Solve 4(1+k)^2-8(1+k)+3

$$4 ( 1 + k ) ^ { 2 } - 8 ( 1 + k ) + 3$$

Evaluate

$$4k^{2}-1$$

Solution Steps

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

$$4\left(1+2k+k^{2}\right)-8\left(1+k\right)+3$$

Use the distributive property to multiply 4 by 1+2k+k^{2}.

$$4+8k+4k^{2}-8\left(1+k\right)+3$$

Use the distributive property to multiply -8 by 1+k.

$$4+8k+4k^{2}-8-8k+3$$

Subtract 8 from 4 to get -4.

$$-4+8k+4k^{2}-8k+3$$

Combine 8k and -8k to get 0.

$$-4+4k^{2}+3$$

Add -4 and 3 to get -1.

$$-1+4k^{2}$$

Expand

$$4k^{2}-1$$

Solution Steps

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

$$4\left(1+2k+k^{2}\right)-8\left(1+k\right)+3$$

Use the distributive property to multiply 4 by 1+2k+k^{2}.

$$4+8k+4k^{2}-8\left(1+k\right)+3$$

Use the distributive property to multiply -8 by 1+k.

$$4+8k+4k^{2}-8-8k+3$$

Subtract 8 from 4 to get -4.

$$-4+8k+4k^{2}-8k+3$$

Combine 8k and -8k to get 0.

$$-4+4k^{2}+3$$

Add -4 and 3 to get -1.

$$-1+4k^{2}$$