Factor polynomial $$ 16q^{3}-48q^{2}+32q $$

$$ 16q^{3}-48q^{2}+32q = 16q( q-1 )( q-2 ) $$

Step 1 :

After factoring out $ 16q $ we have:

$$ 16q^{3}-48q^{2}+32q = 16q ( q^{2}-3q+2 ) $$

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:

$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = 2 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ 2 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 2 }$.

PRODUCT = 2
1     2	-1     -2

Step 4: Find out which pair sums up to $\color{blue}{ b = -3 }$

PRODUCT = 2 and SUM = -3
1     2	-1     -2

Step 5: Put -1 and -2 into placeholders to get factored form.

$$ \begin{aligned} q^{2}-3q+2 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ q^{2}-3q+2 & = (x -1)(x -2) \end{aligned} $$

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