How do you factor $d^{2} + 8 d + 7$ $d^2 + 8d +7$ ?

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Answer 1 · 2016-05-20T11:22:24

$(d + 1) (d + 7)$ $color(green)( (d + 1) ( d + 7 )$ is the factorised form of the expression.

$d^{2} + 8 d + 7$ $d^2 + 8d + 7$

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:

$N_{1} \cdot N_{2} = a \cdot c = 1 \cdot 7 = 7$ $N_1*N_2 = a*c = 1* 7 = 7$

AND

$N_{1} + N_{2} = b = 8$ $N_1 +N_2 = b = 8$

After trying out a few numbers we get $N_{1} = 7$ $N_1 = 7$ and $N_{2} = 1$ $N_2 =1$
$7 \cdot 1 = 7$ $7*1 = 7$ , and $7 + 1 = 8$ $7 + 1 = 8$

$d^{2} + 8 d + 7 = d^{2} + (7 d + 1 d) + 7$ $d^2 + color(green)(8d) + 7 = d^2 + color(green)((7d + 1d)) + 7$

$= d (d + 7) + 1 (d + 7)$ $= d ( d + 7 ) + 1 ( d + 7 )$

$(d + 7)$ $(d+7)$ is a common factor to each of the terms

$= (d + 1) (d + 7)$ $=color(green)( (d + 1) ( d + 7 )$

How do you factor d2+8d+7d^2 + 8d +7?