Factor polynomial $$ 8n^{4}+88n^{3}-1216n^{2} $$

$$ 8n^{4}+88n^{3}-1216n^{2} = 8n^{2}( n-8 )( n+19 ) $$

Step 1 :

After factoring out $ 8n^{2} $ we have:

$$ 8n^{4}+88n^{3}-1216n^{2} = 8n^{2} ( n^{2}+11n-152 ) $$

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:

$$ \color{blue}{ b = 11 } ~ \text{ and } ~ \color{red}{ c = -152 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 11 } $ and multiply to $ \color{red}{ -152 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -152 }$.

PRODUCT = -152
-1     152	1     -152
-2     76	2     -76
-4     38	4     -38
-8     19	8     -19

Step 4: Find out which pair sums up to $\color{blue}{ b = 11 }$

PRODUCT = -152 and SUM = 11
-1     152	1     -152
-2     76	2     -76
-4     38	4     -38
-8     19	8     -19

Step 5: Put -8 and 19 into placeholders to get factored form.

$$ \begin{aligned} n^{2}+11n-152 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ n^{2}+11n-152 & = (x -8)(x + 19) \end{aligned} $$

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