Factor polynomial $$ 3x^{3}-18x^{2}+24x $$

$$ 3x^{3}-18x^{2}+24x = 3x( x-2 )( x-4 ) $$

Step 1 :

After factoring out $ 3x $ we have:

$$ 3x^{3}-18x^{2}+24x = 3x ( x^{2}-6x+8 ) $$

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:

$$ \color{blue}{ b = -6 } ~ \text{ and } ~ \color{red}{ c = 8 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -6 } $ and multiply to $ \color{red}{ 8 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 8 }$.

PRODUCT = 8
1     8	-1     -8
2     4	-2     -4

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

PRODUCT = 8 and SUM = -6
1     8	-1     -8
2     4	-2     -4

Step 5: Put -2 and -4 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-6x+8 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-6x+8 & = (x -2)(x -4) \end{aligned} $$

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