Factor polynomial $$ 8y^{2}+10y-25 $$

$$ 8y^{2}+10y-25 = ( 4y-5 )( 2y+5 ) $$

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:

$$ a = 8 , b = 10 ~ \text{ and } ~ c = -25 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = -25} $.

$$ a \cdot c = -200 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -200 $ and add to $ b = 10 $.

Step 4: All pairs of numbers with a product of $ -200 $ are:

PRODUCT = -200
-1     200	1     -200
-2     100	2     -100
-4     50	4     -50
-5     40	5     -40
-8     25	8     -25
-10     20	10     -20

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 10 }$

PRODUCT = -200 and SUM = 10
-1     200	1     -200
-2     100	2     -100
-4     50	4     -50
-5     40	5     -40
-8     25	8     -25
-10     20	10     -20

Step 6: Replace middle term $ 10 x $ with $ 20x-10x $:

$$ 8x^{2}+10x-25 = 8x^{2}+20x-10x-25 $$

Step 7: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 8x^{2}+20x-10x-25 = 4x\left(2x+5\right) -5\left(2x+5\right) = \left(4x-5\right) \left(2x+5\right) $$

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