Factor polynomial $$ 3y^{2}+14y-24 $$

$$ 3y^{2}+14y-24 = ( 3y-4 )( y+6 ) $$

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:

$$ a = 3 , b = 14 ~ \text{ and } ~ c = -24 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -24} $.

$$ a \cdot c = -72 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -72 $ and add to $ b = 14 $.

Step 4: All pairs of numbers with a product of $ -72 $ are:

PRODUCT = -72
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

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

PRODUCT = -72 and SUM = 14
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

Step 6: Replace middle term $ 14 x $ with $ 18x-4x $:

$$ 3x^{2}+14x-24 = 3x^{2}+18x-4x-24 $$

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}+18x-4x-24 = 3x\left(x+6\right) -4\left(x+6\right) = \left(3x-4\right) \left(x+6\right) $$

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