Factor polynomial $$ 8x^{2}-26x+15 $$

$$ 8x^{2}-26x+15 = ( 2x-5 )( 4x-3 ) $$

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:

$$ a = 8 , b = -26 ~ \text{ and } ~ c = 15 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 15} $.

$$ a \cdot c = 120 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 120 $ and add to $ b = -26 $.

Step 4: All pairs of numbers with a product of $ 120 $ are:

PRODUCT = 120
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -26 }$

PRODUCT = 120 and SUM = -26
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 6: Replace middle term $ -26 x $ with $ -6x-20x $:

$$ 8x^{2}-26x+15 = 8x^{2}-6x-20x+15 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 8x^{2}-6x-20x+15 = 2x\left(4x-3\right) -5\left(4x-3\right) = \left(2x-5\right) \left(4x-3\right) $$

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