How do you factor $12d^2 + 4d - 1$ ?

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Answer 1 · 2015-09-23T12:47:02

$color(blue)((2d+1) (6d -1 )$ is the factorised form of the expression.

$12d^2 +4d -1$

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

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

$N_1*N_2 = a*c = 12*-1 = -12$
AND
$N_1 +N_2 = b = 4$

After trying out a few numbers we get $N_1 = -6$ and $N_2 =2$
$(-2)*6 = -12$ , and $6+(-2)= 4$

$12d^2 +color(blue)(4d) -1 = 12d^2 color(blue)(-2d +6d) -1$

$=2d (6d -1 ) +1(6d -1)$

$color(blue)((2d+1) (6d -1 )$ is the factorised form of the expression.

How do you factor 12d^2 + 4d - 112d^2 + 4d - 1?