Step 1 :
After factoring out $ 3 $ we have:
$$ 9x^{2}-18x+12 = 3 ( 3x^{2}-6x+4 ) $$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 = 4} $.
$$ a \cdot c = 12 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 12 $ and add to $ b = -6 $.
Step 5: All pairs of numbers with a product of $ 12 $ are:
| PRODUCT = 12 | |
| 1 12 | -1 -12 |
| 2 6 | -2 -6 |
| 3 4 | -3 -4 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -6 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -6 }$, we conclude the polynomial cannot be factored.