Factor polynomial $$ -4t^{2}+32t-60 $$

$$ -4t^{2}+32t-60 = ( -4 )( t-3 )( t-5 ) $$

Step 1 :

After factoring out $ -4 $ we have:

$$ -4t^{2}+32t-60 = -4 ( t^{2}-8t+15 ) $$

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:

$$ \color{blue}{ b = -8 } ~ \text{ and } ~ \color{red}{ c = 15 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -8 } $ and multiply to $ \color{red}{ 15 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 15 }$.

PRODUCT = 15
1     15	-1     -15
3     5	-3     -5

Step 4: Find out which pair sums up to $\color{blue}{ b = -8 }$

PRODUCT = 15 and SUM = -8
1     15	-1     -15
3     5	-3     -5

Step 5: Put -3 and -5 into placeholders to get factored form.

$$ \begin{aligned} t^{2}-8t+15 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ t^{2}-8t+15 & = (x -3)(x -5) \end{aligned} $$

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