Factor polynomial $$ 9j^{2}+27j-90 $$

$$ 9j^{2}+27j-90 = 9( j-2 )( j+5 ) $$

Step 1 :

After factoring out $ 9 $ we have:

$$ 9j^{2}+27j-90 = 9 ( j^{2}+3j-10 ) $$

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:

$$ \color{blue}{ b = 3 } ~ \text{ and } ~ \color{red}{ c = -10 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 3 } $ and multiply to $ \color{red}{ -10 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -10 }$.

PRODUCT = -10
-1     10	1     -10
-2     5	2     -5

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

PRODUCT = -10 and SUM = 3
-1     10	1     -10
-2     5	2     -5

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

$$ \begin{aligned} j^{2}+3j-10 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ j^{2}+3j-10 & = (x -2)(x + 5) \end{aligned} $$

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