Step 1 :
After factoring out $ -2x $ we have:
$$ -2x^{3}-2x^{2}+40x = -2x ( x^{2}+x-20 ) $$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:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = -20 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ -20 } $.
Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -20 }$.
PRODUCT = -20 | |
-1 20 | 1 -20 |
-2 10 | 2 -10 |
-4 5 | 4 -5 |
Step 4: Find out which pair sums up to $\color{blue}{ b = 1 }$
PRODUCT = -20 and SUM = 1 | |
-1 20 | 1 -20 |
-2 10 | 2 -10 |
-4 5 | 4 -5 |
Step 5: Put -4 and 5 into placeholders to get factored form.
$$ \begin{aligned} x^{2}+x-20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+x-20 & = (x -4)(x + 5) \end{aligned} $$