Factor polynomial $$ 3x^{2}+8x-3 $$

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

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:

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

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 $ 9x-x $:

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

Step 7: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -1 $ out of the last two terms.

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

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