NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.5#
To simplify a rational expression: factor the numerator and the denominator completely, then cancel the factors they share (which is allowed because the denominator is not zero).
Identities used in this exercise:
$$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
$$a^2 - b^2 = (a + b)(a - b)$$
1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: #
(i) $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$
Numerator. We take out 3, then look for two numbers with sum $-1$ and product $-6$: they are $-3$ and $2$. We factor $3p^2 - 3pq - 18q^2$:
$$= 3(p^2 - pq - 6q^2)$$$$= 3(p^2 - 3pq + 2pq - 6q^2)$$
$$= 3\bigl(p(p - 3q) + 2q(p - 3q)\bigr)$$
$$= 3(p - 3q)(p + 2q)$$
Denominator. We need two numbers with sum $3$ and product $-10$: they are $5$ and $-2$. We factor $p^2 + 3pq - 10q^2$:
$$= p^2 + 5pq - 2pq - 10q^2$$$$= p(p + 5q) - 2q(p + 5q)$$
$$= (p + 5q)(p - 2q)$$
So:
$$\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \dfrac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}$$The numerator and denominator have no common factor, so this factorised form is the simplest form.
(ii) $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$
The numerator matches $(a - b)^3$ with $a = n$, $b = m$:
$$n^3 - 3n^2m + 3nm^2 - m^3 = (n - m)^3$$In the denominator we take out 5 and use $(a - b)^2$. We factor $5m^2 - 10mn + 5n^2$:
$$= 5(m^2 - 2mn + n^2)$$$$= 5(m - n)^2$$
$$= 5(n - m)^2$$
(since $(m - n)^2 = (n - m)^2$). We simplify the expression:
$$= \dfrac{(n - m)^3}{5(n - m)^2}$$$$= \dfrac{n - m}{5}$$
(iii) $\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$
Numerator. We write it as $a^3 + b^3 + c^3 - 3abc$ with $a = w$, $b = -v$, $c = x$. Then $-3abc = -3(w)(-v)(x) = 3wvx$ ✓. Using the identity:
$$w^3 - v^3 + x^3 + 3wvx = (w - v + x)\bigl(w^2 + v^2 + x^2 - w(-v) - (-v)x - xw\bigr)$$$$w^3 - v^3 + x^3 + 3wvx = (w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)$$
Denominator. With the same $a = w$, $b = -v$, $c = x$, it matches $(a + b + c)^2$:
$$2ab = -2wv, \quad 2bc = -2vx, \quad 2ca = 2wx$$So:
$$w^2 + v^2 + x^2 - 2wv - 2vx + 2wx = (w - v + x)^2$$We simplify the expression:
$$= \dfrac{(w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x)^2}$$$$= \dfrac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}$$
(iv) $\dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}$
The numerator is $(a - b)^2$ with $a = 5z$, $b = 2y$:
$$4y^2 - 20yz + 25z^2 = (5z - 2y)^2$$The denominator is a difference of squares:
$$25z^2 - 4y^2 = (5z - 2y)(5z + 2y)$$We simplify the expression:
$$= \dfrac{(5z - 2y)^2}{(5z - 2y)(5z + 2y)}$$$$= \dfrac{5z - 2y}{5z + 2y}$$
(v) $\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$
We factor each quadratic:
$$x^2 + x - 6 = (x + 3)(x - 2)$$$$x^2 - 7x + 12 = (x - 3)(x - 4)$$
$$x^2 - 6x + 8 = (x - 2)(x - 4)$$
$$x^2 - 9 = (x - 3)(x + 3)$$
We simplify the expression:
$$= \dfrac{(x + 3)(x - 2)(x - 3)(x - 4)}{(x - 2)(x - 4)(x - 3)(x + 3)}$$$$= 1$$
Every factor cancels, so the expression equals 1 (wherever the denominator is not zero).
(vi) $\dfrac{p^4 - 16}{p^2 - 4p + 4}$
The numerator is a difference of squares, used twice. We factor $p^4 - 16$:
$$= (p^2)^2 - 4^2$$$$= (p^2 - 4)(p^2 + 4)$$
$$= (p - 2)(p + 2)(p^2 + 4)$$
The denominator is $(a - b)^2$:
$$p^2 - 4p + 4 = (p - 2)^2$$We simplify the expression:
$$= \dfrac{(p - 2)(p + 2)(p^2 + 4)}{(p - 2)^2}$$$$= \dfrac{(p + 2)(p^2 + 4)}{p - 2}$$