Class 9 Maths Chapter 4 Exercise 4.5 Solutions – Ganita Manjari

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5 solutions: new algebraic identities and simplifying expressions,

Welcome to Sid Classes, your trusted destination for accurate, reliable, and student-friendly academic resources! In this detailed study guide, we provide complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5. Designed to reflect how top-scoring students write their answers in school examinations, this walkthrough makes simplifying rational expressions, applying polynomial factorizations, and canceling common algebraic factors entirely clear.

🔑 Key Concepts Used in this Exercise:

  • Splitting the Middle Term & Identities: Factorizing numerators and denominators completely using algebraic identities and splitting techniques before simplifying.
  • Rational Expression Reduction: Canceling common non-zero polynomial factors between the numerator and denominator to arrive at the simplest form.

NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5 Solutions

Question 1

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

Solution:

(i) \(\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}\)

Factorizing the numerator by taking out \(3\) and splitting the middle term:
\(3p^2 - 3pq - 18q^2 = 3(p^2 - pq - 6q^2) = 3(p - 3q)(p + 2q)\)

Factorizing the denominator by splitting the middle term (\(5pq\) and \(-2pq\)):
\(p^2 + 3pq - 10q^2 = (p + 5q)(p - 2q)\)

Substituting the factored forms:
\(= \frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}\)
\(= \mathbf{\frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}}\)

(ii) \(\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}\)

Numerator is the cubic identity expansion of \((n - m)^3\):
\(n^3 - 3n^2m + 3nm^2 - m^3 = (n - m)^3\)

Denominator can be factored by taking out \(5\):
\(5m^2 - 10mn + 5n^2 = 5(m^2 - 2mn + n^2) = 5(m - n)^2 = 5(n - m)^2\)

Simplifying the fraction by canceling \((n - m)^2\):
\(= \frac{(n - m)^3}{5(n - m)^2}\)
\(= \mathbf{\frac{n - m}{5}}\)

(iii) \(\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}\)

Using the identity \(a^3 + b^3 + c^3 - 3abc\) with \(a = w, b = -v, c = x\):
Numerator \(= w^3 + (-v)^3 + x^3 - 3w(-v)x = (w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)\)

Denominator is the trinomial square expansion of \((w - v + x)^2\):
Denominator \(= (w - v + x)^2\)

Dividing numerator by denominator:
\(= \frac{(w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x)^2}\)
\(= \mathbf{\frac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}}\)

(iv) \(\frac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2}\)

Numerator is a perfect square: \((2y - 5z)^2\)
Denominator is a difference of squares: \((5z)^2 - (2y)^2 = (5z - 2y)(5z + 2y) = -(2y - 5z)(2y + 5z)\)

Simplifying the expression:
\(= \frac{(2y - 5z)^2}{-(2y - 5z)(2y + 5z)} = -\frac{2y - 5z}{2y + 5z}\)
\(= \mathbf{\frac{5z - 2y}{5z + 2y}}\)

(v) \(\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}\)

Factorizing each quadratic expression completely:
• \(x^2 + x - 6 = (x + 3)(x - 2)\)
• \(x^2 - 7x + 12 = (x - 3)(x - 4)\)
• \(x^2 - 6x + 8 = (x - 2)(x - 4)\)
• \(x^2 - 9 = (x - 3)(x + 3)\)

Substituting the factors into the expression:
\(= \frac{(x + 3)(x - 2)(x - 3)(x - 4)}{(x - 2)(x - 4)(x - 3)(x + 3)}\)
Since all numerator and denominator terms cancel out completely:
\(= \mathbf{1}\)

(vi) \(\frac{p^4 - 16}{p^2 - 4p + 4}\)

Factorizing the numerator using difference of squares twice:
\(p^4 - 16 = (p^2 - 4)(p^2 + 4) = (p - 2)(p + 2)(p^2 + 4)\)

Factorizing the denominator as a perfect square:
\(p^2 - 4p + 4 = (p - 2)^2\)

Simplifying the expression by canceling \((p - 2)\):
\(= \frac{(p - 2)(p + 2)(p^2 + 4)}{(p - 2)^2}\)
\(= \mathbf{\frac{(p + 2)(p^2 + 4)}{p - 2}}\)

Frequently Asked Questions (FAQs)

Q1: How do you simplify rational algebraic expressions?
A: Factorize both the numerator and the denominator completely into linear or irreducible quadratic factors, then cancel out any common factors shared between them.

Q2: What should you do when terms look like opposites, such as \((2y - 5z)\) and \((5z - 2y)\)?
A: Recognize that \((5z - 2y) = -(2y - 5z)\), which helps correctly handle negative signs during cancellation.

Q3: Can all rational expressions be reduced to constants like in Question 1(v)?
A: Not always, but when all factors in the numerator match all factors in the denominator exactly, the entire expression simplifies to 1 for all valid domain values.

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