Factor polynomial $$ 9x^{2}+90x-99 $$

$$ 9x^{2}+90x-99 = 9( x-1 )( x+11 ) $$

Step 1 :

After factoring out $ 9 $ we have:

$$ 9x^{2}+90x-99 = 9 ( x^{2}+10x-11 ) $$

Step 2 :

Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = 10 } ~ \text{ and } ~ \color{red}{ c = -11 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 10 } $ and multiply to $ \color{red}{ -11 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -11 }$.

PRODUCT = -11
-1     11	1     -11

Step 4: Find out which pair sums up to $\color{blue}{ b = 10 }$

PRODUCT = -11 and SUM = 10
-1     11	1     -11

Step 5: Put -1 and 11 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+10x-11 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+10x-11 & = (x -1)(x + 11) \end{aligned} $$

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