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) \(\quad-6,-2\)