Question

Quick algebra 1 question for 10 points!

Answer 1

$y = 2x {}^{2} - 4x - 16 \\ \delta = b {}^{2} - 4ac \\ \delta = ( - 4) {}^{2} - 4(2)( - 16) \\ \delta = 16 + 128 \\ \delta = 144 > 0$

$x = \frac{ - b + \sqrt{ \delta} }{2a} = \frac{4 + 12}{2 \times 2} = \frac{16}{4} = 4$

$x = \frac{ - b - \sqrt{ \delta} }{2a} = \frac{4 - 12}{2 \times 2} \frac{ - 8}{4} = - 2$

$x = 4 \: \: \: \: \: or \: \: \: \: x = - 2 \\ (x - 4) = 0 \: \: \: \: \: or \: \: \: \: \: (x + 2) = 0$

$f(x) = \alpha (x - 4)(x + 2) \\ f(x) = \alpha x { }^{2} + 2 \alpha x - 4 \alpha x - 8 \alpha \\ f(x) = \alpha x {}^{2} - 2 \alpha x - 8 \alpha \\ by \: comparison \\ \alpha = 2$

Factors are:

1) 2

2) x - 4

3) x + 2