# How do you find the zeros by rewriting the function y=2x^2-578 in intercept form?

Jul 20, 2017

$x = \pm 17$

#### Explanation:

$\text{take out a "color(blue)"common factor"" of 2}$

$\Rightarrow y = 2 \left({x}^{2} - 289\right)$

${x}^{2} - 289 \text{ is a "color(blue)"difference of squares}$

$\left[\text{note } 289 = {17}^{2}\right]$

$\Rightarrow y = 2 \left(x - 17\right) \left(x + 17\right) \leftarrow \textcolor{red}{\text{in-intercept form}}$

$\text{to obtain the zeros set y = 0}$

$\Rightarrow 2 \left(x - 17\right) \left(x + 17\right) = 0$

$\text{set each factor to zero and solve for x}$

$x - 17 = 0 \Rightarrow x = 17$

$x + 17 = 0 \Rightarrow x = - 17$

$\Rightarrow \text{ the zeros are } x = \pm 17$