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

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

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Answer 1 · 2016-05-06T21:31:12

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

Explanation:

Use an AC method:

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

The pair $45, 8$ $45, 8$ works.

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

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

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

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

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

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

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Footnote

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

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

The prime factorisation of $360$ $360$ is:

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

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

$1 \cdot 360$ $1 * 360$

$3 \cdot 120$ $3 * 120$

$5 \cdot 72$ $5 * 72$

$9 \cdot 40$ $9 * 40$

$15 \cdot 24$ $15 * 24$

$45 \cdot 8$ $45 * 8$

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

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