Factor polynomial $$ 3x^{2}+18x-120 $$

$$ 3x^{2}+18x-120 = 3( x-4 )( x+10 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 3x^{2}+18x-120 = 3 ( x^{2}+6x-40 ) $$

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:

$$ \color{blue}{ b = 6 } ~ \text{ and } ~ \color{red}{ c = -40 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 6 } $ and multiply to $ \color{red}{ -40 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -40 }$.

PRODUCT = -40
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 4: Find out which pair sums up to $\color{blue}{ b = 6 }$

PRODUCT = -40 and SUM = 6
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 5: Put -4 and 10 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+6x-40 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+6x-40 & = (x -4)(x + 10) \end{aligned} $$

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