Factor polynomial $$ 3y^{5}+36y^{4}-84y^{3} $$

$$ 3y^{5}+36y^{4}-84y^{3} = 3y^{3}( y-2 )( y+14 ) $$

Step 1 :

After factoring out $ 3y^{3} $ we have:

$$ 3y^{5}+36y^{4}-84y^{3} = 3y^{3} ( y^{2}+12y-28 ) $$

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:

$$ \color{blue}{ b = 12 } ~ \text{ and } ~ \color{red}{ c = -28 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 12 } $ and multiply to $ \color{red}{ -28 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -28 }$.

PRODUCT = -28
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 4: Find out which pair sums up to $\color{blue}{ b = 12 }$

PRODUCT = -28 and SUM = 12
-1     28	1     -28
-2     14	2     -14
-4     7	4     -7

Step 5: Put -2 and 14 into placeholders to get factored form.

$$ \begin{aligned} y^{2}+12y-28 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ y^{2}+12y-28 & = (x -2)(x + 14) \end{aligned} $$

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