How do you factor completely $x^2-x-72$ ?

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Answer 1 · 2016-04-13T16:23:11

$color(blue)( (x + 8 ) ( x - 9)$ is the factorised form of the expression.

$x^2 - x - 72$

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:

$N_1*N_2 = a*c = 1*(-72) = -72$

AND

$N_1 +N_2 = b = -1$

After trying out a few numbers we get $N_1 = -9$ and $N_2 =8$

$8*(-9) = -72$ , and $8+(-9)= -1$

$x^2 - x - 72 = x^2 - 9x + 8x - 72$

$= x ( x - 9) + 8 ( x - 9 )$

$(x - 9 )$ is a common factor to each of the terms

$= (x + 8 ) ( x - 9)$

$color(blue)( (x + 8 ) ( x - 9)$ is the factorised form of the expression.

How do you factor completely x^2-x-72 x^2-x-72 ?