Factor polynomial $$ 14x^{2}-x-3 $$

$$ 14x^{2}-x-3 = ( 2x-1 )( 7x+3 ) $$

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:

$$ a = 14 , b = -1 ~ \text{ and } ~ c = -3 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 14 }$ by the constant term $\color{blue}{c = -3} $.

$$ a \cdot c = -42 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -42 $ and add to $ b = -1 $.

Step 4: All pairs of numbers with a product of $ -42 $ are:

PRODUCT = -42
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -1 }$

PRODUCT = -42 and SUM = -1
-1     42	1     -42
-2     21	2     -21
-3     14	3     -14
-6     7	6     -7

Step 6: Replace middle term $ -1 x $ with $ 6x-7x $:

$$ 14x^{2}-x-3 = 14x^{2}+6x-7x-3 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 14x^{2}+6x-7x-3 = 2x\left(7x+3\right) -1\left(7x+3\right) = \left(2x-1\right) \left(7x+3\right) $$

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