Chapter 4: Factorization and Algebraic Manipulation

Mathematics

Class 92025–26 SNC20 Questions SolvedEnglish Medium

Exercise 4.3 — Class 9 Mathematics Solutions

All 20 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 4.3

6 Questions (14 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

Find HCF by factorization method.

Solution

This question consists of 6 sub-parts. Follow the step-by-step factorization for each sub-part below to find the highest common factor (HCF).

Step by Step

Q1Part (i)

EasyApproved by Miss Aniya • Sep 8, 2026

$21 x^{2} y, 35 x y^{2}$

Quick Solution

Under Review

\begin{matrix} 21 x^{2} y & = 3 \times 7 \times x \times x \times y \\ 35 x y^{2} & = 5 \times 7 \times x \times y \times y \\ HCF & = 7 \times x \times y \\ = 7 x y \end{matrix}

Step by Step

Q1Part (ii)

EasyApproved by Miss Aniya • Sep 8, 2026

$4 x^{2} - 9 y^{2}, 2 x^{2} - 3 x y$

Quick Solution

Under Review

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

Step by Step

Q1Part (iii)

EasyApproved by Miss Aniya • Sep 8, 2026

$x^{3} - 1, x^{2} + x + 1$

Quick Solution

Under Review

\begin{matrix} x^{3} - 1 & = (x)^{3} - (1)^{3} \\ = (x - 1) (x^{2} + x + 1) \\ x^{2} + x + 1 & = (x^{2} + x + 1) \\ HCF & = x^{2} + x + 1 \end{matrix}

Step by Step

Q1Part (iv)

MediumApproved by Miss Aniya • Sep 8, 2026

$a^{3} + 2 a^{2} - 3 a, 2 a^{3} + 5 a^{2} - 3 a$

Quick Solution

Under Review

\begin{matrix} a^{3} + 2 a^{2} - 3 a & = a (a^{2} + 2 a - 3) \\ = a (a^{2} + 3 a - a - 3) \\ = a (a + 3) (a - 1) \\ 2 a^{3} + 5 a^{2} - 3 a & = a (2 a^{2} + 5 a - 3) \\ = a (2 a^{2} + 6 a - a - 3) \\ = a (a + 3) (2 a - 1) \\ HCF & = a (a + 3) \end{matrix}

Step by Step

Q1Part (v)

Medium

$t^{2} - 3 t - 4, t^{2} + 5 t + 4, t^{2} - 1$

Quick Solution

Under Review

\begin{matrix} t^{2} - 3 t - 4 & = t^{2} - 4 t + t - 4 \\ = (t - 4) (t + 1) \\ t^{2} + 5 t + 4 & = t^{2} + 4 t + t + 4 \\ = (t + 4) (t + 1) \\ t^{2} - 1 & = (t + 1) (t - 1) \\ HCF & = t + 1 \end{matrix}

Step by Step

Q1Part (vi)

Medium

$x^{2} + 15 x + 56, x^{2} + 5 x - 24, x^{2} + 8 x$

Quick Solution

Under Review

\begin{matrix} x^{2} + 15 x + 56 & = x^{2} + 7 x + 8 x + 56 \\ = (x + 7) (x + 8) \\ x^{2} + 5 x - 24 & = x^{2} + 8 x - 3 x - 24 \\ = (x + 8) (x - 3) \\ x^{2} + 8 x & = x (x + 8) \\ HCF & = x + 8 \end{matrix}

Step by Step

Question 2

HardApproved by Miss Aniya • Sep 8, 2026

Find HCF of the following expressions by using division method:

Solution

This question consists of 4 sub-parts. Each pair or set of polynomials is solved by applying the polynomial division algorithm for Highest Common Factor (HCF). Any common numerical or monomial factors across all given polynomials are extracted first, non-common scalar multiples in intermediate remainders are discarded, and the final divisor producing a remainder of zero is multiplied back with the extracted common factors to yield the HCF.

Step by Step

Q2Part (ii)

Medium

$x^{3} - 9 x^{2} + 23 x - 15, x^{2} - 4 x + 3$

Quick Solution

Under Review

\begin{matrix} Divide x^{3} - 9 x^{2} + 23 x - 15 by x^{2} - 4 x + 3 : \\ 1st remainder & = - 5 x^{2} + 20 x - 15 \\ 2nd remainder & = 0 \\ Quotient & = x - 5 \\ HCF & = x^{2} - 4 x + 3 \end{matrix}

Step by Step

Q2Part (iii)

Hard

$2 x^{3} + 2 x^{2} + 2 x + 2, 6 x^{3} + 12 x^{2} + 6 x + 12$

Quick Solution

Under Review

\begin{matrix} 2 x^{3} + 2 x^{2} + 2 x + 2 & = 2 (x^{3} + x^{2} + x + 1) \\ 6 x^{3} + 12 x^{2} + 6 x + 12 & = 6 (x^{3} + 2 x^{2} + x + 2) \\ Divide x^{3} + 2 x^{2} + x + 2 by x^{3} + x^{2} + x + 1 : \\ Remainder & = x^{2} + 1 \\ Divide x^{3} + x^{2} + x + 1 by x^{2} + 1 : \\ Remainder & = 0 \\ HCF & = 2 (x^{2} + 1) \end{matrix}

Step by Step

Q2Part (iv)

Hard

$2 x^{3} - 4 x^{2} - 16 x, x^{3} - 4 x, 3 x^{2} - 6 x$

Quick Solution

Under Review

\begin{matrix} 3 x^{2} - 6 x & = 3 (x^{2} - 2 x) \\ Divide x^{3} - 4 x by x^{2} - 2 x : \\ Remainder & = 0 \\ {HCF}_{1} & = x^{2} - 2 x \\ Divide 2 x^{3} - 4 x^{2} - 16 x by x^{2} - 2 x : \\ Remainder & = - 16 x \to x \\ Divide x^{2} - 2 x by x : \\ Remainder & = 0 \\ HCF & = x \end{matrix}

Step by Step

Question 3

MediumApproved by Miss Aniya • Sep 8, 2026

Find LCM of the following expressions by using prime factorization method.

