Chapter 4: Factorization and Algebraic Manipulation

Mathematics

Class 92025–26 SNC25 Questions SolvedEnglish Medium

Exercise 4.2 — Class 9 Mathematics Solutions

All 25 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 4.2

4 Questions (21 Sub-parts)

Quick Solution Step by Step

Just the working. Switch to Step by Step for the full explanation.

Question 1

MediumApproved by Miss Aniya • Sep 8, 2026

Factorize each of the following expressions:

Solution

This question consists of 6 sub-parts. Each algebraic expression is factorized by completing the square to express it in the form of a difference of two squares: $a^{2} - b^{2} = (a + b) (a - b)$

Step by Step

Q1Part (i)

MediumApproved by Miss Aniya • Sep 8, 2026

$4 x^{4} + 81 y^{4}$

Quick Solution

Under Review

\begin{matrix} 4 x^{4} + 81 y^{4} & = (2 x^{2})^{2} + (9 y^{2})^{2} + 2 (2 x^{2}) (9 y^{2}) - 2 (2 x^{2}) (9 y^{2}) \\ = (2 x^{2} + 9 y^{2})^{2} - 36 x^{2} y^{2} \\ = (2 x^{2} + 9 y^{2})^{2} - (6 x y)^{2} \\ = (2 x^{2} + 9 y^{2} + 6 x y) (2 x^{2} + 9 y^{2} - 6 x y) \\ = (2 x^{2} + 6 x y + 9 y^{2}) (2 x^{2} - 6 x y + 9 y^{2}) \end{matrix}

Step by Step

Q1Part (ii)

MediumApproved by Miss Aniya • Sep 8, 2026

$a^{4} + 64 b^{4}$

Quick Solution

Under Review

\begin{matrix} a^{4} + 64 b^{4} & = (a^{2})^{2} + (8 b^{2})^{2} + 2 (a^{2}) (8 b^{2}) - 2 (a^{2}) (8 b^{2}) \\ = (a^{2} + 8 b^{2})^{2} - 16 a^{2} b^{2} \\ = (a^{2} + 8 b^{2})^{2} - (4 a b)^{2} \\ = (a^{2} + 8 b^{2} + 4 a b) (a^{2} + 8 b^{2} - 4 a b) \\ = (a^{2} + 4 a b + 8 b^{2}) (a^{2} - 4 a b + 8 b^{2}) \end{matrix}

Step by Step

Q1Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

$x^{4} + 4 x^{2} + 16$

Quick Solution

Under Review

\begin{matrix} x^{4} + 4 x^{2} + 16 & = (x^{2})^{2} + 4^{2} + 2 (x^{2}) (4) - 2 (x^{2}) (4) + 4 x^{2} \\ = (x^{2} + 4)^{2} - 8 x^{2} + 4 x^{2} \\ = (x^{2} + 4)^{2} - 4 x^{2} \\ = (x^{2} + 4)^{2} - (2 x)^{2} \\ = (x^{2} + 4 + 2 x) (x^{2} + 4 - 2 x) \\ = (x^{2} + 2 x + 4) (x^{2} - 2 x + 4) \end{matrix}

Step by Step

Q1Part (iv)

MediumApproved by Miss Aniya • Sep 8, 2026

$x^{4} - 14 x^{2} + 1$

Quick Solution

Under Review

\begin{matrix} x^{4} - 14 x^{2} + 1 & = (x^{2})^{2} + 1^{2} + 2 (x^{2}) (1) - 2 (x^{2}) (1) - 14 x^{2} \\ = (x^{2} + 1)^{2} - 2 x^{2} - 14 x^{2} \\ = (x^{2} + 1)^{2} - 16 x^{2} \\ = (x^{2} + 1)^{2} - (4 x)^{2} \\ = (x^{2} + 1 + 4 x) (x^{2} + 1 - 4 x) \\ = (x^{2} + 4 x + 1) (x^{2} - 4 x + 1) \end{matrix}

Step by Step

Q1Part (v)

