Factor polynomial $$ 5y^{2}-14y-3 $$

$$ 5y^{2}-14y-3 = ( y-3 )( 5y+1 ) $$

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:

$$ a = 5 , b = -14 ~ \text{ and } ~ c = -3 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = -3} $.

$$ a \cdot c = -15 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -15 $ and add to $ b = -14 $.

Step 4: All pairs of numbers with a product of $ -15 $ are:

PRODUCT = -15
-1     15	1     -15
-3     5	3     -5

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -14 }$

PRODUCT = -15 and SUM = -14
-1     15	1     -15
-3     5	3     -5

Step 6: Replace middle term $ -14 x $ with $ x-15x $:

$$ 5x^{2}-14x-3 = 5x^{2}+x-15x-3 $$

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

$$ 5x^{2}+x-15x-3 = x\left(5x+1\right) -3\left(5x+1\right) = \left(x-3\right) \left(5x+1\right) $$

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