Solve c(-b)=(a+b)^2+3(a+b)+i

$$c ( - b ) = ( a + b ) ^ { 2 } + 3 ( a + b ) + i$$

Evaluate

$$\left(a+b\right)^{2}+3a+3b+i$$

Solution Steps

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

$$a^{2}+2ab+b^{2}+3\left(a+b\right)+i$$

Use the distributive property to multiply 3 by a+b.

$$a^{2}+2ab+b^{2}+3a+3b+i$$

Expand

$$a^{2}+2ab+3a+b^{2}+3b+i$$

Solution Steps

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

$$a^{2}+2ab+b^{2}+3\left(a+b\right)+i$$

Use the distributive property to multiply 3 by a+b.

$$a^{2}+2ab+b^{2}+3a+3b+i$$