MediumApproved by Miss Aniya • Sep 8, 2026

$x^{4} - 30 x^{2} y^{2} + 9 y^{4}$

Quick Solution

Under Review

\begin{matrix} x^{4} - 30 x^{2} y^{2} + 9 y^{4} & = (x^{2})^{2} + (3 y^{2})^{2} + 2 (x^{2}) (3 y^{2}) - 2 (x^{2}) (3 y^{2}) - 30 x^{2} y^{2} \\ = (x^{2} + 3 y^{2})^{2} - 6 x^{2} y^{2} - 30 x^{2} y^{2} \\ = (x^{2} + 3 y^{2})^{2} - 36 x^{2} y^{2} \\ = (x^{2} + 3 y^{2})^{2} - (6 x y)^{2} \\ = (x^{2} + 3 y^{2} + 6 x y) (x^{2} + 3 y^{2} - 6 x y) \\ = (x^{2} + 6 x y + 3 y^{2}) (x^{2} - 6 x y + 3 y^{2}) \end{matrix}

Step by Step

Question 2

HardApproved by Miss Aniya • Sep 8, 2026

Factorize each of the following expressions:

Solution

This question consists of 6 sub-parts. In each problem, we pair linear factors or quadratic terms having matching coefficients, apply a temporary substitution $y$ , factorize the resulting quadratic equation, and back-substitute to complete the factorization.

Step by Step

Q2Part (i)

HardApproved by Miss Aniya • Sep 8, 2026

$(x + 1) (x + 2) (x + 3) (x + 4) + 1$

Quick Solution

Under Review

\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 4) + 1 & = [(x + 1) (x + 4)] [(x + 2) (x + 3)] + 1 \\ = (x^{2} + 5 x + 4) (x^{2} + 5 x + 6) + 1 \\ Let y & = x^{2} + 5 x \\ (y + 4) (y + 6) + 1 & = y^{2} + 6 y + 4 y + 24 + 1 \\ = y^{2} + 10 y + 25 \\ = (y + 5)^{2} \\ = (x^{2} + 5 x + 5)^{2} \end{matrix}

Step by Step

Q2Part (ii)

HardApproved by Miss Aniya • Sep 8, 2026

$(x + 2) (x - 7) (x - 4) (x - 1) + 17$

Quick Solution

Under Review

\begin{matrix} (x + 2) (x - 7) (x - 4) (x - 1) + 17 & = [(x + 2) (x - 7)] [(x - 4) (x - 1)] + 17 \\ = (x^{2} - 5 x - 14) (x^{2} - 5 x + 4) + 17 \\ Let y & = x^{2} - 5 x \\ (y - 14) (y + 4) + 17 & = y^{2} + 4 y - 14 y - 56 + 17 \\ = y^{2} - 10 y - 39 \\ = y^{2} - 13 y + 3 y - 39 \\ = y (y - 13) + 3 (y - 13) \\ = (y - 13) (y + 3) \\ = (x^{2} - 5 x - 13) (x^{2} - 5 x + 3) \end{matrix}

Step by Step

Q2Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

$(2 x^{2} + 7 x + 3) (2 x^{2} + 7 x + 5) + 1$

Quick Solution

Under Review

\begin{matrix} Let y & = 2 x^{2} + 7 x \\ (2 x^{2} + 7 x + 3) (2 x^{2} + 7 x + 5) + 1 & = (y + 3) (y + 5) + 1 \\ = y^{2} + 5 y + 3 y + 15 + 1 \\ = y^{2} + 8 y + 16 \\ = (y + 4)^{2} \\ = (2 x^{2} + 7 x + 4)^{2} \end{matrix}

Step by Step

Q2Part (iv)

