Step 1 :
After factoring out $ 4 $ we have:
$$ 8x^{2}+60x+112 = 4 ( 2x^{2}+15x+28 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 28} $.
$$ a \cdot c = 56 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 56 $ and add to $ b = 15 $.
Step 5: All pairs of numbers with a product of $ 56 $ are:
PRODUCT = 56 | |
1 56 | -1 -56 |
2 28 | -2 -28 |
4 14 | -4 -14 |
7 8 | -7 -8 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 15 }$
PRODUCT = 56 and SUM = 15 | |
1 56 | -1 -56 |
2 28 | -2 -28 |
4 14 | -4 -14 |
7 8 | -7 -8 |
Step 7: Replace middle term $ 15 x $ with $ 8x+7x $:
$$ 2x^{2}+15x+28 = 2x^{2}+8x+7x+28 $$Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ 7 $ out of the last two terms.
$$ 2x^{2}+8x+7x+28 = 2x\left(x+4\right) + 7\left(x+4\right) = \left(2x+7\right) \left(x+4\right) $$