Step 1 :
After factoring out $ -16 $ we have:
$$ -16t^{2}+64t+336 = -16 ( t^{2}-4t-21 ) $$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:
$$ \color{blue}{ b = -4 } ~ \text{ and } ~ \color{red}{ c = -21 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -4 } $ and multiply to $ \color{red}{ -21 } $.
Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -21 }$.
PRODUCT = -21 | |
-1 21 | 1 -21 |
-3 7 | 3 -7 |
Step 4: Find out which pair sums up to $\color{blue}{ b = -4 }$
PRODUCT = -21 and SUM = -4 | |
-1 21 | 1 -21 |
-3 7 | 3 -7 |
Step 5: Put 3 and -7 into placeholders to get factored form.
$$ \begin{aligned} t^{2}-4t-21 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ t^{2}-4t-21 & = (x + 3)(x -7) \end{aligned} $$