Solve left(a^2-2aright)^2-23left(a^2-2aright)+120

$${ \left( { a }^{ 2 } -2a \right) }^{ 2 } -23 \left( { a }^{ 2 } -2a \right) +120$$

Evaluate

$$\left(a-5\right)\left(a-4\right)\left(a+2\right)\left(a+3\right)$$

Solution Steps

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

$$\left(a^{2}\right)^{2}-4a^{2}a+4a^{2}-23\left(a^{2}-2a\right)+120$$

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

$$a^{4}-4a^{2}a+4a^{2}-23\left(a^{2}-2a\right)+120$$

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

$$a^{4}-4a^{3}+4a^{2}-23\left(a^{2}-2a\right)+120$$

Use the distributive property to multiply -23 by a^{2}-2a.

$$a^{4}-4a^{3}+4a^{2}-23a^{2}+46a+120$$

Combine 4a^{2} and -23a^{2} to get -19a^{2}.

$$a^{4}-4a^{3}-19a^{2}+46a+120$$

Expand

$$a^{4}-4a^{3}-19a^{2}+46a+120$$

Solution Steps

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

$$\left(a^{2}\right)^{2}-4a^{2}a+4a^{2}-23\left(a^{2}-2a\right)+120$$

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

$$a^{4}-4a^{2}a+4a^{2}-23\left(a^{2}-2a\right)+120$$

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

$$a^{4}-4a^{3}+4a^{2}-23\left(a^{2}-2a\right)+120$$

Use the distributive property to multiply -23 by a^{2}-2a.

$$a^{4}-4a^{3}+4a^{2}-23a^{2}+46a+120$$

Combine 4a^{2} and -23a^{2} to get -19a^{2}.

$$a^{4}-4a^{3}-19a^{2}+46a+120$$