Solution

This question consists of 5 sub-parts. To find the Least Common Multiple (LCM) of algebraic expressions using prime factorization:

Factorize each expression completely into prime/irreducible factors.
Identify:
- Factors common to all expressions.
- Factors common to any two expressions (when three or more expressions are given).
- Remaining non-common factors.
Compute the LCM using the formula:

LCM = (Product of common factors) \times (Product of non-common factors)

Step by Step

Q3Part (i)

EasyApproved by Miss Aniya • Sep 8, 2026

$2 a^{2} b, 4 a b^{2}, 6 a b$

Quick Solution

Under Review

\begin{matrix} 2 a^{2} b & = 2 \times a \times a \times b \\ 4 a b^{2} & = 2 \times 2 \times a \times b \times b \\ 6 a b & = 2 \times 3 \times a \times b \\ LCM & = (2 a b) \times (a \times 2 b \times 3) \\ = 12 a^{2} b^{2} \end{matrix}

Step by Step

Q3Part (ii)

EasyApproved by Miss Aniya • Sep 8, 2026

$x^{2} + x, x^{3} + x^{2}$

Quick Solution

Under Review

\begin{matrix} x^{2} + x & = x (x + 1) \\ x^{3} + x^{2} & = x^{2} (x + 1) \\ LCM & = x (x + 1) \times x \\ = x^{2} (x + 1) \end{matrix}

Step by Step

Q3Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

$a^{2} - 4 a + 4, a^{2} - 2 a$

Quick Solution

Under Review

\begin{matrix} a^{2} - 4 a + 4 & = (a - 2)^{2} \\ a^{2} - 2 a & = a (a - 2) \\ LCM & = (a - 2) \times a (a - 2) \\ = a (a - 2)^{2} \end{matrix}

Step by Step

Q3Part (iv)

MediumApproved by Miss Aniya • Sep 9, 2026

$x^{4} - 16, x^{3} - 4 x$

Quick Solution

Under Review

\begin{matrix} x^{4} - 16 & = (x^{2} + 4) (x^{2} - 4) \\ = (x^{2} + 4) (x + 2) (x - 2) \\ x^{3} - 4 x & = x (x^{2} - 4) \\ = x (x + 2) (x - 2) \\ LCM & = (x^{2} - 4) \times x (x^{2} + 4) \\ = x (x^{4} - 16) \end{matrix}

Step by Step

Q3Part (v)

MediumApproved by Miss Aniya • Sep 9, 2026

$16 - 4 x^{2}, x^{2} + x - 6, 4 - x^{2}$

Quick Solution

Under Review

\begin{matrix} 16 - 4 x^{2} & = 4 (4 - x^{2}) \\ = - 4 (x + 2) (x - 2) \\ x^{2} + x - 6 & = x^{2} + 3 x - 2 x - 6 \\ = (x + 3) (x - 2) \\ 4 - x^{2} & = - (x + 2) (x - 2) \\ LCM & = (4 - x^{2}) \times 4 (x + 3) \\ = 4 (4 - x^{2}) (x + 3) \end{matrix}

Step by Step

Question 4

MediumApproved by Miss Aniya • Sep 10, 2026

The HCF of two polynomials is $y - 7$ and their LCM is $y^{3} - 10 y^{2} + 11 y + 70$ . If one of the polynomials is $y^{2} - 5 y - 14$ , find the other.

Quick Solution

Under Review

\begin{matrix} q (y) & = \frac{HCF \times LCM}{p (y)} \\ = \frac{(y - 7) (y^{3} - 10 y^{2} + 11 y + 70)}{(y - 7) (y + 2)} \\ = \frac{y^{3} - 10 y^{2} + 11 y + 70}{y + 2} \\ = y^{2} - 12 y + 35 \\ = (y - 5) (y - 7) \end{matrix}

Step by Step

Question 5

HardApproved by Miss Aniya • Sep 9, 2026

The LCM and HCF of two polynomials $p (x)$ and $q (x)$ are $36 x^{3} (x + a) (x^{3} - a^{3})$ and $x^{2} (x - a)$ respectively. If $p (x) = 4 x^{2} (x^{2} - a^{2})$ , find $q (x)$ .

Quick Solution

Under Review

\begin{matrix} q (x) & = \frac{HCF \times LCM}{p (x)} \\ = \frac{x^{2} (x - a) \times 36 x^{3} (x + a) (x^{3} - a^{3})}{4 x^{2} (x^{2} - a^{2})} \\ = \frac{36 x^{5} (x^{2} - a^{2}) (x^{3} - a^{3})}{4 x^{2} (x^{2} - a^{2})} \\ = 9 x^{3} (x^{3} - a^{3}) \end{matrix}

Step by Step

Question 6

Medium

The HCF and LCM of two polynomials are $(x + a)$ and $12 x^{2} (x + a) (x^{2} - a^{2})$ respectively. Find the product of the two polynomials.

Quick Solution

Under Review

\begin{matrix} p (x) \times q (x) & = HCF \times LCM \\ = (x + a) \times 12 x^{2} (x + a) (x^{2} - a^{2}) \\ = 12 x^{2} (x + a)^{2} (x^{2} - a^{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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