# How do you factor completely 9x^2 + 53x +40?

May 6, 2016

$9 {x}^{2} + 53 x + 40 = \left(9 x + 8\right) \left(x + 5\right)$

#### Explanation:

Use an AC method:

Find a pair of factors of $A C = 9 \cdot 40 = 360$ with sum $B = 53$

The pair $45 , 8$ works.

Use this pair to split the middle term and factor by grouping:

$9 {x}^{2} + 53 x + 40$

$= 9 {x}^{2} + 45 x + 8 x + 40$

$= \left(9 {x}^{2} + 45 x\right) + \left(8 x + 40\right)$

$= 9 x \left(x + 5\right) + 8 \left(x + 5\right)$

$= \left(9 x + 8\right) \left(x + 5\right)$

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Footnote

How did I find the pair $45 , 8$?

Note that $53$ is odd, so as a sum of two numbers one must be odd and the other even.

The prime factorisation of $360$ is:

$360 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5$

So the only possible splits into two factors which put all the $2$'s on the right hand side are:

$1 \cdot 360$

$3 \cdot 120$

$5 \cdot 72$

$9 \cdot 40$

$15 \cdot 24$

$45 \cdot 8$

The last of these works, in that $45 + 8 = 53$