How do you solve $x^2 - x - 12 = 0$ by factoring?

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Answer 1 · 2015-08-16T17:16:21

The solutions are
$color(blue)(x=-3$
$color(blue)(x=4$

$x^2−x−12=0$

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

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

$N_1*N_2 = a*c = 1*-12 = -12$
AND
$N_1 +N_2 = b = -1$

After trying out a few numbers we get $N_1 = 3$ and $N_2 =-4$
$3*-4 = -12$ , and $3+(-4)= -1$

$x^2−x−12=x^2−4x+3x−12$

$x(x-4) +3(x−4)=0$

$(x+3)(x-4)=0$

Now we equate the factors to zero.
$x+3 = 0 , color(blue)(x=-3$
$x-4 = 0 , color(blue)(x=4$

How do you solve x^2 - x - 12 = 0x^2 - x - 12 = 0 by factoring?