How do you factor completely: $8x^2 - 8x - 16$ ?

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Answer 1 · 2015-07-17T07:24:29

$color(blue)(8(x+1)(x−2)$

$8x^2−8x−16$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 8*(-16) = -128$
and
$N_1 +N_2 = b = -8$

After trying out a few numbers we get $N_1 = -16$ and $N_2 =8$

$(-16)*8 = -128$ , and $-16+8=-8$

$8x^2−color(blue)(8x)−16 = 8x^2−color(blue)(16x +8x)−16$

$= 8x(x−2) +8(x−2)$

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

$= color(blue)(8(x+1)(x−2)$ ,which is the factorised form.

How do you factor completely: 8x^2 - 8x - 168x^2 - 8x - 16?