Factor polynomial $$ 8x^{2}+11x-10 $$

$$ 8x^{2}+11x-10 = ( 8x-5 )( x+2 ) $$

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:

$$ a = 8 , b = 11 ~ \text{ and } ~ c = -10 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -80 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -80 $ and add to $ b = 11 $.

Step 4: All pairs of numbers with a product of $ -80 $ are:

PRODUCT = -80
-1     80	1     -80
-2     40	2     -40
-4     20	4     -20
-5     16	5     -16
-8     10	8     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 11 }$

PRODUCT = -80 and SUM = 11
-1     80	1     -80
-2     40	2     -40
-4     20	4     -20
-5     16	5     -16
-8     10	8     -10

Step 6: Replace middle term $ 11 x $ with $ 16x-5x $:

$$ 8x^{2}+11x-10 = 8x^{2}+16x-5x-10 $$

Step 7: Apply factoring by grouping. Factor $ 8x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 8x^{2}+16x-5x-10 = 8x\left(x+2\right) -5\left(x+2\right) = \left(8x-5\right) \left(x+2\right) $$

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