How To Factor A Cubic Polynomial: 12 Steps (with Pictures) - WikiHow

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Terms of Use wikiHow is where trusted research and expert knowledge come together. Learn why people trust wikiHow How to Factor a Cubic Polynomial PDF download Download Article Co-authored by Joseph Quinones

Last Updated: October 5, 2025 Fact Checked

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  • Factoring By Grouping
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  • Factoring Using the Free Term
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  • Practice Problems and Answers
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This article was co-authored by Joseph Quinones. Joseph Quinones is a Physics Teacher working at South Bronx Community Charter High School. Joseph specializes in astronomy and astrophysics and is interested in science education and science outreach, currently practicing ways to make physics accessible to more students with the goal of bringing more students of color into the STEM fields. He has experience working on Astrophysics research projects at the Museum of Natural History (AMNH). Joseph recieved his Bachelor's degree in Physics from Lehman College and his Masters in Physics Education from City College of New York (CCNY). He is also a member of a network called New York City Men Teach. There are 9 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 2,890,050 times.

This is an article about how to factorize a 3rd degree polynomial. We will explore how to factor using grouping as well as using the factors of the free term.

Steps

Part 1 Part 1 of 2:

Factoring By Grouping

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  1. Step 1 Group the polynomial into two sections. 1 Group the polynomial into two sections. Grouping the polynomial into two sections will let you attack each section individually.[1]
    • Say we're working with the polynomial x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)
  2. Step 2 Find what's the common in each section. 2 Find what's the common in each section.
    • Looking at (x3 + 3x2), we can see that x2 is common.
    • Looking at (- 6x - 18), we can see that -6 is common.
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  3. Step 3 Factor the commonalities out of the two terms. 3 Factor the commonalities out of the two terms.[2]
    • Factoring out x2 from the first section, we get x2(x + 3).
    • Factoring out -6 from the second section, you'll get -6(x + 3).
  4. Step 4 If each of the two terms contains the same factor, you can combine the factors together. 4 If each of the two terms contains the same factor, you can combine the factors together.[3]
    • This gives you (x + 3)(x2 - 6).
  5. Step 5 Find the solution by looking at the roots. 5 Find the solution by looking at the roots.[4] If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.[5]
    • The solutions are -3, √6 and -√6.
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Part 2 Part 2 of 2:

Factoring Using the Free Term

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  1. Step 1 Rearrange the expression so it's in the form of ax3+bx2+cx+d. 1 Rearrange the expression so it's in the form of ax3+bx2+cx+d.[6]
    • Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.
  2. Step 2 Find the all of the factors of "d". 2 Find the all of the factors of "d". The constant "d" is going to be the number that doesn't have any variables, such as "x," next to it.
    • Factors are the numbers you can multiply together to get another number. In your case, the factors of 10, or "d," are: 1, 2, 5, and 10.
  3. Step 3 Find one factor that causes the polynomial to equal to zero. 3 Find one factor that causes the polynomial to equal to zero. We want to determine which factor makes the polynomial equal zero when we substitute the factor for each "x" in the equation.[7]
    • Start by using your first factor, 1. Substitute "1" for each "x" in the equation: (1)3 - 4(1)2 - 7(1) + 10 = 0
    • This gives you: 1 - 4 - 7 + 10 = 0.
    • Because 0 = 0 is a true statement, you know that x = 1 is a solution.
  4. Step 4 Do a little rearranging. 4 Do a little rearranging. If x = 1, you can rearrange the statement to look a bit different without changing what it means.[8]
    • "x = 1" is the same thing as "x - 1 = 0" or "(x - 1)". You've just subtracted a "1" from each side of the equation.
  5. Step 5 Factor your root out of the rest of the equation. 5 Factor your root out of the rest of the equation. "(x - 1)" is our root. See if you can factor it out of the rest of the equation. Take it one polynomial at a time.[9]
    • Can you factor (x - 1) out of the x3? No you can't. But you can borrow a -x2 from the second variable; then factor it: x2(x - 1) = x3 - x2.
    • Can you factor (x - 1) out of what remains from your second variable? No, again you can't. You need to borrow another little bit from the third variable. You need to borrow a 3x from -7x. This gives you -3x(x - 1) = -3x2 + 3x.
    • Since you took a 3x from -7x, our third variable is now -10x and our constant is 10. Can you factor this? You can! -10(x - 1) = -10x + 10.
    • What you did was rearrange the variables so that you could factor out a (x - 1) out of the entire equation. Your rearranged equation looks like this: x3 - x2 - 3x2 + 3x - 10x + 10 = 0, but it's still the same thing as x3 - 4x2 - 7x + 10 = 0.
  6. Step 6 Continue to substitute by the factors of the free term. 6 Continue to substitute by the factors of the free term. Look at the numbers that you factored out using the (x - 1) in Step 5:
    • x2(x - 1) - 3x(x - 1) - 10(x - 1) = 0. You can rearrange this to be a lot easier to factor one more time: (x - 1)(x2 - 3x - 10) = 0.
    • You're only trying to factor (x2 - 3x - 10) here. This factors down into (x + 2)(x - 5).
  7. Step 7 Your solutions will be the factored roots. 7 Your solutions will be the factored roots. You can check whether your solutions actually work by plugging each one, individually, back into the original equation.[10]
    • (x - 1)(x + 2)(x - 5) = 0 This gives you solutions of 1, -2, and 5.
    • Plug -2 back into the equation: (-2)3 - 4(-2)2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.
    • Plug 5 back into the equation: (5)3 - 4(5)2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.
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Practice Problems and Answers

