Solve (2z+1)^2-9(2z+1)-22

$$( 2 z + 1 ) ^ { 2 } - 9 ( 2 z + 1 ) - 22$$

Evaluate

$$2\left(z-5\right)\left(2z+3\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2z+1\right)^{2}.

$$4z^{2}+4z+1-9\left(2z+1\right)-22$$

Use the distributive property to multiply -9 by 2z+1.

$$4z^{2}+4z+1-18z-9-22$$

Combine 4z and -18z to get -14z.

$$4z^{2}-14z+1-9-22$$

Subtract 9 from 1 to get -8.

$$4z^{2}-14z-8-22$$

Subtract 22 from -8 to get -30.

$$4z^{2}-14z-30$$

Expand

$$4z^{2}-14z-30$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2z+1\right)^{2}.

$$4z^{2}+4z+1-9\left(2z+1\right)-22$$

Use the distributive property to multiply -9 by 2z+1.

$$4z^{2}+4z+1-18z-9-22$$

Combine 4z and -18z to get -14z.

$$4z^{2}-14z+1-9-22$$

Subtract 9 from 1 to get -8.

$$4z^{2}-14z-8-22$$

Subtract 22 from -8 to get -30.

$$4z^{2}-14z-30$$