Factor polynomial $$ 3x^{3}-54x^{2}+240x $$

$$ 3x^{3}-54x^{2}+240x = 3x( x-8 )( x-10 ) $$

Step 1 :

After factoring out $ 3x $ we have:

$$ 3x^{3}-54x^{2}+240x = 3x ( x^{2}-18x+80 ) $$

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:

$$ \color{blue}{ b = -18 } ~ \text{ and } ~ \color{red}{ c = 80 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -18 } $ and multiply to $ \color{red}{ 80 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 80 }$.

PRODUCT = 80
1     80	-1     -80
2     40	-2     -40
4     20	-4     -20
5     16	-5     -16
8     10	-8     -10

Step 4: Find out which pair sums up to $\color{blue}{ b = -18 }$

PRODUCT = 80 and SUM = -18
1     80	-1     -80
2     40	-2     -40
4     20	-4     -20
5     16	-5     -16
8     10	-8     -10

Step 5: Put -8 and -10 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-18x+80 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-18x+80 & = (x -8)(x -10) \end{aligned} $$

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