Step 1 :
After factoring out $ 2 $ we have:
$$ 8x^{2}+38x-60 = 2 ( 4x^{2}+19x-30 ) $$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 $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -30} $.
$$ a \cdot c = -120 $$Step 4: Find out two numbers that multiply to $ a \cdot c = -120 $ and add to $ b = 19 $.
Step 5: All pairs of numbers with a product of $ -120 $ are:
PRODUCT = -120 | |
-1 120 | 1 -120 |
-2 60 | 2 -60 |
-3 40 | 3 -40 |
-4 30 | 4 -30 |
-5 24 | 5 -24 |
-6 20 | 6 -20 |
-8 15 | 8 -15 |
-10 12 | 10 -12 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 19 }$
PRODUCT = -120 and SUM = 19 | |
-1 120 | 1 -120 |
-2 60 | 2 -60 |
-3 40 | 3 -40 |
-4 30 | 4 -30 |
-5 24 | 5 -24 |
-6 20 | 6 -20 |
-8 15 | 8 -15 |
-10 12 | 10 -12 |
Step 7: Replace middle term $ 19 x $ with $ 24x-5x $:
$$ 4x^{2}+19x-30 = 4x^{2}+24x-5x-30 $$Step 8: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -5 $ out of the last two terms.
$$ 4x^{2}+24x-5x-30 = 4x\left(x+6\right) -5\left(x+6\right) = \left(4x-5\right) \left(x+6\right) $$