Step 1 :
After factoring out $ 2t $ we have:
$$ 6t^{3}+26t^{2}-20t = 2t ( 3t^{2}+13t-10 ) $$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 $ 15x-2x $:
$$ 3x^{2}+13x-10 = 3x^{2}+15x-2x-10 $$Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -2 $ out of the last two terms.
$$ 3x^{2}+15x-2x-10 = 3x\left(x+5\right) -2\left(x+5\right) = \left(3x-2\right) \left(x+5\right) $$