Factor polynomial $$ 2b^{2}-15b+27 $$

$$ 2b^{2}-15b+27 = ( 2b-9 )( b-3 ) $$

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:

$$ a = 2 , b = -15 ~ \text{ and } ~ c = 27 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 27} $.

$$ a \cdot c = 54 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 54 $ and add to $ b = -15 $.

Step 4: All pairs of numbers with a product of $ 54 $ are:

PRODUCT = 54
1     54	-1     -54
2     27	-2     -27
3     18	-3     -18
6     9	-6     -9

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -15 }$

PRODUCT = 54 and SUM = -15
1     54	-1     -54
2     27	-2     -27
3     18	-3     -18
6     9	-6     -9

Step 6: Replace middle term $ -15 x $ with $ -6x-9x $:

$$ 2x^{2}-15x+27 = 2x^{2}-6x-9x+27 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -9 $ out of the last two terms.

$$ 2x^{2}-6x-9x+27 = 2x\left(x-3\right) -9\left(x-3\right) = \left(2x-9\right) \left(x-3\right) $$

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