Step 1 :
After factoring out $ -3x $ we have:
$$ -3x^{3}+27x^{2}-60x = -3x ( x^{2}-9x+20 ) $$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:
$$ \color{blue}{ b = -9 } ~ \text{ and } ~ \color{red}{ c = 20 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -9 } $ 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 $\color{blue}{ b = -9 }$
PRODUCT = 20 and SUM = -9 | |
1 20 | -1 -20 |
2 10 | -2 -10 |
4 5 | -4 -5 |
Step 5: Put -4 and -5 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-9x+20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-9x+20 & = (x -4)(x -5) \end{aligned} $$