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

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Answer 1 · 2015-09-17T17:01:58

$(d + 4) (d + 8)$ $color(blue)((d+4)(d+8)$ 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.

Answer 2 · 2015-09-17T18:48:25

Factor: $d^{2} + 12 d + 32$ $d^2 + 12d + 32$

Ans: $(x + 4) (x + 8)$ $(x + 4)(x + 8)$

I use the new AC Method (Socratic Search)

$y = d^{2} + 12 d + 32 = (d + p) (d + q)$ $y = d^2 + 12d + 32 = (d + p)(d + q)$

Factor pairs of $(32) \to (2, 16) (4, 8)$ $(32) -> (2, 16)(4, 8)$ . This sum is

$4 + 8 = 12 = b$ $4 + 8 = 12 = b$

Then $p = 4$ $p = 4$ and $q = 8$ $q = 8$

Factored form: $y = (d + 4) (d + 8)$ $y = (d + 4)(d + 8)$

NOTE . This new AC Method shows a systematic way to find the 2 numbers p and q. It also avoids the lengthy factoring by grouping.

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