How do you factor $x^{2} + 5 x - 36$ $x^2 + 5x - 36$ ?

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How do you factor $x^{2} + 5 x - 36$ $x^2 + 5x - 36$ ?

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Answer 1 · 2016-05-23T21:05:29

$x^{2} + 5 x - 36 = (x + 9) (x - 4)$ $x^2+5x-36=(x+9)(x-4)$

Explanation:

Find a pair of factors of $36$ $36$ which differ by $5$ $5$ .

The pair $9, 4$ $9, 4$ works in that $9 \times 4 = 36$ $9xx4=36$ and $9 - 4 = 5$ $9-4=5$

Hence we find:

$x^{2} + 5 x - 36 = (x + 9) (x - 4)$ $x^2+5x-36=(x+9)(x-4)$

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Alternative method

Alternatively, we can complete the square and use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

with $a = (2 x + 5)$ $a=(2x+5)$ and $b = 13$ $b=13$ as follows.

First multiply by $2^{2} = 4$ $2^2 = 4$ to cut down on arithmetic involving fractions. Remember to divide by it at the end...

$4 (x^{2} + 5 x - 36)$ $4(x^2+5x-36)$

$= 4 x^{2} + 20 x - 144$ $=4x^2+20x-144$

$= {(2 x)}^{2} + 2 (2 x) (5) - 144$ $=(2x)^2+2(2x)(5)-144$

$= {(2 x + 5)}^{2} - 25 - 144$ $=(2x+5)^2-25-144$

$= {(2 x + 5)}^{2} - 169$ $=(2x+5)^2-169$

$= {(2 x + 5)}^{2} - 13^{2}$ $=(2x+5)^2-13^2$

$= ((2 x + 5) - 13) ((2 x + 5) + 13)$ $=((2x+5)-13)((2x+5)+13)$

$= (2 x - 8) (2 x + 18)$ $=(2x-8)(2x+18)$

$= (2 (x - 4)) (2 (x + 9))$ $=(2(x-4))(2(x+9))$

$= 4 (x - 4) (x + 9)$ $=4(x-4)(x+9)$

Dividing both ends by $4$ $4$ we find:

$x^{2} + 5 x - 36 = (x - 4) (x + 9)$ $x^2+5x-36 = (x-4)(x+9)$

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How do you factor $x^{2} + 5 x - 36$ $x^2 + 5x - 36$ ?

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