Exercise 1
Factor completely:
(i)
(ii)
(iii)
(iv)
(v) (vi) (Hint: was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
Recognising a perfect square trinomial. An expression is \(\displaystyle (a+b)^{2}\) in disguise when the first and last terms are perfect squares, say \(\displaystyle a^{2}\) and \(\displaystyle b^{2}\), and the middle term is exactly \(\displaystyle 2ab\). So for each part: take the square roots of the outer two terms, then check that twice their product gives the middle term. Checking matters — without it you can factor something that is not actually a square.(i) \(\displaystyle 9x^{2}+24xy+16y^{2}\). Here \(\displaystyle 9x^{2}=(3x)^{2}\) and \(\displaystyle 16y^{2}=(4y)^{2}\). Check the middle: \(\displaystyle 2(3x)(4y)=24xy\). It matches.
\[9x^{2}+24xy+16y^{2}=(3x+4y)^{2} \](ii) \(\displaystyle 4s^{2}+20st+25t^{2}\). Roots \(\displaystyle 2s\) and \(\displaystyle 5t\); check \(\displaystyle 2(2s)(5t)=20st\). It matches.
\[4s^{2}+20st+25t^{2}=(2s+5t)^{2} \](iii) \(\displaystyle 49x^{2}+28xy+4y^{2}\). Roots \(\displaystyle 7x\) and \(\displaystyle 2y\); check \(\displaystyle 2(7x)(2y)=28xy\). It matches.
\[49x^{2}+28xy+4y^{2}=(7x+2y)^{2} \](iv) \(\displaystyle 64p^{2}+\frac{32}{3}pq+\frac{4}{9}q^{2}\). Roots \(\displaystyle 8p\) and \(\displaystyle \frac{2}{3}q\); check \(\displaystyle 2(8p)\left(\frac{2}{3}q\right)=\frac{32}{3}pq\). It matches.
\[64p^{2}+\tfrac{32}{3}pq+\tfrac{4}{9}q^{2}=\left(8p+\tfrac{2}{3}q\right)^{2}=\tfrac{1}{9}(24p+2q)^{2} \](v) \(\displaystyle 3a^{2}+4ab+\frac{4}{3}b^{2}\) (starred). Now \(\displaystyle 3a^{2}\) is not the square of a rational expression, so the identity cannot be applied as it stands. This is where the hint comes in: in Example $\displaystyle 7$ a common factor of \(\displaystyle 2\) was pulled out first. Do the same here and pull out \(\displaystyle 3\):
\[3a^{2}+4ab+\tfrac{4}{3}b^{2}=3\left(a^{2}+\tfrac{4}{3}ab+\tfrac{4}{9}b^{2}\right). \]
Inside the bracket the outer terms are \(\displaystyle a^{2}\) and \(\displaystyle \left(\frac{2}{3}b\right)^{2}\), and \(\displaystyle 2(a)\left(\frac{2}{3}b\right)=\frac{4}{3}ab\) — a genuine perfect square.
\[3a^{2}+4ab+\tfrac{4}{3}b^{2}=3\left(a+\tfrac{2}{3}b\right)^{2}=\tfrac{1}{3}(3a+2b)^{2} \]
(Check by expanding: \(\displaystyle \frac{1}{3}(9a^{2}+12ab+4b^{2})=3a^{2}+4ab+\frac{4}{3}b^{2}\).)(vi) \(\displaystyle \frac{9}{5}s^{2}+6sv+5v^{2}\) (starred). Again neither outer term is a perfect square of something rational. Pulling out \(\displaystyle \frac{1}{5}\) clears both denominators at once:
\[\tfrac{9}{5}s^{2}+6sv+5v^{2}=\tfrac{1}{5}\left(9s^{2}+30sv+25v^{2}\right). \]
Inside, \(\displaystyle 9s^{2}=(3s)^{2}\), \(\displaystyle 25v^{2}=(5v)^{2}\) and \(\displaystyle 2(3s)(5v)=30sv\). It matches.
\[\tfrac{9}{5}s^{2}+6sv+5v^{2}=\tfrac{1}{5}(3s+5v)^{2} \]
(Check: \(\displaystyle \frac{1}{5}(9s^{2}+30sv+25v^{2})=\frac{9}{5}s^{2}+6sv+5v^{2}\).)So the answer to the hint's question is yes: whenever the coefficients are awkward, taking out a suitable common factor — a whole number in (v), a fraction in (vi) — can turn the leftover expression into a perfect square.Answers: (i) \(\displaystyle (3x+4y)^{2}\); (ii) \(\displaystyle (2s+5t)^{2}\); (iii) \(\displaystyle (7x+2y)^{2}\); (iv) \(\displaystyle \left(8p+\frac{2}{3}q\right)^{2}\); (v) \(\displaystyle 3\left(a+\frac{2}{3}b\right)^{2}=\frac{1}{3}(3a+2b)^{2}\); (vi) \(\displaystyle \frac{1}{5}(3s+5v)^{2}\).