Factor polynomial $$ 28b^{2}+92b+72 $$

$$ 28b^{2}+92b+72 = 4( 7b+9 )( b+2 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 28b^{2}+92b+72 = 4 ( 7b^{2}+23b+18 ) $$

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:

$$ a = 7 , b = 23 ~ \text{ and } ~ c = 18 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 18} $.

$$ a \cdot c = 126 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 126 $ and add to $ b = 23 $.

Step 5: All pairs of numbers with a product of $ 126 $ are:

PRODUCT = 126
1     126	-1     -126
2     63	-2     -63
3     42	-3     -42
6     21	-6     -21
7     18	-7     -18
9     14	-9     -14

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 23 }$

PRODUCT = 126 and SUM = 23
1     126	-1     -126
2     63	-2     -63
3     42	-3     -42
6     21	-6     -21
7     18	-7     -18
9     14	-9     -14

Step 7: Replace middle term $ 23 x $ with $ 14x+9x $:

$$ 7x^{2}+23x+18 = 7x^{2}+14x+9x+18 $$

Step 8: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ 9 $ out of the last two terms.

$$ 7x^{2}+14x+9x+18 = 7x\left(x+2\right) + 9\left(x+2\right) = \left(7x+9\right) \left(x+2\right) $$

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