Solve a^2(4a-3)^2-a(4a-3)

$$a ^ { 2 } ( 4 a - 3 ) ^ { 2 } - a ( 4 a - 3 )$$

Evaluate

$$a\left(a-1\right)\left(4a-3\right)\left(4a+1\right)$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-3\right)^{2}.

$$a^{2}\left(16a^{2}-24a+9\right)-a\left(4a-3\right)$$

Use the distributive property to multiply a^{2} by 16a^{2}-24a+9.

$$16a^{4}-24a^{3}+9a^{2}-a\left(4a-3\right)$$

Use the distributive property to multiply a by 4a-3.

$$16a^{4}-24a^{3}+9a^{2}-\left(4a^{2}-3a\right)$$

To find the opposite of 4a^{2}-3a, find the opposite of each term.

$$16a^{4}-24a^{3}+9a^{2}-4a^{2}+3a$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$16a^{4}-24a^{3}+5a^{2}+3a$$

Expand

$$16a^{4}-24a^{3}+5a^{2}+3a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-3\right)^{2}.

$$a^{2}\left(16a^{2}-24a+9\right)-a\left(4a-3\right)$$

Use the distributive property to multiply a^{2} by 16a^{2}-24a+9.

$$16a^{4}-24a^{3}+9a^{2}-a\left(4a-3\right)$$

Use the distributive property to multiply a by 4a-3.

$$16a^{4}-24a^{3}+9a^{2}-\left(4a^{2}-3a\right)$$

To find the opposite of 4a^{2}-3a, find the opposite of each term.

$$16a^{4}-24a^{3}+9a^{2}-4a^{2}+3a$$

Combine 9a^{2} and -4a^{2} to get 5a^{2}.

$$16a^{4}-24a^{3}+5a^{2}+3a$$