Practice Problems and Answers to Factor a Cubic Polynomial

Community Q&A

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  • Question What if the constant term is zero? Orangejews Orangejews Community Answer Then x is one of its factors. Factoring that out reduces the rest to a quadratic. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 86 Helpful 244
  • Question What if the third degree polynomial does not have the constant term? Community Answer Community Answer You just need to factor out the x. Let's say you're given: x^3+3x^2+2x=0. Then x(x^2+3x+2)=0. Roots: x -> 0; x^2+3x+1 -> (x+2)(x+1) -> -2, -1. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 179 Helpful 282
  • Question Can all three roots be imaginary? Orangejews Orangejews Community Answer No, not if all coefficients are real. Complex roots always come in conjugate pairs and polynomials always have exactly as many roots as its degree, so a cubic might have 3 real roots, or 1 real root and 2 complex roots. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 74 Helpful 130
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  • The cubic polynomial is a product of three first-degree polynomials or a product of one first-degree polynomial and another unfactorable second-degree polynomial. In this last case you use long division after finding the first-degree polynomial to get the second-degree polynomial. Thanks Helpful 129 Not Helpful 71
  • There are no unfactorable cubic polynomials over the real numbers because every cubic must have a real root. Cubics such as x^3 + x + 1 that have an irrational real root cannot be factored into polynomials with integer or rational coefficients. While it can be factored with the cubic formula, it is irreducible as an integer polynomial. Thanks Helpful 3 Not Helpful 2
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Expert Interview

Thanks for reading our article! If you’d like to learn more about physics, check out our in-depth interview with Joseph Quinones.

References

  1. ↑ https://web.math.ucsb.edu/~vtkala/2016/S/4B/FactoringCubicPolynomials.pdf
  2. ↑ https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/7.10/primary/lesson/factorization-by-grouping-alg-i-hnrs/
  3. ↑ https://www.sciencing.com/solve-cubic-polynomials-2409/
  4. ↑ Joseph Quinones. Physics Teacher. Expert Interview
  5. ↑ https://www.mathsisfun.com/algebra/polynomials-solving.html
  6. ↑ https://www.geeksforgeeks.org/maths/solving-cubic-equations/
  7. ↑ https://www.cuemath.com/algebra/factoring-cubic-polynomials/
  8. ↑ https://commons.nmu.edu/cgi/viewcontent.cgi?article=1376&context=facwork_journalarticles
  9. ↑ https://commons.nmu.edu/cgi/viewcontent.cgi?article=1376&context=facwork_journalarticles
More References (1)
  1. ↑ https://www.calculatorsoup.com/calculators/algebra/cubicequation.php

About This Article

Joseph Quinones Co-authored by: Joseph Quinones Physics Teacher This article was co-authored by Joseph Quinones. Joseph Quinones is a Physics Teacher working at South Bronx Community Charter High School. Joseph specializes in astronomy and astrophysics and is interested in science education and science outreach, currently practicing ways to make physics accessible to more students with the goal of bringing more students of color into the STEM fields. He has experience working on Astrophysics research projects at the Museum of Natural History (AMNH). Joseph recieved his Bachelor's degree in Physics from Lehman College and his Masters in Physics Education from City College of New York (CCNY). He is also a member of a network called New York City Men Teach. This article has been viewed 2,890,050 times. 29 votes - 66% Co-authors: 33 Updated: October 5, 2025 Views: 2,890,050 Categories: Algebra Article SummaryX

To factor a cubic polynomial, start by grouping it into 2 sections. Then, find what's common between the terms in each group, and factor the commonalities out of the terms. If each of the 2 terms contains the same factor, combine them. Finally, solve for the variable in the roots to get your solutions. To learn how to factor a cubic polynomial using the free form, scroll down! Did this summary help you?YesNo

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Yes No Did this article help you? Say thanks with a small tip. icon $1 $3 $5 Leave a Tip Support our mission to help everyone in the world learn how to do anything. Advertisement Cookies make wikiHow better. By continuing to use our site, you agree to our cookie policy. Joseph Quinones Co-authored by: Joseph Quinones Physics Teacher 29 votes - 66% Click a star to vote Co-authors: 33 Updated: October 5, 2025 Views: 2,890,050 Rich Aversa

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"Nice format. Should be careful with terminology: for instance, in step 5 of Factoring Using the Free Term, (x-1) is..." more Anonymous

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