Factor polynomial $$ 14x^{2}-23x+8 $$

$$ 14x^{2}-23x+8 = ( 7x-8 )( 2x-1 ) $$

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:

$$ a = 14 , b = -23 ~ \text{ and } ~ c = 8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 14 }$ by the constant term $\color{blue}{c = 8} $.

$$ a \cdot c = 112 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 112 $ and add to $ b = -23 $.

Step 4: All pairs of numbers with a product of $ 112 $ are:

PRODUCT = 112
1     112	-1     -112
2     56	-2     -56
4     28	-4     -28
7     16	-7     -16
8     14	-8     -14

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -23 }$

PRODUCT = 112 and SUM = -23
1     112	-1     -112
2     56	-2     -56
4     28	-4     -28
7     16	-7     -16
8     14	-8     -14

Step 6: Replace middle term $ -23 x $ with $ -7x-16x $:

$$ 14x^{2}-23x+8 = 14x^{2}-7x-16x+8 $$

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

$$ 14x^{2}-7x-16x+8 = 7x\left(2x-1\right) -8\left(2x-1\right) = \left(7x-8\right) \left(2x-1\right) $$

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