Factor polynomial $$ 21b^{2}+87b-90 $$

$$ 21b^{2}+87b-90 = 3( 7b-6 )( b+5 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 21b^{2}+87b-90 = 3 ( 7b^{2}+29b-30 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 7 , b = 29 ~ \text{ and } ~ c = -30 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -30} $.

$$ a \cdot c = -210 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -210 $ and add to $ b = 29 $.

Step 5: All pairs of numbers with a product of $ -210 $ are:

PRODUCT = -210
-1     210	1     -210
-2     105	2     -105
-3     70	3     -70
-5     42	5     -42
-6     35	6     -35
-7     30	7     -30
-10     21	10     -21
-14     15	14     -15

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 29 }$

PRODUCT = -210 and SUM = 29
-1     210	1     -210
-2     105	2     -105
-3     70	3     -70
-5     42	5     -42
-6     35	6     -35
-7     30	7     -30
-10     21	10     -21
-14     15	14     -15

Step 7: Replace middle term $ 29 x $ with $ 35x-6x $:

$$ 7x^{2}+29x-30 = 7x^{2}+35x-6x-30 $$

Step 8: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -6 $ out of the last two terms.

$$ 7x^{2}+35x-6x-30 = 7x\left(x+5\right) -6\left(x+5\right) = \left(7x-6\right) \left(x+5\right) $$

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