Factor polynomial $$ 2x^{2}+9x-5 $$

$$ 2x^{2}+9x-5 = ( 2x-1 )( x+5 ) $$

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:

$$ a = 2 , b = 9 ~ \text{ and } ~ c = -5 $$

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

$$ a \cdot c = -10 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -10 $ and add to $ b = 9 $.

Step 4: All pairs of numbers with a product of $ -10 $ are:

PRODUCT = -10
-1     10	1     -10
-2     5	2     -5

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 9 }$

PRODUCT = -10 and SUM = 9
-1     10	1     -10
-2     5	2     -5

Step 6: Replace middle term $ 9 x $ with $ 10x-x $:

$$ 2x^{2}+9x-5 = 2x^{2}+10x-x-5 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 2x^{2}+10x-x-5 = 2x\left(x+5\right) -1\left(x+5\right) = \left(2x-1\right) \left(x+5\right) $$

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