2.5: Finding Factors from Roots

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) $\quad-6,-2$