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

Sep 17, 2015

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

#### Explanation:

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

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

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

${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot 32 = 32$
and
${N}_{1} + {N}_{2} = b = 12$

After trying out a few numbers we get ${N}_{1} = 8$ and ${N}_{2} = 4$
$8 \cdot 4 = 32$, and $8 + 4 = 12$

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

$d \left(d + 8\right) + 4 \left(d + 8\right)$

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

Sep 17, 2015

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

Ans: $\left(x + 4\right) \left(x + 8\right)$

#### Explanation:

I use the new AC Method (Socratic Search)

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

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

$4 + 8 = 12 = b$

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

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

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.