NCERT Solutions Class 9 Maths Exercise 4.5 — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, Exercise Set 4.5. All six parts solved.

A rational expression is one algebraic expression divided by another, and simplifying one works exactly as with numerical fractions: factorise the top, factorise the bottom, and cancel whatever they share.

The whole difficulty is the factorising, and the identities of this chapter do most of it:

$$a^2 – b^2 = (a+b)(a-b) \qquad a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$ $$a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Throughout, the denominators are assumed non-zero — cancelling a factor is only valid when that factor is not $0$.

Key insight. Cancel factors, never terms. $\frac{x^2 + 4}{x^2}$ is not $\frac{4}{1}$, because $x^2$ is being added in the numerator, not multiplied. Only after both top and bottom are written as products can anything be struck out — which is why factorising is the whole job and cancelling is the easy last step.

Question 1

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

Solution.

(i) $\dfrac{3p^2 – 3pq – 18q^2}{p^2 + 3pq – 10q^2}$

Factorise both. The numerator has a common factor of $3$:

$$3p^2 – 3pq – 18q^2 = 3(p^2 – pq – 6q^2) = 3(p – 3q)(p + 2q)$$

(two terms multiplying to $-6q^2$ and adding to $-q$: namely $-3q$ and $+2q$).

For the denominator, two terms multiplying to $-10q^2$ and adding to $3q$: namely $5q$ and $-2q$.

$$p^2 + 3pq – 10q^2 = (p + 5q)(p – 2q)$$

So the expression is

$$\frac{3(p – 3q)(p + 2q)}{(p + 5q)(p – 2q)}$$

Now compare the factors: the numerator has $(p – 3q)$ and $(p + 2q)$; the denominator has $(p + 5q)$ and $(p – 2q)$. None of them match — note in particular that $(p + 2q)$ and $(p – 2q)$ are different factors, and so are $(p – 3q)$ and $(p + 5q)$.

So nothing cancels, and the expression is already in its lowest terms. The factored form is the simplified answer; there is no further reduction to find.

(ii) $\dfrac{n^3 – 3n^2m + 3nm^2 – m^3}{5m^2 – 10mn + 5n^2}$

The numerator is the expansion of a cube. Comparing with $(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$ with $a = n$, $b = m$:

$$n^3 – 3n^2m + 3nm^2 – m^3 = (n – m)^3$$

The denominator has a common factor of $5$, and what remains is a square:

$$5m^2 – 10mn + 5n^2 = 5(m^2 – 2mn + n^2) = 5(m – n)^2$$

Now $(m – n)^2 = (n – m)^2$, since squaring removes the sign — so both are powers of the same quantity:

$$\frac{(n-m)^3}{5(n-m)^2} = \frac{n – m}{5}$$

(iii) $\dfrac{w^3 – v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 – 2wv – 2vx + 2wx}$

The numerator matches $a^3 + b^3 + c^3 – 3abc$ with $a = w$, $b = -v$, $c = x$ — check the last term: $-3abc = -3(w)(-v)(x) = +3wvx$ ✓. So

$$\text{numerator} = (w – v + x)\left( w^2 + v^2 + x^2 + wv + vx – wx \right)$$

using $a^2 + b^2 + c^2 – ab – bc – ca$ with the same substitutions.

The denominator is a three-term square. With $w$, $-v$, $x$: $2(w)(-v) = -2wv$ ✓, $2(-v)(x) = -2vx$ ✓, $2(x)(w) = 2wx$ ✓

$$\text{denominator} = (w – v + x)^2$$

Cancel one factor of $(w – v + x)$:

$$= \frac{w^2 + v^2 + x^2 + wv + vx – wx}{w – v + x}$$

(iv) $\dfrac{4y^2 – 20yz + 25z^2}{25z^2 – 4y^2}$

Numerator: a perfect square with $2y$ and $5z$, since $2(2y)(5z) = 20yz$:

$$4y^2 – 20yz + 25z^2 = (2y – 5z)^2 = (5z – 2y)^2$$

Denominator: a difference of two squares:

$$25z^2 – 4y^2 = (5z – 2y)(5z + 2y)$$

Cancel one $(5z – 2y)$:

$$= \frac{5z – 2y}{5z + 2y}$$

(v) $\dfrac{(x^2 + x – 6)(x^2 – 7x + 12)}{(x^2 – 6x + 8)(x^2 – 9)}$

Factorise all four quadratics:

$$x^2 + x – 6 = (x + 3)(x – 2) \qquad x^2 – 7x + 12 = (x – 3)(x – 4)$$ $$x^2 – 6x + 8 = (x – 2)(x – 4) \qquad x^2 – 9 = (x – 3)(x + 3)$$

So the expression is

$$\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}$$

Every factor on top appears on the bottom, so everything cancels:

$$= 1$$

(vi) $\dfrac{p^4 – 16}{p^2 – 4p + 4}$

The numerator is a difference of squares, and the first factor is itself a difference of squares:

$$p^4 – 16 = (p^2)^2 – 4^2 = (p^2 – 4)(p^2 + 4) = (p – 2)(p + 2)(p^2 + 4)$$

($p^2 + 4$ is a sum of squares and does not factorise further with real numbers.)

The denominator is a perfect square:

$$p^2 – 4p + 4 = (p – 2)^2$$

Cancel one factor of $(p – 2)$:

$$= \frac{(p + 2)(p^2 + 4)}{p – 2}$$

Common mistakes

• Cancelling terms instead of factors. In part (vi), striking the $4$ in $p^4 – 16$ against the $4$ in $p^2 – 4p + 4$ is meaningless: they are terms inside sums, not factors of the whole. Nothing can be cancelled until both top and bottom are single products.
• Part (i), forcing a cancellation. $(p + 2q)$ and $(p – 2q)$ look similar and are different factors, as are $(p – 3q)$ and $(p + 5q)$. When nothing matches, the factored form is the answer; inventing a cancellation gives a different expression altogether.
• Part (ii), treating $(m-n)^2$ and $(n-m)^2$ as different. They are equal, because squaring removes the sign. But $(m-n)^3 = -(n-m)^3$, so the same move on an odd power does introduce a minus sign — watch which power you are dealing with.
• Part (iii), mis-signing the identity. With $b = -v$, the term $-3abc$ becomes $+3wvx$, which is what makes the numerator match. Substituting $b = v$ instead gives $w^3 + v^3 + x^3 – 3wvx$, a different expression.
• Part (iv), cancelling $25z^2$ against $25z^2$. Both appear, but as terms in sums, not as factors. Factorise first and the genuine common factor $(5z – 2y)$ appears.
• Part (v), stopping at a partial cancellation. All four factors cancel here; the answer is exactly $1$, not $\frac{x+3}{x+3}$ or similar.
• Part (vi), trying to factor $p^2 + 4$. A sum of two squares has no real factorisation. Only a difference does.
• Forgetting the non-zero condition. Cancelling $(x-2)$ is only valid when $x \neq 2$. The question states the assumption, and it is worth recording with the answer.

Practise next

• The End-of-Chapter Exercises, Question 4 — three more rational expressions to simplify, using the sum and difference of cubes.
• Exercise Set 4.4 — worth revisiting if the factorising here was the hard part, since every simplification in this exercise depends on it.