Factor polynomial $$ -3x^{2}-16x-5 $$

$$ -3x^{2}-16x-5 = -( 3x+1 )( x+5 ) $$

Step 1 :

After factoring out $ -1 $ we have:

$$ -3x^{2}-16x-5 = - ~ ( 3x^{2}+16x+5 ) $$

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:

$$ a = 3 , b = 16 ~ \text{ and } ~ c = 5 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = 5} $.

$$ a \cdot c = 15 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 15 $ and add to $ b = 16 $.

Step 5: All pairs of numbers with a product of $ 15 $ are:

PRODUCT = 15
1     15	-1     -15
3     5	-3     -5

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 16 }$

PRODUCT = 15 and SUM = 16
1     15	-1     -15
3     5	-3     -5

Step 7: Replace middle term $ 16 x $ with $ 15x+x $:

$$ 3x^{2}+16x+5 = 3x^{2}+15x+x+5 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ 1 $ out of the last two terms.

$$ 3x^{2}+15x+x+5 = 3x\left(x+5\right) + 1\left(x+5\right) = \left(3x+1\right) \left(x+5\right) $$

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