# 2.5: Finding Factors from Roots

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One method of solving equations involves finding the factors of the polynomial expression in the equation and then setting each factor equal to zero.
$\begin{array}{c} x^{2}+8 x+15=0 \\ (x+5)(x+3)=0 \\ x+5=0 \quad x+3=0 \\ x=-5 \quad x=-3 \end{array}$
In this process, the reasoning is that if $$(x+5)$$ times $$(x+3)$$ equals zero, then one of those expressions must be equal to zero. In setting them equal to zero, we find the solutions of $$x=-5,-3 .$$ Plugging them back into the factored expression we see the following:
$(-5+5)(-5+3)=0 *-2=0$
and
$(-3+5)(-3+3)=2 * 0=0$
This process works in reverse as well. In other words, if we know a root of the function, we can find factors for the expression.
Example
Find a quadratic equation that has roots of -2 and +3
$\begin{array}{cc} x=-2 & x=3 \\ x+2=0 & x-3=0 \\ (x+2)(x-3)=0 \\ x^{2}-x-6=0 \end{array}$

Roots that are fractions are a little trickier, but really no more difficult:
Example
Find a quadratic equation that has roots of -5 and $$\frac{2}{3}$$
$\begin{array}{cc} x=-5 & x=\frac{2}{3} \\ x+5=0 & 3 x=2 \\ x+5=0 & 3 x-2=0 \\ (x+5)(3 x-2)=0 \\ 3 x^{2}+13 x-10=0 \end{array}$

Exercises 2.5
Find a quadratic equation that has the indicated roots.
1) $$\quad 4,-1$$
2) $$\quad -2,7$$
3) $$\quad \frac{3}{2}, 1$$
4) $$\quad-\frac{1}{5}, \frac{2}{3}$$
5) $$\quad \frac{1}{3}, 3$$
6) $$\quad-4, \frac{2}{5}$$
7) $$\quad \frac{1}{2},-\frac{7}{2}$$
8) $$\quad-1, \frac{3}{5}$$
9) $$\quad-\frac{2}{3},-3$$
10) $$\quad-\frac{2}{3},-\frac{3}{4}$$
11) $$\quad-\frac{5}{2}, 3$$
12) $$\quad-6,-2$$

This page titled 2.5: Finding Factors from Roots is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Richard W. Beveridge.