Factor polynomial $$ 9t^{3}-90t^{2}+144t $$

$$ 9t^{3}-90t^{2}+144t = 9t( t-2 )( t-8 ) $$

Step 1 :

After factoring out $ 9t $ we have:

$$ 9t^{3}-90t^{2}+144t = 9t ( t^{2}-10t+16 ) $$

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:

$$ \color{blue}{ b = -10 } ~ \text{ and } ~ \color{red}{ c = 16 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -10 } $ and multiply to $ \color{red}{ 16 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 16 }$.

PRODUCT = 16
1     16	-1     -16
2     8	-2     -8
4     4	-4     -4

Step 4: Find out which pair sums up to $\color{blue}{ b = -10 }$

PRODUCT = 16 and SUM = -10
1     16	-1     -16
2     8	-2     -8
4     4	-4     -4

Step 5: Put -2 and -8 into placeholders to get factored form.

$$ \begin{aligned} t^{2}-10t+16 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ t^{2}-10t+16 & = (x -2)(x -8) \end{aligned} $$

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