How to Factor a Cubic Polynomial

Two Parts:Factoring By GroupingFactoring Using the Free Term

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
Factoring By Grouping

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    Group the polynomial into two sections. Grouping the polynomial into two sections will let you attack each section individually.
    • Say, "We're working with the polynomial." x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)
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    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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    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).
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    If each of the two terms contains the same factor, you can combine the factors together.
    • This gives you (x + 3)(x2 - 6).
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    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.
    • The solutions are -3, √6 and -√6.

Part 2
Factoring Using the Free Term

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    Rearrange the expression so it's in the form of aX3+bX2+cX+d.
    • Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.
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    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.
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    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.
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    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.
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    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.
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    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).
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    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.

Tips

  • 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.
  • 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.

Article Info

Categories: Algebra