MediumApproved by Miss Aniya • Sep 10, 2026

$(3 x^{2} + 5 x + 3) (3 x^{2} + 5 x + 5) - 3$

Quick Solution

Under Review

\begin{matrix} Let y & = 3 x^{2} + 5 x \\ (3 x^{2} + 5 x + 3) (3 x^{2} + 5 x + 5) - 3 & = (y + 3) (y + 5) - 3 \\ = y^{2} + 5 y + 3 y + 15 - 3 \\ = y^{2} + 8 y + 12 \\ = y^{2} + 6 y + 2 y + 12 \\ = y (y + 6) + 2 (y + 6) \\ = (y + 6) (y + 2) \\ = (3 x^{2} + 5 x + 6) (3 x^{2} + 5 x + 2) \end{matrix}

Step by Step

Q2Part (v)

HardApproved by Miss Aniya • Sep 8, 2026

$(x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2}$

Quick Solution

Under Review

\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2} & = [(x + 1) (x + 6)] [(x + 2) (x + 3)] - 3 x^{2} \\ = [(x^{2} + 6) + 7 x] [(x^{2} + 6) + 5 x] - 3 x^{2} \\ Let y & = x^{2} + 6 \\ (y + 7 x) (y + 5 x) - 3 x^{2} & = y^{2} + 5 x y + 7 x y + 35 x^{2} - 3 x^{2} \\ = y^{2} + 12 x y + 32 x^{2} \\ = y^{2} + 8 x y + 4 x y + 32 x^{2} \\ = y (y + 8 x) + 4 x (y + 8 x) \\ = (y + 8 x) (y + 4 x) \\ = (x^{2} + 6 + 8 x) (x^{2} + 6 + 4 x) \\ = (x^{2} + 8 x + 6) (x^{2} + 4 x + 6) \end{matrix}

Step by Step

Q2Part (vi)

Hard

$(x + 1) (x - 1) (x + 2) (x - 2) + 13 x^{2}$

Quick Solution

Under Review

\begin{matrix} (x + 1) (x - 1) (x + 2) (x - 2) + 13 x^{2} & = (x^{2} - 1) (x^{2} - 4) + 13 x^{2} \\ = x^{4} - 4 x^{2} - x^{2} + 4 + 13 x^{2} \\ = x^{4} + 8 x^{2} + 4 \\ = (x^{2} + 2)^{2} + (2 x)^{2} \end{matrix}

Sum of squares: cannot be factorized (textbook misprint). With the intended $+ 5 x^{2}$ :

\begin{matrix} (x + 1) (x - 1) (x + 2) (x - 2) + 5 x^{2} & = (x^{2} - 1) (x^{2} - 4) + 5 x^{2} \\ = x^{4} - 5 x^{2} + 4 + 5 x^{2} \\ = x^{4} + 4 \\ = (x^{2})^{2} + (2)^{2} + 2 (x^{2}) (2) - 2 (x^{2}) (2) \\ = (x^{2} + 2)^{2} - 4 x^{2} \\ = (x^{2} + 2)^{2} - (2 x)^{2} \\ = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2) \end{matrix}

Step by Step

Question 3

MediumUnder Review

Factorize:

Solution

This question consists of 4 sub-parts. Follow the step-by-step factorization using the algebraic cube-of-a-binomial identities $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ and $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ for each sub-part below.

Step by Step

Q3Part (i)

MediumApproved by Miss Aniya • Sep 8, 2026

$8 x^{3} + 12 x^{2} + 6 x + 1$

Quick Solution

Under Review

\begin{matrix} 8 x^{3} + 12 x^{2} + 6 x + 1 & = (2 x)^{3} + 3 (2 x)^{2} (1) + 3 (2 x) (1)^{2} + (1)^{3} \\ = (2 x + 1)^{3} \end{matrix}

Step by Step

Q3Part (ii)

MediumApproved by Miss Aniya • Sep 8, 2026

$27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3}$

Quick Solution

Under Review

\begin{matrix} 27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3} & = (3 a)^{3} + 3 (3 a)^{2} (4 b) + 3 (3 a) (4 b)^{2} + (4 b)^{3} \\ = (3 a + 4 b)^{3} \end{matrix}

Step by Step

Q3Part (iii)

