Answer:
[tex](x-2)(8x+4)[/tex]
Step-by-step explanation:
we have
[tex]8x^{2} -12x-8[/tex]
Equate to zero and find the roots of the quadratic equation
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]8x^{2} -12x-8=0[/tex]
so
[tex]a=8\\b=-12\\c=-8[/tex]
substitute in the formula
[tex]x=\frac{-(-12)(+/-)\sqrt{-12^{2}-4(8)(-8)}} {2(8)}[/tex]
[tex]x=\frac{12(+/-)\sqrt{400}} {16}[/tex]
[tex]x=\frac{12(+/-)20} {16}[/tex]
[tex]x_1=\frac{12(+)20} {16}=2[/tex]
[tex]x_2=\frac{12(-)20} {16}=-\frac{1}{2}[/tex]
therefore
[tex]8x^{2} -12x-8=8(x-2)(x+\frac{1}{2})=(x-2)(8x+4)[/tex]
