Solve (3a-1)^2-3(2-5a+3a^2)

$$( 3 a - 1 ) ^ { 2 } - 3 ( 2 - 5 a + 3 a ^ { 2 } )$$

Evaluate

$$9a-5$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-1\right)^{2}.

$$9a^{2}-6a+1-3\left(2-5a+3a^{2}\right)$$

Use the distributive property to multiply -3 by 2-5a+3a^{2}.

$$9a^{2}-6a+1-6+15a-9a^{2}$$

Subtract 6 from 1 to get -5.

$$9a^{2}-6a-5+15a-9a^{2}$$

Combine -6a and 15a to get 9a.

$$9a^{2}+9a-5-9a^{2}$$

Combine 9a^{2} and -9a^{2} to get 0.

$$9a-5$$

Expand

$$9a-5$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a-1\right)^{2}.

$$9a^{2}-6a+1-3\left(2-5a+3a^{2}\right)$$

Use the distributive property to multiply -3 by 2-5a+3a^{2}.

$$9a^{2}-6a+1-6+15a-9a^{2}$$

Subtract 6 from 1 to get -5.

$$9a^{2}-6a-5+15a-9a^{2}$$

Combine -6a and 15a to get 9a.

$$9a^{2}+9a-5-9a^{2}$$

Combine 9a^{2} and -9a^{2} to get 0.

$$9a-5$$