Factor polynomial $$ 2t^{4}+10t^{3}-28t^{2} $$

$$ 2t^{4}+10t^{3}-28t^{2} = 2t^{2}( t-2 )( t+7 ) $$

Step 1 :

After factoring out $ 2t^{2} $ we have:

$$ 2t^{4}+10t^{3}-28t^{2} = 2t^{2} ( t^{2}+5t-14 ) $$

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:

$$ \color{blue}{ b = 5 } ~ \text{ and } ~ \color{red}{ c = -14 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 5 } $ and multiply to $ \color{red}{ -14 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -14 }$.

PRODUCT = -14
-1     14	1     -14
-2     7	2     -7

Step 4: Find out which pair sums up to $\color{blue}{ b = 5 }$

PRODUCT = -14 and SUM = 5
-1     14	1     -14
-2     7	2     -7

Step 5: Put -2 and 7 into placeholders to get factored form.

$$ \begin{aligned} t^{2}+5t-14 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ t^{2}+5t-14 & = (x -2)(x + 7) \end{aligned} $$

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