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

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

Step 1 :

After factoring out $ -1 $ we have:

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

Step 2 :

Step 2: 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 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 45} $.

$$ a \cdot c = 90 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 90 $ and add to $ b = 19 $.

Step 5: 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 6: 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 7: Replace middle term $ 19 x $ with $ 10x+9x $:

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

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ 9 $ out of the last two terms.

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

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