Solve (4p+q)(4p-q)-(3p-q)^2

$$( 4 p + q ) ( 4 p - q ) - ( 3 p - q ) ^ { 2 }$$

Evaluate

$$7p^{2}+6pq-2q^{2}$$

Solution Steps

Consider \left(4p+q\right)\left(4p-q\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(4p\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16p^{2}-q^{2}-\left(3p-q\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3p-q\right)^{2}.

$$16p^{2}-q^{2}-\left(9p^{2}-6pq+q^{2}\right)$$

To find the opposite of 9p^{2}-6pq+q^{2}, find the opposite of each term.

$$16p^{2}-q^{2}-9p^{2}+6pq-q^{2}$$

Combine 16p^{2} and -9p^{2} to get 7p^{2}.

$$7p^{2}-q^{2}+6pq-q^{2}$$

Combine -q^{2} and -q^{2} to get -2q^{2}.

$$7p^{2}-2q^{2}+6pq$$

Expand

$$7p^{2}+6pq-2q^{2}$$

Solution Steps

Consider \left(4p+q\right)\left(4p-q\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(4p\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16p^{2}-q^{2}-\left(3p-q\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3p-q\right)^{2}.

$$16p^{2}-q^{2}-\left(9p^{2}-6pq+q^{2}\right)$$

To find the opposite of 9p^{2}-6pq+q^{2}, find the opposite of each term.

$$16p^{2}-q^{2}-9p^{2}+6pq-q^{2}$$

Combine 16p^{2} and -9p^{2} to get 7p^{2}.

$$7p^{2}-q^{2}+6pq-q^{2}$$

Combine -q^{2} and -q^{2} to get -2q^{2}.

$$7p^{2}-2q^{2}+6pq$$