Factor polynomial $$ 96x^{2}+82x+12 $$

$$ 96x^{2}+82x+12 = 2( 16x+3 )( 3x+2 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 96x^{2}+82x+12 = 2 ( 48x^{2}+41x+6 ) $$

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:

$$ a = 48 , b = 41 ~ \text{ and } ~ c = 6 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 48 }$ by the constant term $\color{blue}{c = 6} $.

$$ a \cdot c = 288 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 288 $ and add to $ b = 41 $.

Step 5: All pairs of numbers with a product of $ 288 $ are:

PRODUCT = 288
1     288	-1     -288
2     144	-2     -144
3     96	-3     -96
4     72	-4     -72
6     48	-6     -48
8     36	-8     -36
9     32	-9     -32
12     24	-12     -24
16     18	-16     -18

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 41 }$

PRODUCT = 288 and SUM = 41
1     288	-1     -288
2     144	-2     -144
3     96	-3     -96
4     72	-4     -72
6     48	-6     -48
8     36	-8     -36
9     32	-9     -32
12     24	-12     -24
16     18	-16     -18

Step 7: Replace middle term $ 41 x $ with $ 32x+9x $:

$$ 48x^{2}+41x+6 = 48x^{2}+32x+9x+6 $$

Step 8: Apply factoring by grouping. Factor $ 16x $ out of the first two terms and $ 3 $ out of the last two terms.

$$ 48x^{2}+32x+9x+6 = 16x\left(3x+2\right) + 3\left(3x+2\right) = \left(16x+3\right) \left(3x+2\right) $$

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