Factor polynomial $$ 7a^{3}-14a^{2}-21a $$

$$ 7a^{3}-14a^{2}-21a = 7a( a+1 )( a-3 ) $$

Step 1 :

After factoring out $ 7a $ we have:

$$ 7a^{3}-14a^{2}-21a = 7a ( a^{2}-2a-3 ) $$

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:

$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = -3 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ -3 } $.

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

PRODUCT = -3
-1     3	1     -3

Step 4: Find out which pair sums up to $\color{blue}{ b = -2 }$

PRODUCT = -3 and SUM = -2
-1     3	1     -3

Step 5: Put 1 and -3 into placeholders to get factored form.

$$ \begin{aligned} a^{2}-2a-3 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ a^{2}-2a-3 & = (x + 1)(x -3) \end{aligned} $$

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