Chapter 4: Factorization and Algebraic Manipulation

Mathematics

Class 92025–26 SNC16 Questions SolvedEnglish Medium

Exercise 4.4 — Class 9 Mathematics Solutions

All 16 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 4.4

6 Questions (10 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 9, 2026

Find the square root of the following polynomials by factorization method:

Solution

This question consists of 6 sub-parts. For each polynomial, we first check for and factor out any common numerical factor (as in parts (iv), (v), and (vi)), express the trinomial as a perfect square using the identity $a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2}$ , and then extract the square root using $\sqrt{P} = \pm \sqrt{(a \pm b)^{2}} = \pm (a \pm b)$ .

Step by Step

Q1Part (i)

EasyApproved by Miss Aniya • Sep 9, 2026

$x^{2} - 8 x + 16$

Quick Solution

Under Review

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

Step by Step

Q1Part (ii)

EasyApproved by Miss Aniya • Sep 9, 2026

$9 x^{2} + 12 x + 4$

Quick Solution

Under Review

\begin{matrix} 9 x^{2} + 12 x + 4 & = (3 x)^{2} + 2 (3 x) (2) + (2)^{2} \\ = (3 x + 2)^{2} \\ \sqrt{9 x^{2} + 12 x + 4} & = \pm (3 x + 2) \end{matrix}

Step by Step

Q1Part (iii)

MediumApproved by Miss Aniya • Sep 9, 2026

$36 a^{2} + 84 a + 49$

Quick Solution

Under Review

\begin{matrix} 36 a^{2} + 84 a + 49 & = (6 a)^{2} + 2 (6 a) (7) + (7)^{2} \\ = (6 a + 7)^{2} \\ \sqrt{36 a^{2} + 84 a + 49} & = \pm (6 a + 7) \end{matrix}

Step by Step

Q1Part (iv)

Medium

$64 y^{2} - 32 y + 4$

Quick Solution

Under Review

\begin{matrix} 64 y^{2} - 32 y + 4 & = 4 (16 y^{2} - 8 y + 1) \\ = 4 [(4 y)^{2} - 2 (4 y) (1) + (1)^{2}] \\ = 4 (4 y - 1)^{2} \\ \sqrt{64 y^{2} - 32 y + 4} & = \pm 2 (4 y - 1) \\ = \pm (8 y - 2) \end{matrix}

Step by Step

Q1Part (v)

MediumApproved by Miss Alisha • Sep 11, 2026

$200 t^{2} - 120 t + 18$

Quick Solution

Under Review

\begin{matrix} 200 t^{2} - 120 t + 18 & = 2 (100 t^{2} - 60 t + 9) \\ = 2 [(10 t)^{2} - 2 (10 t) (3) + (3)^{2}] \\ = 2 (10 t - 3)^{2} \\ \sqrt{200 t^{2} - 120 t + 18} & = \pm \sqrt{2} (10 t - 3) \end{matrix}

Step by Step

Q1Part (vi)

MediumApproved by Miss Aniya • Sep 9, 2026

$40 x^{2} + 120 x + 90$

Quick Solution

Under Review

\begin{matrix} 40 x^{2} + 120 x + 90 & = 10 (4 x^{2} + 12 x + 9) \\ = 10 [(2 x)^{2} + 2 (2 x) (3) + (3)^{2}] \\ = 10 (2 x + 3)^{2} \\ \sqrt{40 x^{2} + 120 x + 90} & = \pm \sqrt{10} (2 x + 3) \end{matrix}

Step by Step

Question 2

HardApproved by Miss Aniya • Sep 9, 2026

Find the square root of the following polynomials by division method:

Solution

This question consists of 4 sub-parts. For each polynomial, the division method for polynomial square roots is applied systematically: arrange terms in descending order, find the first term of the square root from the leading term, double the current quotient to determine subsequent terms and form updated divisors, and subtract until the remainder is zero.

Step by Step

Q2Part (i)

Hard

$4 x^{4} - 28 x^{3} + 37 x^{2} + 42 x + 9$

Quick Solution

Under Review

\begin{matrix} \sqrt{4 x^{4}} & = 2 x^{2} \\ (4 x^{2} - 7 x) (- 7 x) & = - 28 x^{3} + 49 x^{2} \\ Remainder & = - 12 x^{2} + 42 x + 9 \\ (4 x^{2} - 14 x - 3) (- 3) & = - 12 x^{2} + 42 x + 9 \\ Remainder & = 0 \\ Square root & = \pm (2 x^{2} - 7 x - 3) \end{matrix}

Step by Step

Q2Part (ii)

