Factor polynomial $$ 5n^{2}-13n+8 $$

$$ 5n^{2}-13n+8 = ( 5n-8 )( n-1 ) $$

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:

$$ a = 5 , b = -13 ~ \text{ and } ~ c = 8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 8} $.

$$ a \cdot c = 40 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 40 $ and add to $ b = -13 $.

Step 4: All pairs of numbers with a product of $ 40 $ are:

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

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = 40 and SUM = -13
1     40	-1     -40
2     20	-2     -20
4     10	-4     -10
5     8	-5     -8

Step 6: Replace middle term $ -13 x $ with $ -5x-8x $:

$$ 5x^{2}-13x+8 = 5x^{2}-5x-8x+8 $$

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

$$ 5x^{2}-5x-8x+8 = 5x\left(x-1\right) -8\left(x-1\right) = \left(5x-8\right) \left(x-1\right) $$

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