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Factoring Polynomials (page 2 of 2)

Since factors and zeroes are so closely related, you can use solving techniques to factor: If a polynomial has a zero (a "solution") of x = a, then it also has a factor of x a. So algebraic tools that help you find "solutions" of polynomials can also help you factor polynomials.

About the only differences between "factoring" and "solving" exercises are that, for factoring, you have to keep track of any numbers that you divide off (like the 2 that I divided out of the quadratic in the previous problem) and you may need to use the difference- or sum-of-cubes formula. On the "plus" side, though, the polynomials for "factoring" exercises generally involve nicer numbers, without the complex values or square roots common in "solving" exercises.

  • Completely factor the following polynomial:
    30x5 166x4 542x3 + 2838x2 + 1520x 800
  • All the coefficients are even, so I can factor out a 2. But this isn't an equation, so I can't just "divide off" the 2; I'm going to have to remember to include that factor of 2 in my final factored form. Copyright Elizabeth Stapel 2005-2011 All Rights Reserved




    Even after dividing out the 2, I'm still stuck with some pretty big numbers: 15x5 83x4 271x3 + 1419x2 + 760x 400. The Rational Roots Test gives me a staggeringly-long list of possible roots, including loads of fractions, so I'm going to just stick with whole-number factors of 400, at least at first. I'll figure out the whole long list later, if I have to, but not before....

    Since the solutions to polynomials are usually pretty "close to the middle", I won't bother testing solutions like x = 400 in my synthetic division, at least not initially. I'll keep my first guesses small. And I'll check the graph right away to see which guesses seem most promising.

(If you don't have a graphing calculator, you've got a rough road ahead of you. Apply the Rational Roots Test. Start testing the smaller whole-number values it gives you (usually factors of the constant term), and work out from there. Keep in mind that a "solution" of "x = a" means you have a factor of "x a". In this example, you'd probably want to start with values like x = 1, 2, and 5. Be sure to use the synethetic division trick for knowing when to stop moving to higher or lower test values.)

graph of y = 15x^5 - 83x^4 - 271x^3 + 1419x^2 + 760x - 400

    I can tell from the graph that x = 5 looks like it'll be a zero, and a repeated one at that, since the graph just touches the axis instead of going through it. So if x = 5 divides out evenly, then the first thing I'll do is try x = 5 again.

      synthetic division with x = 5

    This means that x 5 is a factor twice, and the division leaves me with 15x3 + 67x2 + 24x 16. (I've taken out two zeroes two factors so the degree of the polynomial has gone from 5 down to 3.) I'll do another quick graph to see what I'm left working with now:

      graph of y = 15x^3 + 67x^2 + 24x - 16

    Judging from the graph, I'll try x = 4.

      synthetic division with x = -4

    Taking out the x + 4 factor leaves me with 15x2 + 7x 4. I can use regular factoring methods to finish the problem: 15x2 + 7x 4 = (5x + 4)(3x 1). Remembering the initial 2 that I factored out, the complete factorization is:

      2(x 5)(x 5)(x + 4)(5x + 4)(3x 1)

        = 2(x 5)2(x + 4)(5x + 4)(3x 1)

The factors don't have to be listed in any particular order, by the way, since order doesn't matter for multiplication. But any constant term (the "2", in this case) should go in front. And many books prefer repeated factors to be written in exponential form, like the "(x 5)2" in the second form of the final answer above.

A variant of this sort of question is where they give you a factor or a zero (usually something messy), and want you to find the rest of the factors or zeroes. There are examples of how this works in the last page of the synthetic division lesson.

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Cite this article as:

Stapel, Elizabeth. "Factoring Polynomials." Purplemath. Available from Accessed


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