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

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Answer 1 · 2016-05-06T21:31:12

$9x^2+53x+40=(9x+8)(x+5)$

Use an AC method:

Find a pair of factors of $AC=9*40 = 360$ with sum $B=53$

The pair $45, 8$ works.

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

$9x^2+53x+40$

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

$=(9x^2+45x)+(8x+40)$

$=9x(x+5)+8(x+5)$

$=(9x+8)(x+5)$

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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:

$360 = 2 * 2 * 2 * 3 * 3 * 5$

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

$1 * 360$

$3 * 120$

$5 * 72$

$9 * 40$

$15 * 24$

$45 * 8$

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

How do you factor completely 9x^2 + 53x +409x^2 + 53x +40?