It seems that $ 7x^{2}-6x+16 $ cannot be factored out.
Step 1: 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 2: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 16} $.
$$ a \cdot c = 112 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 112 $ and add to $ b = -6 $.
Step 4: All pairs of numbers with a product of $ 112 $ are:
| PRODUCT = 112 | |
| 1 112 | -1 -112 |
| 2 56 | -2 -56 |
| 4 28 | -4 -28 |
| 7 16 | -7 -16 |
| 8 14 | -8 -14 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -6 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -6 }$, we conclude the polynomial cannot be factored.