Digging Up Polynomial Roots with Factoring
When solving for roots (xintercepts of a polynomial), you usually need to factor the function rule and set it equal to 0. The factorization can be simple and obvious or complicated and obscure. You always hope for the simple and obvious, move to challenging and doable, and resort to the “big guns” when the factors are more obscure.
Before you go to those lengths, though, you need to exhaust other methods. The other methods of factoring include

Dividing out a greatest common factor (GCF)

Factoring a perfect square binomial

Factoring by grouping

Factoring quadraticlike trinomials
Sample questions

Find the roots (solutions) of the polynomial x^{7} – 82x^{5} + 81x^{3} = 0.
x= 0, 0, 0, 9, –9, 1, –1. This equation technically has seven solutions, but the 0 is a multiple root, so you end up with only five different numbers. To find these solutions, you first factor x^{3} out of each term to get x^{3}(x^{4} – 82x^{2} + 81) =
The two binomials are both the difference of perfect squares, so you can factor them into the difference and sum of the roots of the terms. You get x^{3}(x – 9)(x + 9)(x – 1)(x + 1) = 0. Setting each of the factors equal to 0, you find the roots.

Find the roots (solutions) of the polynomial x^{3} – 16x^{2} + 100x – 1,600 = 0.
x = 16. The polynomial doesn’t have a common factor in the four terms, but you can group the terms for the pairs of common factors. You get x^{2}(x – 16) + 100(x – 16) = 0, which factors into (x – 16)(x^{2} + 100) = 0. The second binomial is the sum of squares, which doesn’t factor.
Setting these two factors equal to 0, you get x = 16 from the first factor, but the second factor doesn’t produce any real answers. Even though you started with a thirddegree polynomial, which can yield up to three solutions, this polynomial has only one real root. The solution is just x = 16.
Practice questions

Find the roots (solutions) of the polynomial 3x^{4} – 12x^{3} – 27x^{2} + 108x = 0.

Find the roots (solutions) of the polynomial x^{5} – 16x^{3} + x^{2} – 16 = 0.

Find the roots (solutions) of the polynomial x^{6} + 9x^{3} + 8 = 0.

Find the roots (solutions) of the polynomial 36x^{5} – 13x^{3} + x = 0.
Following are answers to the practice questions:

The answer is x= 0, 3, –3, 4.
First factor 3x out of each term to get 3x(x^{3} – 4x^{2} – 9x + 36) = 0. You can then factor the terms in the parentheses by grouping: 3x[x^{2}(x – 4)–9(x – 4)] = 3x[(x – 4)(x^{2} – 9)] = 3x[(x – 4)(x – 3)(x + 3)] = 0. Set each factor equal to 0 to solve for the roots.

The answer is x= 4, –4, –1.
The polynomial factors by grouping: x^{3}(x^{2} – 16) + 1(x^{2} – 16) = (x^{2} – 16)(x^{3} + 1) = (x – 4)(x + 4)(x + 1)(x^{2} – x + 1) = 0. The first three factors give you the real roots. The last factor is a quadratic that has no real solution when you set it equal to 0.
You don’t really have to factor x^{3} + 1 in problem 6 to find the root. If you just set x^{3} + 1 equal to 0, you get x^{3} = –1, and taking the cube root of both sides gives you the solution –1. The factored form simply shows you how this problem could have had five roots, but not all are real numbers in this case.

The answer is x= –2, –1.
The polynomial is quadraticlike. It factors into (x^{3} + 8)(x^{3} + 1) = 0. Setting each factor equal to 0, you get the two roots.

The answer is
First, factor x out of each term to get x(36x^{4} – 13x^{2} + 1) = 0. The quadraticlike trinomial factors, giving you x(9x^{2} – 1)(4x^{2} – 1) = 0. Each binomial is the difference of squares, so both binomials factor. For the final factorization, you end up with x(3x – 1)(3x + 1)(2x – 1)(2x + 1) = 0. Setting each factor equal to 0 gives you the five different solutions.