Factor polynomial $$ 8x^{2}+71x+56 $$

$$ 8x^{2}+71x+56 = ( 8x+7 )( x+8 ) $$

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:

$$ a = 8 , b = 71 ~ \text{ and } ~ c = 56 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 56} $.

$$ a \cdot c = 448 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 448 $ and add to $ b = 71 $.

Step 4: All pairs of numbers with a product of $ 448 $ are:

PRODUCT = 448
1     448	-1     -448
2     224	-2     -224
4     112	-4     -112
7     64	-7     -64
8     56	-8     -56
14     32	-14     -32
16     28	-16     -28

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 71 }$

PRODUCT = 448 and SUM = 71
1     448	-1     -448
2     224	-2     -224
4     112	-4     -112
7     64	-7     -64
8     56	-8     -56
14     32	-14     -32
16     28	-16     -28

Step 6: Replace middle term $ 71 x $ with $ 64x+7x $:

$$ 8x^{2}+71x+56 = 8x^{2}+64x+7x+56 $$

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

$$ 8x^{2}+64x+7x+56 = 8x\left(x+8\right) + 7\left(x+8\right) = \left(8x+7\right) \left(x+8\right) $$

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