Step 1 :
After factoring out $ -1 $ we have:
$$ -4x^{2}-41x-10 = - ~ ( 4x^{2}+41x+10 ) $$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 $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 10} $.
$$ a \cdot c = 40 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 40 $ and add to $ b = 41 $.
Step 5: All pairs of numbers with a product of $ 40 $ are:
PRODUCT = 40 | |
1 40 | -1 -40 |
2 20 | -2 -20 |
4 10 | -4 -10 |
5 8 | -5 -8 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 41 }$
PRODUCT = 40 and SUM = 41 | |
1 40 | -1 -40 |
2 20 | -2 -20 |
4 10 | -4 -10 |
5 8 | -5 -8 |
Step 7: Replace middle term $ 41 x $ with $ 40x+x $:
$$ 4x^{2}+41x+10 = 4x^{2}+40x+x+10 $$Step 8: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ 1 $ out of the last two terms.
$$ 4x^{2}+40x+x+10 = 4x\left(x+10\right) + 1\left(x+10\right) = \left(4x+1\right) \left(x+10\right) $$