Solve -8p-4*(p+q)^2-3(p+q)-28

$$- 8 p - 4 \cdot ( p + q ) ^ { 2 } - 3 ( p + q ) - 28$$

Evaluate

$$-4\left(p+q\right)^{2}-3q-11p-28$$

Solution Steps

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

$$-8p-4\left(p^{2}+2pq+q^{2}\right)-3\left(p+q\right)-28$$

Use the distributive property to multiply -4 by p^{2}+2pq+q^{2}.

$$-8p-4p^{2}-8pq-4q^{2}-3\left(p+q\right)-28$$

Use the distributive property to multiply -3 by p+q.

$$-8p-4p^{2}-8pq-4q^{2}-3p-3q-28$$

Combine -8p and -3p to get -11p.

$$-11p-4p^{2}-8pq-4q^{2}-3q-28$$

Expand

$$-4p^{2}-8pq-11p-4q^{2}-3q-28$$

Solution Steps

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

$$-8p-4\left(p^{2}+2pq+q^{2}\right)-3\left(p+q\right)-28$$

Use the distributive property to multiply -4 by p^{2}+2pq+q^{2}.

$$-8p-4p^{2}-8pq-4q^{2}-3\left(p+q\right)-28$$

Use the distributive property to multiply -3 by p+q.

$$-8p-4p^{2}-8pq-4q^{2}-3p-3q-28$$

Combine -8p and -3p to get -11p.

$$-11p-4p^{2}-8pq-4q^{2}-3q-28$$