Question

Factor the trinomial 8x^2-12x-8

Answer 1

Answer:

$(x-2)(8x+4)$

Step-by-step explanation:

we have

$8x^{2} -12x-8$

Equate to zero and find the roots of the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$8x^{2} -12x-8=0$

so

$a=8\\b=-12\\c=-8$

substitute in the formula

$x=\frac{-(-12)(+/-)\sqrt{-12^{2}-4(8)(-8)}} {2(8)}$

$x=\frac{12(+/-)\sqrt{400}} {16}$

$x=\frac{12(+/-)20} {16}$

$x_1=\frac{12(+)20} {16}=2$

$x_2=\frac{12(-)20} {16}=-\frac{1}{2}$

therefore

$8x^{2} -12x-8=8(x-2)(x+\frac{1}{2})=(x-2)(8x+4)$