Factor polynomial $$ 4a^{4}+32a^{3}-36a^{2} $$

$$ 4a^{4}+32a^{3}-36a^{2} = 4a^{2}( a-1 )( a+9 ) $$

Step 1 :

After factoring out $ 4a^{2} $ we have:

$$ 4a^{4}+32a^{3}-36a^{2} = 4a^{2} ( a^{2}+8a-9 ) $$

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:

$$ \color{blue}{ b = 8 } ~ \text{ and } ~ \color{red}{ c = -9 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 8 } $ and multiply to $ \color{red}{ -9 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -9 }$.

PRODUCT = -9
-1     9	1     -9
-3     3	3     -3

Step 4: Find out which pair sums up to $\color{blue}{ b = 8 }$

PRODUCT = -9 and SUM = 8
-1     9	1     -9
-3     3	3     -3

Step 5: Put -1 and 9 into placeholders to get factored form.

$$ \begin{aligned} a^{2}+8a-9 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ a^{2}+8a-9 & = (x -1)(x + 9) \end{aligned} $$

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