# How do you factor y=x^2 + 8x - 9 ?

Apr 2, 2016

$y = {x}^{2} + 8 x - 9 \text{ "->" } y = \left(x + 9\right) \left(x - 1\right)$

#### Explanation:

As there is no coefficient of ${x}^{2}$ so we are only looking for factors of 9 where their difference is 8.

Consider the factors 1 and 9

We know that $1 \times 9 = 9$. So that is ok!

We also know that $9 - 1 = 8$ So that is ok as well!

So we have the general form of (x+?)(x+?)

The 8 from $8 x$ is positive so for the difference we need 9 to be positive and the 1 to be negative so that we end up with $9 - 1 = 8$

color(blue)((x+9)color(brown)((x-1))

Just as a check lets multiply out the brackets

$\textcolor{b r o w n}{\textcolor{b l u e}{x} \left(x - 1\right) \textcolor{b l u e}{+ 9} \left(x - 1\right)}$

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

${x}^{2} + 8 x - 9 \textcolor{red}{\text{ Works ok!}}$
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$y = {x}^{2} + 8 x - 9 \text{ "->" } y = \left(x + 9\right) \left(x - 1\right)$