Factor polynomial $$ 14a^{2}+7a-7 $$

$$ 14a^{2}+7a-7 = 7( 2a-1 )( a+1 ) $$

Step 1 :

After factoring out $ 7 $ we have:

$$ 14a^{2}+7a-7 = 7 ( 2a^{2}+a-1 ) $$

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:

$$ a = 2 , b = 1 ~ \text{ and } ~ c = -1 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -1} $.

$$ a \cdot c = -2 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -2 $ and add to $ b = 1 $.

Step 5: All pairs of numbers with a product of $ -2 $ are:

PRODUCT = -2
-1     2	1     -2

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 1 }$

PRODUCT = -2 and SUM = 1
-1     2	1     -2

Step 7: Replace middle term $ 1 x $ with $ 2x-x $:

$$ 2x^{2}+x-1 = 2x^{2}+2x-x-1 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 2x^{2}+2x-x-1 = 2x\left(x+1\right) -1\left(x+1\right) = \left(2x-1\right) \left(x+1\right) $$

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