It seems that $ 8x^{2}-5x+40 $ 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 $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = 40} $.
$$ a \cdot c = 320 $$Step 3: Find out two numbers that multiply to $ a \cdot c = 320 $ and add to $ b = -5 $.
Step 4: All pairs of numbers with a product of $ 320 $ are:
PRODUCT = 320 | |
1 320 | -1 -320 |
2 160 | -2 -160 |
4 80 | -4 -80 |
5 64 | -5 -64 |
8 40 | -8 -40 |
10 32 | -10 -32 |
16 20 | -16 -20 |
Step 5: Find out which factor pair sums up to $\color{blue}{ b = -5 }$
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ -5 }$, we conclude the polynomial cannot be factored.