 Ishaan Mcneil

2022-02-09

How do you multiply $\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)$? Amubbemibren3k

The first terms in each parentheses are multiplied by each other. Then the ‘outer’ and ‘inner’ pairs are multiplied and added to each other. Finally, the last terms in each parentheses are multiplied together. The final result may be simplified.
$\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)$
Step 1. ${a}^{2}\cdot -{a}^{2}=-{a}^{4}$
Step 2. $\left({a}^{2}\cdot 2a\right)+\left(3a\cdot -{a}^{2}\right)=2{a}^{3}-3{a}^{3};-{a}^{3}$
Step 3. $3a\cdot 2a=6{a}^{2}$
Step 4. $-{a}^{4}-{a}^{3}+6{a}^{2}$ Multiply using F.O.I.L
F: $\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)\to -{a}^{4}$
O: $\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)\to 2{a}^{3}$
I: $\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)\to -3{a}^{3}$
L: $\left({a}^{2}+3a\right)\left(-{a}^{2}+2a\right)\to 6{a}^{2}$
Now by combining like terms we end up with:
$-{a}^{4}-{a}^{3}+6{a}^{2}$

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