# Finding the Roots of a Factored Equation

In pre-calculus, you can use the zero-product property to find the roots of a factored equation. After you factor a polynomial into its different pieces, you can set each piece equal to zero to solve for the roots with the zero-product property. The *zero-product property* says that if several factors are multiplying to give you zero, at least one of them has to be zero. Your job is to find all the values of *x* that make the polynomial equal to zero. This task is much easier if the polynomial is factored because you can set each factor equal to zero and solve for *x.*

Factoring *x*^{2} + 3*x* – 10 = 0 gives you (*x* + 5)(*x* – 2). Moving forward is easy because each factor is linear (first degree). The term *x* + 5 = 0 gives you one solution, *x* = –5, and *x* – 2 = 0 gives you the other solution, *x* = 2.

These solutions each become an *x*-intercept on the graph of the polynomial.

Sometimes after you’ve factored, one or both of the two factors can be factored again, in which case you should continue factoring. In other cases, they may be unfactorable. If one of these factors is a quadratic, you can find the roots only by using the quadratic formula. For example, 6*x*^{4} – 12*x*^{3} + 4*x*^{2} = 0 factors to 2*x*^{2}(3*x*^{2} – 6*x* + 2) = 0. The first term, 2*x*^{2} = 0, is solvable using algebra, but the second factor, 3*x*^{2} – 6*x* + 2 = 0, is unfactorable and requires the quadratic formula.

In other cases, they may unfactorable, in which case you can solve them only by using the quadratic formula.