9/4) -(3/2) + 4 = -(16/4)+4 = -4+4 = 0$. So you certainly made a mistake, but since you did not show your arithmetic, I don't know where you made your evaluation mistake. $\endgroup$ –  Arturo Magidin Commented Sep 4, 2011 at 3:51
• $\begingroup$ I had 2(−0.5)^3−9(−0.5)^2+3(−0.5)+4 in my calculator ... hmm ... ya... working manually worked ... $\endgroup$ –  Jiew Meng Commented Sep 4, 2011 at 3:59

3 Answers 3

Of course there is :-) A technique for solving cubics has been known since the 16th century.

Let's start with $x^3+ax+b=0$, which is your case. We'll see later that it's always possible to eliminate the $x^2$ term, so this is in fact the only case we really need to worry about.

The trick is to make the problem initially look harder by setting $x=p+q$. Instead of a single variable we now have two; we hope to use the extra freedom to gain something, and we shall.

Substituting, we get

$(p+q)^3+a(p+q)+b=0$

$p^3+3p^2q+3pq^2+q^3+a(p+q)+b=0$

and now we observe that $3p^2q+3pq^2$ can be rearranged:

$p^3+q^3+3pq(p+q)+a(p+q)+b=0$

and finally, we observe the two terms with a common $p+q$:

$p^3+q^3+(3pq+a)(p+q)=0$.

This seems about as bad as where we started - however, we have this extra degree of freedom: we can force $3pq+a$ to be 0. If we do that, by setting $q = -\frac{a}{3p}$, we get

$p^3 - \frac{a^3}{27p^3} + b = 0$

and we're basically done: renaming $p^3$ as $z$, this is a quadratic equation in $z$. Solve it, determine $p$, calculate $q$, and the solution to the original equation is $x=p+q$ (of course there could be multiple solutions).

All of this can be recast as explicit equations for the solution, but I never remember those; the only way I can remember this is by recalling the $x=p+q$ trick.

Finally, if you have

$x^3+ax^2+bx+c=0$

you can make a simple linear change of variables $x=y+r$, whereby this becomes

$y^3+(3r+a)y^2+\ldots=0$

so choosing $r = -a/3$ makes the $y^2$ term vanish. Solve for $y$ as before and remember the shift $x=y-a/3$ to get the solution for the original equation.

• 4 $\begingroup$ There's a nice geometric interpretation of the "depression" of a cubic: by Vieta, we know that the mean of the three roots is $-a/3$, so performing the translation implied by the depression corresponds to shifting the roots such that the shape formed by the three roots in the complex plane is "centered" on the origin. $\endgroup$ –  J. M. ain't a mathematician Commented Sep 4, 2011 at 5:40

I think "trial and error" is probably referring to the Rational Roots Test . To use it, try all fractions of the form $\pm \frac{p}{q}$ where $p$ is a factor of the constant term and $q$ is a factor of the highest degree term.

In your example, we would evaluate each of the following in the function and see if any of them are roots: $\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{2}{1}, \pm \frac{2}{2}, \pm\frac{4}{1}, \pm \frac{4}{2}$. (Several of these are redundant, but I include them just so you can see how the list is formed.)

Note that the vast majority of numbers in this list are not roots. All the Rational Roots Test claims is that if there is a rational root of your function, then it will appear somewhere on the list. Most of the numbers on the list are not roots, however.

Yes there is, but it won't be much use in an exam:

Given the cubic equation:

For the general cubic equation (1) with real coefficients, the general formula for the roots, in terms of the coefficients, is as follows if $(2 b^3-9 a b c+27 a^2 d)^2-4 (b^2-3 a c)^3=-27 a^2 \Delta>0$, i.e. if there are two non real roots: However, this formula is wrong if the operand of the square root is negative or if the coefficients belong to a field which is not contained in the field of the real numbers: When this operand is real and positive, the cubic roots are real and well defined. In the other case, the square root is not real and one has to choose, once for all a determination for it, for example the one with positive imaginary part. For extracting the cubic roots we have also to choose a determination for the cubic roots, and this gives nine possible values for the first root of an equation which has only three roots. A correct solution may be obtained by remarking that the proof of above formula shows that the product of the two cubic roots is rational. This gives the following formula in which $\sqrt{ }$ or $\sqrt[3]{ }$ stand for any determination of the square or cubic root, if $b^2-3ac \mbox{ } \neq \mbox{ } 0$. If $Q \mbox{ } \neq \mbox{ } 0$ and $b^2 − 3ac = 0$, the sign of Q has to be chosen for having $C \mbox{ } \neq \mbox{ } 0$. If $Q = 0$ and $b^2 − 3ac = 0$, the three roots are equal: $$x_1=x_2=x_3=-\frac{b}{3a}$$. If $Q = 0$ and $b^2-3ac \mbox{ } \neq \mbox{ } 0$, above expression for the roots is correct but misleading, hiding the fact that no radical is needed to represent the roots. In fact, in this case, there is a double root $$x_1=x_2=\frac{bc-9ad}{2(3ac-b^2)}$$. and a simple root $$x_3=\frac{9a^2d-4abc+b^3}{a(3ac-b^2)}$$. (continued) source: Wikipedia

• $\begingroup$ In view of the extraordinary length of the quotation, it could well be suggested that you simply link that page, and then explain what you want to express??Per chance, I am making some unconscious mistake, but I really see no reason to post such a long quotation, instead of one simple link(as you already did), without the content; in any case, what is the link for?? $\endgroup$ –  awllower Commented Sep 4, 2011 at 9:24
• 3 $\begingroup$ @awllower: it's a standard practice in Stack Exchange to quote the most important bits of the source site due to the possibility of link rot (although with a popular article in Wikipedia, the chance of that happening is probably closer to nil). The link was there since I originally intended for a (much) shorter quote, but ended up converting the whole section. $\endgroup$ –  Lie Ryan Commented Sep 4, 2011 at 9:51

You must log in to answer this question.

