Factor polynomial $$ 9b^{2}+65b+14 $$

$$ 9b^{2}+65b+14 = ( 9b+2 )( b+7 ) $$

Step 1: 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:

$$ a = 9 , b = 65 ~ \text{ and } ~ c = 14 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 9 }$ by the constant term $\color{blue}{c = 14} $.

$$ a \cdot c = 126 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 126 $ and add to $ b = 65 $.

Step 4: 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 5: Find out which factor pair sums up to $\color{blue}{ b = 65 }$

PRODUCT = 126 and SUM = 65
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: Replace middle term $ 65 x $ with $ 63x+2x $:

$$ 9x^{2}+65x+14 = 9x^{2}+63x+2x+14 $$

Step 7: Apply factoring by grouping. Factor $ 9x $ out of the first two terms and $ 2 $ out of the last two terms.

$$ 9x^{2}+63x+2x+14 = 9x\left(x+7\right) + 2\left(x+7\right) = \left(9x+2\right) \left(x+7\right) $$

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