Factor polynomial $$ 3a^{2}-8a-3 $$

$$ 3a^{2}-8a-3 = ( a-3 )( 3a+1 ) $$

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:

$$ a = 3 , b = -8 ~ \text{ and } ~ c = -3 $$

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

$$ a \cdot c = -9 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -9 $ and add to $ b = -8 $.

Step 4: All pairs of numbers with a product of $ -9 $ are:

PRODUCT = -9
-1     9	1     -9
-3     3	3     -3

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -8 }$

PRODUCT = -9 and SUM = -8
-1     9	1     -9
-3     3	3     -3

Step 6: Replace middle term $ -8 x $ with $ x-9x $:

$$ 3x^{2}-8x-3 = 3x^{2}+x-9x-3 $$

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

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

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