# How do you find all the zeros of f(x) = 4(x + 3)^2 - 1?

Mar 5, 2016

#### Answer:

Set $f \left(x\right) = 0$ and solve for $x$ to find that $f \left(x\right)$ has zeroes at $x = - \frac{7}{2}$ and $x = - \frac{5}{2}$

#### Explanation:

$4 {\left(x + 3\right)}^{2} - 1 = 0$

$\implies 4 {\left(x + 3\right)}^{2} = 1$

$\implies {\left(x + 3\right)}^{2} = \frac{1}{4}$

$\implies x + 3 = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$

$\implies x = - 3 \pm \frac{1}{2}$

$\therefore x = - \frac{7}{2}$ or $x = - \frac{5}{2}$