Question

$\text{Factorise:}{p}^2+q^2+2(pq+qr+rp).$

• A. $(p+q)(p+2q+r)$
• B. $(q+r)(p+q+2r)$
• C. $(r+p)(p+q+2r)$
• D. $(p+q)(p+q+2r)$

Answer 1

The correct option is D $(p+q)(p+q+2r)$

$\text{Given expression:}{p}^2+q^2+2(pq+qr+rp)$

$\text{The above expression can be}\text{rewritten as,}=(p^2+q^2+2pq)+2(qr+rp)$

$\text{Now, applying the identity}(a+b{)}^2=a^2+2ab+b^2,\text{we get}{p}^2+q^2+2pq=(p+q{)}^2$

So,
$(p^2+q^2+2pq)+2(qr+rp)$
$=(p+q{)}^2+2r(p+q)=(p+q\left)\right(p+q)+2r(p+q)$

$\text{Taking the common factor out,}\text{we get,}=(p+q)(p+q+2r)$