How do you factor $x^{2} - 2 x - 8$ $x^2 - 2x - 8$ ?

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How do you factor $x^{2} - 2 x - 8$ $x^2 - 2x - 8$ ?

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Answer 1 · 2015-06-10T17:36:42

$(x + 2) (x - 4)$ $color(blue)((x+2)(x - 4)$

Explanation:

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

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

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

$N_{1} \cdot N_{2} = a \cdot c = 1 \cdot (- 8) = - 8$ $N_1*N_2 = a*c = 1*(-8) = -8$
and
$N_{1} + N_{2} = b = - 2$ $N_1 +N_2 = b = -2$

After trying out a few numbers we get $N_{1} = - 4$ $N_1 = -4$ and $N_{2} = 2$ $N_2 =2$
$- 4 \cdot 2 = - 8$ $-4*2 = -8$ , and $2 + (- 4) = - 2$ $2+(-4)= -2$

$x^{2} - 2 x - 8 = x^{2} - 4 x + 2 x - 8$ $x^2 - 2x - 8 = x^2 - 4x +2x - 8$
$= x (x - 4) + 2 (x - 4)$ $=x(x - 4) +2(x - 4)$
$= (x + 2) (x - 4)$ $=color(blue)((x+2)(x - 4)$

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How do you factor $x^{2} - 2 x - 8$ $x^2 - 2x - 8$ ?

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