Factor polynomial $$ 21y^{2}-2y-3 $$

$$ 21y^{2}-2y-3 = ( 7y-3 )( 3y+1 ) $$

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:

$$ a = 21 , b = -2 ~ \text{ and } ~ c = -3 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 21 }$ by the constant term $\color{blue}{c = -3} $.

$$ a \cdot c = -63 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -63 $ and add to $ b = -2 $.

Step 4: All pairs of numbers with a product of $ -63 $ are:

PRODUCT = -63
-1     63	1     -63
-3     21	3     -21
-7     9	7     -9

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -2 }$

PRODUCT = -63 and SUM = -2
-1     63	1     -63
-3     21	3     -21
-7     9	7     -9

Step 6: Replace middle term $ -2 x $ with $ 7x-9x $:

$$ 21x^{2}-2x-3 = 21x^{2}+7x-9x-3 $$

Step 7: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 21x^{2}+7x-9x-3 = 7x\left(3x+1\right) -3\left(3x+1\right) = \left(7x-3\right) \left(3x+1\right) $$

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