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This is an article about how to factorize a 3 rd degree polynomial. We will explore how to factor using grouping as well as using the factors of the free term.
Steps
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Group the polynomial into two sections. Grouping the polynomial into two sections will let you attack each section individually. [1] X Research source- Say we're working with the polynomial x 3 + 3x 2 - 6x - 18 = 0. Let's group it into (x 3 + 3x 2 ) and (- 6x - 18)
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Find what's the common in each section.- Looking at (x 3 + 3x 2 ), we can see that x 2 is common.
- Looking at (- 6x - 18), we can see that -6 is common.
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If each of the two terms contains the same factor, you can combine the factors together. [3] X Research source- This gives you (x + 3)(x 2 - 6).
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Find the solution by looking at the roots. If you have an x 2 in your roots , remember that both negative and positive numbers fulfill that equation. [4] X Research source- The solutions are -3, √6 and -√6.
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Rearrange the expression so it's in the form of ax 3 +bx 2 +cx +d. [5] X Research source- Let's say you're working with the equation: x 3 - 4x 2 - 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.
- Start by using your first factor, 1. Substitute "1" for each "x" in the equation:
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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. [6] X Research source- "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. [7] X Research source- Can you factor (x - 1) out of the x 3 ? No you can't. But you can borrow a -x 2 from the second variable; then factor it: x 2 (x - 1) = x 3 - x 2 .
- 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) = -3x 2 + 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: x 3 - x 2 - 3x 2 + 3x - 10x + 10 = 0, but it's still the same thing as x 3 - 4x 2 - 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:- x 2 (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)(x 2 - 3x - 10) = 0.
- You're only trying to factor (x 2 - 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. [8] X Research source- (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
Community Q&A
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QuestionWhat if the constant term is zero?OrangejewsCommunity AnswerThen x is one of its factors. Factoring that out reduces the rest to a quadratic.
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QuestionWhat if the third degree polynomial does not have the constant term?Community AnswerYou 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.
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QuestionCan all three roots be imaginary?OrangejewsCommunity AnswerNo, 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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Tips
- 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
- 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
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References
- ↑ https://web.math.ucsb.edu/~vtkala/2016/S/4B/FactoringCubicPolynomials.pdf
- ↑ https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/7.10/primary/lesson/factorization-by-grouping-alg-i-hnrs/
- ↑ https://sciencing.com/solve-cubic-polynomials-2409.html
- ↑ https://www.mathsisfun.com/algebra/polynomials-solving.html
- ↑ https://www.geeksforgeeks.org/solving-cubic-equations/
- ↑ https://commons.nmu.edu/cgi/viewcontent.cgi?article=1376&context=facwork_journalarticles
- ↑ https://commons.nmu.edu/cgi/viewcontent.cgi?article=1376&context=facwork_journalarticles
- ↑ https://www.calculatorsoup.com/calculators/algebra/cubicequation.php
About This Article
Article Summary
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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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Reader Success Stories
- "Nice format. Should be careful with terminology: for instance, in step 5 of Factoring Using the Free Term, (x-1) is not a "root", it is just a key factor (the root is x=1). Also in step 5 we are not factoring "one polynomial at a time", we are factoring one /term/ at a time (the polynomial is the whole set of terms).To some this may seem like semantics, but to others trying to learn a new skill these little errors cause more confusion and the contradictions make the topic harder to understand." ..." more
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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 not a "root", it is just a key factor (the root is x=1). Also in step 5 we are not factoring "one polynomial at a time", we are factoring one /term/ at a time (the polynomial is the whole set of terms).To some this may seem like semantics, but to others trying to learn a new skill these little errors cause more confusion and the contradictions make the topic harder to understand."
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"I have my math exam tomorrow and, though I can do all the horribly, hard things, I am not able to factorize a cubic polynomial! This article helped me so, so much. With all the examples, it gave me a clear idea of how to do it. Thank you so much. :D"
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"I tried to find out how to use factoring for cubes using grouping. Every other article shows us that we need to pull one of the factors. Yours was the first one to show us how to get to the first factor so we knew what to pull out. Thank you."
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"This answered exactly the math question I asked, with color-coded step-by-step images and helpful, clear descriptions. I came away knowing 2 different ways to factor a cubic polynomial, just as the article intended and I desired."
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"The pictures are really helpful. I don't even have to read the entire thing to understand; I just view the images and the titles of the steps and I am good to go. It's much faster and easier this way!"
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