Factor polynomial $$ 2x^{2}-19x+45 $$

$$ 2x^{2}-19x+45 = ( x-5 )( 2x-9 ) $$

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:

$$ a = 2 , b = -19 ~ \text{ and } ~ c = 45 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 45} $.

$$ a \cdot c = 90 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 90 $ and add to $ b = -19 $.

Step 4: All pairs of numbers with a product of $ 90 $ are:

PRODUCT = 90
1     90	-1     -90
2     45	-2     -45
3     30	-3     -30
5     18	-5     -18
6     15	-6     -15
9     10	-9     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -19 }$

PRODUCT = 90 and SUM = -19
1     90	-1     -90
2     45	-2     -45
3     30	-3     -30
5     18	-5     -18
6     15	-6     -15
9     10	-9     -10

Step 6: Replace middle term $ -19 x $ with $ -9x-10x $:

$$ 2x^{2}-19x+45 = 2x^{2}-9x-10x+45 $$

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

$$ 2x^{2}-9x-10x+45 = x\left(2x-9\right) -5\left(2x-9\right) = \left(x-5\right) \left(2x-9\right) $$

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