# How do you factor completely 9x^2+9x-10?

Apr 17, 2016

$9 {x}^{2} + 9 x - 10 = \left(3 x - 2\right) \left(3 x + 5\right)$

#### Explanation:

One method involves completing the square, then using the difference of squares identity:

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

First multiply through by $4$ to make the arithmetic simpler, not forgetting to divide through by $4$ at the end.

$4 \left(9 {x}^{2} + 9 x - 10\right)$

$= 36 {x}^{2} + 36 x - 40$

$= {\left(6 x + 3\right)}^{2} - 9 - 40$

$= {\left(6 x + 3\right)}^{2} - {7}^{2}$

$= \left(\left(6 x + 3\right) - 7\right) \left(\left(6 x + 3\right) + 7\right)$

$= \left(6 x - 4\right) \left(6 x + 10\right)$

$= \left(2 \left(3 x - 2\right)\right) \left(2 \left(3 x + 5\right)\right)$

$= 4 \left(3 x - 2\right) \left(3 x + 5\right)$

Dividing both ends by $4$ we get:

$9 {x}^{2} + 9 x - 10 = \left(3 x - 2\right) \left(3 x + 5\right)$

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Why did I multiply by $4$ to start?

The middle term of the original quadratic expression is odd, so if you do not multiply through by some multiple of $4$ first then you end up doing arithmetic with $\frac{1}{2}$'s and $\frac{1}{4}$'s. It was not necessary to multiply through by anything else to avoid fractions since the leading term $9 {x}^{2} = {\left(3 x\right)}^{2}$ was already a perfect square.

Without premultiplying, it looks like this:

$9 {x}^{2} + 9 x - 10$

$= {\left(3 x + \frac{3}{2}\right)}^{2} - \frac{9}{4} - 10$

$= {\left(3 x + \frac{3}{2}\right)}^{2} - \frac{49}{4}$

$= {\left(3 x + \frac{3}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}$

$= \left(\left(3 x + \frac{3}{2}\right) - \frac{7}{2}\right) \left(\left(3 x + \frac{3}{2}\right) + \frac{7}{2}\right)$

$= \left(3 x - 2\right) \left(3 x + 5\right)$