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.

Part 1
Part 1 of 2:

Factoring By Grouping

  1. 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. 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. 3
    Factor the commonalities out of the two terms.
    • 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. 4
    If each of the two terms contains the same factor, you can combine the factors together.[2]
    • This gives you (x + 3)(x2 - 6).
  5. 5
    Find the solution by looking at the roots. If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.[3]
    • The solutions are -3, √6 and -√6.
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Part 2
Part 2 of 2:

Factoring Using the Free Term

  1. 1
    Rearrange the expression so it's in the form of ax3+bx2+cx+d.[4]
    • Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.
  2. 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. 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.
    • 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. 4
    Do a little rearranging. If x = 1, you can rearrange the statement to look a bit different without changing what it means.
    • "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. 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.
    • 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. 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. 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.
    • (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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Community Q&A

  • 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.
  • 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.
  • 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.
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About This Article

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 32 people, some anonymous, worked to edit and improve it over time. This article has been viewed 2,606,334 times.
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Co-authors: 32
Updated: February 10, 2023
Views: 2,606,334
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!

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