How do you write $d^{2} - 12 d + 32$ $d^2-12d+32$ in factored form?

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Answer 1 · 2015-09-20T16:39:25

$(d - 4) (d - 8)$ $color(blue)((d-4)(d-8)$ is the factorised form of the expression.

$d^{2} - 12 d + 32$ $d^2−12d+32$

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

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

$N_{1} \cdot N_{2} = a \cdot c = 1 \cdot 32 = 32$ $N_1*N_2 = a*c = 1* 32 = 32$

AND

$N_{1} + N_{2} = b = - 12$ $N_1 +N_2 = b = -12$

After trying out a few numbers we get $N_{1} = - 8$ $N_1 = -8$ and $N_{2} = - 4$ $N_2 =-4$

$(- 8) \cdot (- 4) = 32$ $(-8)*(-4) = 32$ and $(- 8) + (- 4) = - 12$ $(-8)+ (- 4)= -12$

$d^{2} - 12 d + 32 = d^{2} - 8 d - 4 d + 32$ $d^2−12d+32 = d^2−8d -4d+32$

$= d (d - 8) - 4 (d - 8)$ $=d(d-8) -4(d-8)$

$(d - 4) (d - 8)$ $color(blue)((d-4)(d-8)$ is the factorised form of the expression.

How do you write d2−12d+32d^2-12d+32 in factored form?