# How do you list all possible roots and find all factors of 4x^2-9?

Sep 28, 2016

Roots: $\pm \frac{3}{2}$ Factors: $\left(2 x + 3\right) \left(2 x - 3\right)$

#### Explanation:

$f \left(x\right) = 4 {x}^{2} - 9$

The roots of $f \left(x\right)$ are the values of $x$ for which $f \left(x\right) = 0$

Since $f \left(x\right)$ is of 2nd degree, we know that the number of roots is at most 2.

Notice that both $4 {x}^{2}$ and $9$ are square values and remember that: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

In this example $a = 2 x$ and $b = 3$

$\therefore f \left(x\right) = \left(2 x + 3\right) \left(2 x - 3\right)$

$f \left(x\right) = 0$ when either $\left(2 x + 3\right) = 0$ or $\left(2 x - 3\right) = 0$

Hence the roots of $f \left(x\right)$ are $\pm \frac{3}{2}$