Hard

$121 x^{4} - 198 x^{3} - 183 x^{2} + 216 x + 144$

Quick Solution

Under Review

\begin{matrix} \sqrt{121 x^{4}} & = 11 x^{2} \\ (22 x^{2} - 9 x) (- 9 x) & = - 198 x^{3} + 81 x^{2} \\ Remainder & = - 264 x^{2} + 216 x + 144 \\ (22 x^{2} - 18 x - 12) (- 12) & = - 264 x^{2} + 216 x + 144 \\ Remainder & = 0 \\ Square root & = \pm (11 x^{2} - 9 x - 12) \end{matrix}

Step by Step

Q2Part (iii)

Hard

$x^{4} - 10 x^{3} y + 27 x^{2} y^{2} - 10 x y^{3} + y^{4}$

Quick Solution

Under Review

\begin{matrix} \sqrt{x^{4}} & = x^{2} \\ (2 x^{2} - 5 x y) (- 5 x y) & = - 10 x^{3} y + 25 x^{2} y^{2} \\ Remainder & = 2 x^{2} y^{2} - 10 x y^{3} + y^{4} \\ (2 x^{2} - 10 x y + y^{2}) (y^{2}) & = 2 x^{2} y^{2} - 10 x y^{3} + y^{4} \\ Remainder & = 0 \\ Square root & = \pm (x^{2} - 5 x y + y^{2}) \end{matrix}

Step by Step

Q2Part (iv)

Hard

$4 x^{4} - 12 x^{3} + 37 x^{2} - 42 x + 49$

Quick Solution

Under Review

\begin{matrix} \sqrt{4 x^{4}} & = 2 x^{2} \\ (4 x^{2} - 3 x) (- 3 x) & = - 12 x^{3} + 9 x^{2} \\ Remainder & = 28 x^{2} - 42 x + 49 \\ (4 x^{2} - 6 x + 7) (7) & = 28 x^{2} - 42 x + 49 \\ Remainder & = 0 \\ Square root & = \pm (2 x^{2} - 3 x + 7) \end{matrix}

Step by Step

Question 3

MediumApproved by Miss Aniya • Sep 9, 2026

An investor's return $R (x)$ in rupees after investing $x$ thousand rupees is given by the quadratic expression: $R (x) = - x^{2} + 6 x - 8$ Factorize the expression and find the investment levels that result in zero return.

Quick Solution

Under Review

\begin{matrix} R (x) & = - (x^{2} - 6 x + 8) \\ = - (x^{2} - 2 x - 4 x + 8) \\ = - (x - 2) (x - 4) \\ R (x) & = 0 \\ ⟹ (x - 2) (x - 4) & = 0 \\ ⟹ x & = 2 or 4 \\ i.e. Rs 2,000 or Rs 4,000 \end{matrix}

Step by Step

Question 4

MediumApproved by Miss Aniya • Sep 9, 2026

A company's profit $P (x)$ in rupees from selling $x$ units of a product is modeled by the cubic expression: $P (x) = x^{3} - 15 x^{2} + 75 x - 125$ Find the break-even point(s), where the profit is zero.

Quick Solution

Under Review

\begin{matrix} P (x) & = (x)^{3} - 3 (x)^{2} (5) + 3 (x) (5)^{2} - (5)^{3} \\ = (x - 5)^{3} \\ P (x) & = 0 \\ ⟹ (x - 5)^{3} & = 0 \\ ⟹ x & = 5 units \end{matrix}

Step by Step

Question 5

MediumApproved by Miss Aniya • Sep 9, 2026

The potential energy $V (x)$ in an electric field varies as a cubic function of distance $x$ , given by: $V (x) = 2 x^{3} - 6 x^{2} + 4 x$ Determine where the potential energy is zero.

Quick Solution

Under Review

\begin{matrix} V (x) & = 2 x (x^{2} - 3 x + 2) \\ = 2 x (x^{2} - x - 2 x + 2) \\ = 2 x (x - 1) (x - 2) \\ V (x) & = 0 \\ ⟹ 2 x (x - 1) (x - 2) & = 0 \\ ⟹ x & = 0, 1, 2 \end{matrix}

Step by Step

Question 6

MediumApproved by Miss Aniya • Sep 9, 2026

In structural engineering, the deflection $y (x)$ of a beam is given by: $y (x) = 2 x^{2} - 8 x + 6$ This equation gives the vertical deflection at any point $x$ along the beam. Find the points of zero deflection.

Quick Solution

Under Review

\begin{matrix} y (x) & = 2 (x^{2} - 4 x + 3) \\ = 2 (x^{2} - 3 x - x + 3) \\ = 2 (x - 1) (x - 3) \\ y (x) & = 0 \\ ⟹ (x - 1) (x - 3) & = 0 \\ ⟹ x & = 1, 3 \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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