Step 1 :
After factoring out $ 3 $ we have:
$$ 6x^{2}-33x+45 = 3 ( 2x^{2}-11x+15 ) $$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 $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 15} $.
$$ a \cdot c = 30 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 30 $ and add to $ b = -11 $.
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 $\color{blue}{ b = -11 }$
PRODUCT = 30 and SUM = -11 | |
1 30 | -1 -30 |
2 15 | -2 -15 |
3 10 | -3 -10 |
5 6 | -5 -6 |
Step 7: Replace middle term $ -11 x $ with $ -5x-6x $:
$$ 2x^{2}-11x+15 = 2x^{2}-5x-6x+15 $$Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -3 $ out of the last two terms.
$$ 2x^{2}-5x-6x+15 = x\left(2x-5\right) -3\left(2x-5\right) = \left(x-3\right) \left(2x-5\right) $$