Step 1 :
After factoring out $ 2 $ we have:
$$ 6x^{2}-26x-20 = 2 ( 3x^{2}-13x-10 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -10} $.
$$ a \cdot c = -30 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -30 $ and add to $ b = -13 $.
Step 5: All pairs of numbers with a product of $ -30 $ are:
PRODUCT = -30 | |
-1 30 | 1 -30 |
-2 15 | 2 -15 |
-3 10 | 3 -10 |
-5 6 | 5 -6 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -13 }$
PRODUCT = -30 and SUM = -13 | |
-1 30 | 1 -30 |
-2 15 | 2 -15 |
-3 10 | 3 -10 |
-5 6 | 5 -6 |
Step 7: Replace middle term $ -13 x $ with $ 2x-15x $:
$$ 3x^{2}-13x-10 = 3x^{2}+2x-15x-10 $$Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 3x^{2}+2x-15x-10 = x\left(3x+2\right) -5\left(3x+2\right) = \left(x-5\right) \left(3x+2\right) $$