Question

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

Answer 1

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

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(+/-)√(400)} {16}`

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

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

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

therefore

`8x^(2) -12x-8=8(x-2)(x+(1)/(2))=(x-2)(8x+4)`