Question

$15lm(2p^2-2q^2)÷3l(p+q)$

• A. $5m(p-q)$
• B. $m(p-q)$
• C. $10m(p-q)$
• D. None

Answer 1

The correct option is C $10m(p-q)$

• Let us first factorize $(2p^2-2q^2)$ as follows:

$2p^2-2q^2=2(p^2-q^2)=2(p+q)(p-q)(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

Since we have to divide $\frac{15lm(2p^2-2q^2)}{3l(p+q)}$, therefore, we will divide $\frac{30lm(p+q)(p-q)}{3l(p+q)}$ as shown below:

$\frac{30lm(p+q)(p-q)}{3l(p+q)}=\frac{30}{3}\times \frac{lm(p+q)(p-q)}{l(p+q)}=\frac{2\times 3\times 5}{3}\times m(p-q)=10m(p-q)$

Hence, $\frac{15lm(2p^2-2q^2)}{3l(p+q)}=10m(p-q)$.