NCERT Solutions · Class 9 Mathematics Exploring Algebraic Identities

Factor the top, factor the bottom, cancel what is common. A rational expression is simplified exactly the way a fraction like \(\displaystyle \frac{18}{24}\) is: write both parts as products, then cancel the factors that appear in both. You may only cancel factors (things being multiplied), never individual terms.(i) \(\displaystyle \dfrac{3p^{2}-3pq-18q^{2}}{p^{2}+3pq-10q^{2}}\). Numerator: take out \(\displaystyle 3\), then split the middle term of \(\displaystyle p^{2}-pq-6q^{2}\) using \(\displaystyle -3\) and \(\displaystyle +2\): \[3p^{2}-3pq-18q^{2}=3(p-3q)(p+2q). \] Denominator: two numbers with product \(\displaystyle -10\) and sum \(\displaystyle +3\) are \(\displaystyle +5\) and \(\displaystyle -2\): \[p^{2}+3pq-10q^{2}=(p+5q)(p-2q). \] So \[\frac{3p^{2}-3pq-18q^{2}}{p^{2}+3pq-10q^{2}}=\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}. \] Compare the four factors: \(\displaystyle p-3q\), \(\displaystyle p+2q\) on top and \(\displaystyle p+5q\), \(\displaystyle p-2q\) below. No factor is common, so nothing cancels and the expression is already in its simplest form. (Note \(\displaystyle p+2q\) and \(\displaystyle p-2q\) are different factors — a sign matters.)A word of caution about this part. Every other question in this set cancels neatly, so as printed this one looks like a misprint: changing the sign of a single middle term would make it cancel. If the numerator were \(\displaystyle 3p^{2}+3pq-18q^{2}=3(p+3q)(p-2q)\), the answer would be \(\displaystyle \frac{3(p+3q)}{p+5q}\); if instead the denominator were \(\displaystyle p^{2}-3pq-10q^{2}=(p-5q)(p+2q)\), the answer would be \(\displaystyle \frac{3(p-3q)}{p-5q}\). Since we cannot tell which was intended, the honest answer to the question as printed is the factored form above.(ii) \(\displaystyle \dfrac{n^{3}-3n^{2}m+3nm^{2}-m^{3}}{5m^{2}-10mn+5n^{2}}\). The numerator is exactly \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\) with \(\displaystyle a=n\), \(\displaystyle b=m\), so it is \(\displaystyle (n-m)^{3}\). The denominator is \(\displaystyle 5(m^{2}-2mn+n^{2})=5(m-n)^{2}\), and \(\displaystyle (m-n)^{2}=(n-m)^{2}\) because squaring removes the sign. \[\frac{(n-m)^{3}}{5(n-m)^{2}}=\frac{n-m}{5} \](iii) \(\displaystyle \dfrac{w^{3}-v^{3}+x^{3}+3wvx}{w^{2}+v^{2}+x^{2}-2wv-2vx+2wx}\). For the numerator use \(\displaystyle a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\) with \(\displaystyle a=w\), \(\displaystyle b=-v\), \(\displaystyle c=x\). Then \(\displaystyle a^{3}+b^{3}+c^{3}=w^{3}-v^{3}+x^{3}\) and \(\displaystyle -3abc=-3(w)(-v)(x)=+3wvx\) — precisely the numerator. So \[w^{3}-v^{3}+x^{3}+3wvx=(w-v+x)\left(w^{2}+v^{2}+x^{2}+wv+vx-wx\right). \] The denominator is the square \(\displaystyle (w-v+x)^{2}\): check \(\displaystyle 2(w)(-v)=-2wv\), \(\displaystyle 2(-v)(x)=-2vx\), \(\displaystyle 2(w)(x)=+2wx\). Cancelling one factor of \(\displaystyle w-v+x\), \[\frac{w^{2}+v^{2}+x^{2}+wv+vx-wx}{w-v+x} \](iv) \(\displaystyle \dfrac{4y^{2}-20yz+25z^{2}}{25z^{2}-4y^{2}}\). Numerator \(\displaystyle =(2y-5z)^{2}=(5z-2y)^{2}\). Denominator \(\displaystyle =(5z)^{2}-(2y)^{2}=(5z-2y)(5z+2y)\). \[\frac{(5z-2y)^{2}}{(5z-2y)(5z+2y)}=\frac{5z-2y}{5z+2y} \](v) \(\displaystyle \dfrac{(x^{2}+x-6)(x^{2}-7x+12)}{(x^{2}-6x+8)(x^{2}-9)}\). Factor all four quadratics: \[x^{2}+x-6=(x+3)(x-2),\quad x^{2}-7x+12=(x-3)(x-4), \] \[x^{2}-6x+8=(x-2)(x-4),\quad x^{2}-9=(x-3)(x+3). \] Top and bottom contain exactly the same four factors, so everything cancels: \[\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}=1 \](vi) \(\displaystyle \dfrac{p^{4}-16}{p^{2}-4p+4}\). Apply the difference of squares twice on top: \(\displaystyle p^{4}-16=(p^{2})^{2}-4^{2}=(p^{2}-4)(p^{2}+4)=(p-2)(p+2)(p^{2}+4)\). The bottom is \(\displaystyle (p-2)^{2}\). \[\frac{(p-2)(p+2)(p^{2}+4)}{(p-2)^{2}}=\frac{(p+2)(p^{2}+4)}{p-2} \] (\(\displaystyle p^{2}+4\) is a sum of squares and does not factor further here.)Answers: (i) \(\displaystyle \frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}\) — nothing cancels, so this is already simplest; (ii) \(\displaystyle \frac{n-m}{5}\); (iii) \(\displaystyle \frac{w^{2}+v^{2}+x^{2}+wv+vx-wx}{w-v+x}\); (iv) \(\displaystyle \frac{5z-2y}{5z+2y}\); (v) \(\displaystyle 1\); (vi) \(\displaystyle \frac{(p+2)(p^{2}+4)}{p-2}\).