Factor polynomial $$ 16m^{2}+m+1 $$

It seems that $ 16m^{2}+m+1 $ 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 $\color{blue}{ a = 16 }$ by the constant term $\color{blue}{c = 1} $.

Step 3: Find out two numbers that multiply to $ a \cdot c = 16 $ and add to $ b = 1 $.

Step 4: All pairs of numbers with a product of $ 16 $ are:

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 1 }$

Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.