Factor polynomial $$ 20x^{2}+55x+30 $$
Answer
Explanation
Step 1 :
After factoring out $ 5 $ we have:
$$ 20x^{2}+55x+30 = 5 ( 4x^{2}+11x+6 ) $$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 = 6} $.
$$ a \cdot c = 24 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 24 $ and add to $ b = 11 $.
Step 5: All pairs of numbers with a product of $ 24 $ are:
| PRODUCT = 24 | |
| 1 24 | -1 -24 |
| 2 12 | -2 -12 |
| 3 8 | -3 -8 |
| 4 6 | -4 -6 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = 11 }$
| PRODUCT = 24 and SUM = 11 | |
| 1 24 | -1 -24 |
| 2 12 | -2 -12 |
| 3 8 | -3 -8 |
| 4 6 | -4 -6 |
Step 7: Replace middle term $ 11 x $ with $ 8x+3x $:
$$ 4x^{2}+11x+6 = 4x^{2}+8x+3x+6 $$Step 8: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ 3 $ out of the last two terms.
$$ 4x^{2}+8x+3x+6 = 4x\left(x+2\right) + 3\left(x+2\right) = \left(4x+3\right) \left(x+2\right) $$