Not the answer you're looking for browse other questions tagged polynomials roots factoring cubics ..

• Upcoming Events
• 2024 Community Moderator Election ends tomorrow
• Featured on Meta
• We've made changes to our Terms of Service & Privacy Policy - July 2024
• Bringing clarity to status tag usage on meta sites
• Upcoming Moderator Election
• 2024 Community Moderator Election

Hot Network Questions

• Which better conducts vibration: wood or metal?
• Creating a deadly "minimum altitude limit" in an airship setting
• Is it possible to prove that any two points of a convex and complete metric space M are connected by some metric segment without the axiom of choice?
• LED lights tripping the breaker when installed in the wrong direction?
• Can science inform philosophy?
• Print lines between two patterns where first pattern appears more than once before second pattern
• Using 尊敬語 and 謙譲語 without 丁寧語?
• Can the plasma jet emitted from a supermassive black hole form a naturally-occurring Tipler cylinder?
• How old were Phineas and Ferb? What year was it?
• GNU grep: This manpage is not compatible with mandoc
• Could a gas giant still be low on the horizon from the equator of a tidally locked moon?
• Why pay repo to finance bond position instead of reverse repo?
• Which aircraft has the simplest folding wing mechanism?
• Is math a bad discipline for teaching jobs or is it just me?
• Does the ship of Theseus have any impact on our perspective of life and death?
• Is it possible to create your own toy DNS root zone?
• How to stop Windows from changing date modified when copying files from network?
• Rounding vertices of polygon to fixed number of decimal places in QGIS
• How can the man remove the string loop hanging from his right hand?
• Font showing in Pages but not in Font book (or other apps)
• How to report the between study variance through tau2 for vaccine effectiveness using R's metafor?
• Is there a way to say "wink wink" or "nudge nudge" in German?
• Why does \appendix reset the counter of the level below the main sectioning level?
• Seasonal variation of aurora and magnetic activity

Game Central

Similar problems from web search.

• Institute of Telecommunications and Global Geoinformation Space of the National Academy of Sciences of Ukraine

What are the actual problems in physics, astrophysics, and cosmology?

Most recent answer.

Top contributors to discussions in this field

• Wu Yi University

• Technion - Israel Institute of Technology

• Texas A&M University-Commerce

• The World Islamic Science and Education University (WISE)

• La Salle Campus Madrid

Get help with your research

Join ResearchGate to ask questions, get input, and advance your work.

All Answers (14)

Similar questions and discussions

• Asked 11 December 2013

• Asked 13 October 2023

• Asked 2 August 2023

• Asked 13 September 2013

• Asked 28 October 2023

• Asked 15 January 2023

• Asked 10 August 2024
• Asked 4 August 2023

• Asked 14 December 2013

Related Publications

• Recruit researchers
• Join for free
• Login Email Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? Keep me logged in Log in or Continue with Google Welcome back! Please log in. Email · Hint Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? Keep me logged in Log in or Continue with Google No account? Sign up

• International
• Education Jobs
• Schools directory
• Resources Education Jobs Schools directory News Search

Trigonometric Equations - Multiple-Choice WS/Practice (20 problems + solutions)

Subject: Mathematics

Age range: 16+

Resource type: Worksheet/Activity

Last updated

18 August 2024

• Share through email
• Share through twitter
• Share through linkedin
• Share through facebook
• Share through pinterest

This is an engaging multiple-choice practice on solving trigonometric equations(basic, multiple angles, of quadratic type, such that can be solved by factoring). There are 5 pages, containing a total of 20 problems each having five optional answers. The first 8 problems are finding all the solutions of each equation given, the next 12 problems are determining the sum of the solutions of an equation on a closed interval. Students have an empty field below each problem where to write down their solution.

Detailed typed solutions are provided.

Tes paid licence How can I reuse this?

Your rating is required to reflect your happiness.

It's good to leave some feedback.

Something went wrong, please try again later.

This resource hasn't been reviewed yet

To ensure quality for our reviews, only customers who have purchased this resource can review it

Report this resource to let us know if it violates our terms and conditions. Our customer service team will review your report and will be in touch.

Not quite what you were looking for? Search by keyword to find the right resource:

Grab your spot at the free arXiv Accessibility Forum

Help | Advanced Search

Computer Science > Machine Learning

Title: nddes: a deep neural network framework for solving forward and inverse problems in delay differential equations.

Abstract: This article proposes a solution framework for delay differential equations (DDEs) based on deep neural networks (DNNs) - the neural delay differential equations (NDDEs), aimed at solving the forward and inverse problems of delay differential equations. This framework embeds the delay differential equations into the neural networks to accommodate the diverse requirements of DDEs in terms of initial conditions, control equations, and known data. NDDEs adjust the network parameters through automatic differentiation and optimization algorithms to minimize the loss function, thereby obtaining numerical solutions to the delay differential equations without the grid dependence and discretization errors typical of traditional numerical methods. In addressing inverse problems, the NDDE framework can utilize observational data to perform precise estimation of single or multiple delay parameters. The results of multiple numerical experiments have shown that NDDEs demonstrate high precision in both forward and inverse problems, proving their effectiveness and promising potential in dealing with delayed differential equation issues.

Submission history

Access paper:.

• HTML (experimental)
• Other Formats

References & Citations

• Google Scholar
• Semantic Scholar

BibTeX formatted citation

Bibliographic and Citation Tools

Code, data and media associated with this article, recommenders and search tools.

• Institution

arXivLabs: experimental projects with community collaborators

arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.

Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.

Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs .