Factor polynomial $$ 2b^{2}-7b+5 $$

$$ 2b^{2}-7b+5 = ( 2b-5 )( b-1 ) $$

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:

$$ a = 2 , b = -7 ~ \text{ and } ~ c = 5 $$

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 = -7 $.

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 $\color{blue}{ b = -7 }$

PRODUCT = 10 and SUM = -7
1     10	-1     -10
2     5	-2     -5

Step 6: Replace middle term $ -7 x $ with $ -2x-5x $:

$$ 2x^{2}-7x+5 = 2x^{2}-2x-5x+5 $$

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

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

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