Solve 3left(a^2+2aright)^2-22left(a^2+2aright)+40

$$3 { \left( { a }^{ 2 } +2a \right) }^{ 2 } -22 \left( { a }^{ 2 } +2a \right) +40$$

Evaluate

$$3\left(a\left(a+2\right)\right)^{2}-22a\left(a+2\right)+40$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+2a\right)^{2}.

$$3\left(\left(a^{2}\right)^{2}+4a^{2}a+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$3\left(a^{4}+4a^{2}a+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$3\left(a^{4}+4a^{3}+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

Use the distributive property to multiply 3 by a^{4}+4a^{3}+4a^{2}.

$$3a^{4}+12a^{3}+12a^{2}-22\left(a^{2}+2a\right)+40$$

Use the distributive property to multiply -22 by a^{2}+2a.

$$3a^{4}+12a^{3}+12a^{2}-22a^{2}-44a+40$$

Combine 12a^{2} and -22a^{2} to get -10a^{2}.

$$3a^{4}+12a^{3}-10a^{2}-44a+40$$

Expand

$$3a^{4}+12a^{3}-10a^{2}-44a+40$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+2a\right)^{2}.

$$3\left(\left(a^{2}\right)^{2}+4a^{2}a+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$3\left(a^{4}+4a^{2}a+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$3\left(a^{4}+4a^{3}+4a^{2}\right)-22\left(a^{2}+2a\right)+40$$

Use the distributive property to multiply 3 by a^{4}+4a^{3}+4a^{2}.

$$3a^{4}+12a^{3}+12a^{2}-22\left(a^{2}+2a\right)+40$$

Use the distributive property to multiply -22 by a^{2}+2a.

$$3a^{4}+12a^{3}+12a^{2}-22a^{2}-44a+40$$

Combine 12a^{2} and -22a^{2} to get -10a^{2}.

$$3a^{4}+12a^{3}-10a^{2}-44a+40$$