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

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

Evaluate

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

Solution Steps

Use the distributive property to multiply 2 by 3a+1.

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

Use the distributive property to multiply 6a+2 by 3a-2 and combine like terms.

$$18a^{2}-6a-4-\left(3a-1\right)^{2}$$

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

$$18a^{2}-6a-4-\left(9a^{2}-6a+1\right)$$

To find the opposite of 9a^{2}-6a+1, find the opposite of each term.

$$18a^{2}-6a-4-9a^{2}+6a-1$$

Combine 18a^{2} and -9a^{2} to get 9a^{2}.

$$9a^{2}-6a-4+6a-1$$

Combine -6a and 6a to get 0.

$$9a^{2}-4-1$$

Subtract 1 from -4 to get -5.

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

Expand

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

Solution Steps

Use the distributive property to multiply 2 by 3a+1.

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

Use the distributive property to multiply 6a+2 by 3a-2 and combine like terms.

$$18a^{2}-6a-4-\left(3a-1\right)^{2}$$

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

$$18a^{2}-6a-4-\left(9a^{2}-6a+1\right)$$

To find the opposite of 9a^{2}-6a+1, find the opposite of each term.

$$18a^{2}-6a-4-9a^{2}+6a-1$$

Combine 18a^{2} and -9a^{2} to get 9a^{2}.

$$9a^{2}-6a-4+6a-1$$

Combine -6a and 6a to get 0.

$$9a^{2}-4-1$$

Subtract 1 from -4 to get -5.

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