Medium

$x^{3} + 48 x^{2} y + 108 x y^{2} + 216 y^{3}$

Quick Solution

Under Review

Intended term $18 x^{2} y$ (book misprints $48 x^{2} y$ ):

\begin{matrix} x^{3} + 18 x^{2} y + 108 x y^{2} + 216 y^{3} & = (x)^{3} + 3 (x)^{2} (6 y) + 3 (x) (6 y)^{2} + (6 y)^{3} \\ = (x + 6 y)^{3} \end{matrix}

Step by Step

Q3Part (iv)

MediumApproved by Miss Aniya • Sep 8, 2026

$8 x^{3} - 125 y^{3} + 150 x y^{2} - 60 x^{2} y$

Quick Solution

Under Review

\begin{matrix} 8 x^{3} - 125 y^{3} + 150 x y^{2} - 60 x^{2} y & = 8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3} \\ = (2 x)^{3} - 3 (2 x)^{2} (5 y) + 3 (2 x) (5 y)^{2} - (5 y)^{3} \\ = (2 x - 5 y)^{3} \end{matrix}

Step by Step

Question 4

MediumApproved by Miss Aniya • Sep 8, 2026

Factorize:

Solution

This question consists of 6 sub-parts. Follow the step-by-step factorization using the sum and difference of cubes formulas for each sub-part below.

Step by Step

Q4Part (i)

EasyApproved by Miss Aniya • Sep 8, 2026

$125 a^{3} - 1$

Quick Solution

Under Review

\begin{matrix} 125 a^{3} - 1 & = (5 a)^{3} - (1)^{3} \\ = (5 a - 1) [(5 a)^{2} + (5 a) (1) + (1)^{2}] \\ = (5 a - 1) (25 a^{2} + 5 a + 1) \end{matrix}

Step by Step

Q4Part (ii)

EasyApproved by Miss Aniya • Sep 8, 2026

$64 x^{3} + 125$

Quick Solution

Under Review

\begin{matrix} 64 x^{3} + 125 & = (4 x)^{3} + (5)^{3} \\ = (4 x + 5) [(4 x)^{2} - (4 x) (5) + (5)^{2}] \\ = (4 x + 5) (16 x^{2} - 20 x + 25) \end{matrix}

Step by Step

Q4Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

$x^{6} - 27$

Quick Solution

Under Review

\begin{matrix} x^{6} - 27 & = (x^{2})^{3} - (3)^{3} \\ = (x^{2} - 3) [(x^{2})^{2} + (x^{2}) (3) + (3)^{2}] \\ = (x^{2} - 3) (x^{4} + 3 x^{2} + 9) \end{matrix}

Step by Step

Q4Part (iv)

EasyApproved by Miss Aniya • Sep 8, 2026

$1000 a^{3} + 1$

Quick Solution

Under Review

\begin{matrix} 1000 a^{3} + 1 & = (10 a)^{3} + (1)^{3} \\ = (10 a + 1) [(10 a)^{2} - (10 a) (1) + (1)^{2}] \\ = (10 a + 1) (100 a^{2} - 10 a + 1) \end{matrix}

Step by Step

Q4Part (v)

MediumApproved by Miss Aniya • Sep 8, 2026

$343 x^{3} + 216$

Quick Solution

Under Review

\begin{matrix} 343 x^{3} + 216 & = (7 x)^{3} + (6)^{3} \\ = (7 x + 6) [(7 x)^{2} - (7 x) (6) + (6)^{2}] \\ = (7 x + 6) (49 x^{2} - 42 x + 36) \end{matrix}

Step by Step

Q4Part (vi)

MediumApproved by Miss Aniya • Sep 8, 2026

$27 - 512 y^{3}$

Quick Solution

Under Review

\begin{matrix} 27 - 512 y^{3} & = (3)^{3} - (8 y)^{3} \\ = (3 - 8 y) [(3)^{2} + (3) (8 y) + (8 y)^{2}] \\ = (3 - 8 y) (9 + 24 y + 64 y^{2}) \end{matrix}

Step by Step

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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