Solve 5(p+q)^2-125(p-q)^2

$$5 ( p + q ) ^ { 2 } - 125 ( p - q ) ^ { 2 }$$

Evaluate

$$20\left(3p-2q\right)\left(3q-2p\right)$$

Solution Steps

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

$$5\left(p^{2}+2pq+q^{2}\right)-125\left(p-q\right)^{2}$$

Use the distributive property to multiply 5 by p^{2}+2pq+q^{2}.

$$5p^{2}+10pq+5q^{2}-125\left(p-q\right)^{2}$$

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

$$5p^{2}+10pq+5q^{2}-125\left(p^{2}-2pq+q^{2}\right)$$

Use the distributive property to multiply -125 by p^{2}-2pq+q^{2}.

$$5p^{2}+10pq+5q^{2}-125p^{2}+250pq-125q^{2}$$

Combine 5p^{2} and -125p^{2} to get -120p^{2}.

$$-120p^{2}+10pq+5q^{2}+250pq-125q^{2}$$

Combine 10pq and 250pq to get 260pq.

$$-120p^{2}+260pq+5q^{2}-125q^{2}$$

Combine 5q^{2} and -125q^{2} to get -120q^{2}.

$$-120p^{2}+260pq-120q^{2}$$

Expand

$$-120p^{2}+260pq-120q^{2}$$

Solution Steps

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

$$5\left(p^{2}+2pq+q^{2}\right)-125\left(p-q\right)^{2}$$

Use the distributive property to multiply 5 by p^{2}+2pq+q^{2}.

$$5p^{2}+10pq+5q^{2}-125\left(p-q\right)^{2}$$

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

$$5p^{2}+10pq+5q^{2}-125\left(p^{2}-2pq+q^{2}\right)$$

Use the distributive property to multiply -125 by p^{2}-2pq+q^{2}.

$$5p^{2}+10pq+5q^{2}-125p^{2}+250pq-125q^{2}$$

Combine 5p^{2} and -125p^{2} to get -120p^{2}.

$$-120p^{2}+10pq+5q^{2}+250pq-125q^{2}$$

Combine 10pq and 250pq to get 260pq.

$$-120p^{2}+260pq+5q^{2}-125q^{2}$$

Combine 5q^{2} and -125q^{2} to get -120q^{2}.

$$-120p^{2}+260pq-120q^{2}$$