How do you factor $8 - 2b - b^2$ ?

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Answer 1 · 2015-06-18T08:21:20

$color(green)(-1(b-2)(b+4)$ is the factorised format.

$8 - 2b - b^2 =- (b^2 + 2b -8)$

We can Split the Middle Term of $b^2 + 2b -8$ 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:

$N_1*N_2 = a*c = 1*-8 = -8$
and
$N_1 +N_2 = b = 2$
After trying out a few numbers we get $N_1 = 4$ and $N_2 =-2$
$4*-2 = -8$ and $4+(-2)= 2$

$b^2 + 2b -8 = b^2 + 4b - 2b -8$
$= b(b+4) - 2(b+4)$
$b+4$ is common to both terms:
$=color(green)( (b-2)(b+4)$

How do you factor 8 - 2b - b^28 - 2b - b^2?