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

Every step explained, with hints and the method used.

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: $a^{2} - b^{2} = (a + b) (a - b)$

Quick Solution

Q1Part (i)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Express as $(2 x^{2})^{2} + (9 y^{2})^{2}$ . Add and subtract $2 (2 x^{2}) (9 y^{2}) = 36 x^{2} y^{2} = (6 x y)^{2}$ to form $(2 x^{2} + 9 y^{2})^{2} - (6 x y)^{2}$ , then factorize using $a^{2} - b^{2} = (a + b) (a - b)$ .

Solution

Express each term as a perfect square.

Recognize the given terms as squares of monomials:
$\begin{matrix} 4 x^{4} & = (2 x^{2})^{2} \\ 81 y^{4} & = (9 y^{2})^{2} \end{matrix}$
Complete the square by adding and subtracting $2 (A) (B)$ .

Here $A = 2 x^{2}$ and $B = 9 y^{2}$ . The required middle term to complete the square is:
$2 (A) (B) = 2 (2 x^{2}) (9 y^{2}) = 36 x^{2} y^{2} = (6 x y)^{2}$
Add and subtract $2 (2 x^{2}) (9 y^{2})$ :
$\begin{matrix} 4 x^{4} + 81 y^{4} & = (2 x^{2})^{2} + (9 y^{2})^{2} \\ = (2 x^{2})^{2} + (9 y^{2})^{2} + 2 (2 x^{2}) (9 y^{2}) - 2 (2 x^{2}) (9 y^{2}) \\ = [(2 x^{2})^{2} + 2 (2 x^{2}) (9 y^{2}) + (9 y^{2})^{2}] - 36 x^{2} y^{2} \\ = (2 x^{2} + 9 y^{2})^{2} - (6 x y)^{2} \end{matrix}$
Apply the difference of two squares identity.

Using $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x^{2} + 9 y^{2}$ and $b = 6 x y$ :
$\begin{matrix} = (2 x^{2} + 9 y^{2} + 6 x y) (2 x^{2} + 9 y^{2} - 6 x y) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = (2 x^{2} + 6 x y + 9 y^{2}) (2 x^{2} - 6 x y + 9 y^{2}) \end{matrix}$

Answer

$(2 x^{2} + 6 x y + 9 y^{2}) (2 x^{2} - 6 x y + 9 y^{2})$

Quick Solution

Q1Part (ii)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Express as $(a^{2})^{2} + (8 b^{2})^{2}$ . Add and subtract $2 (a^{2}) (8 b^{2}) = 16 a^{2} b^{2} = (4 a b)^{2}$ to obtain $(a^{2} + 8 b^{2})^{2} - (4 a b)^{2}$ , then apply $a^{2} - b^{2} = (a + b) (a - b)$ .

Solution

Express each term as a perfect square.

Write the terms as squares of monomials:
$\begin{matrix} a^{4} & = (a^{2})^{2} \\ 64 b^{4} & = (8 b^{2})^{2} \end{matrix}$
Complete the square by adding and subtracting $2 (A) (B)$ .

Here $A = a^{2}$ and $B = 8 b^{2}$ . The required middle term to complete the square is:
$2 (A) (B) = 2 (a^{2}) (8 b^{2}) = 16 a^{2} b^{2} = (4 a b)^{2}$
Add and subtract $2 (a^{2}) (8 b^{2})$ :
$\begin{matrix} a^{4} + 64 b^{4} & = (a^{2})^{2} + (8 b^{2})^{2} \\ = (a^{2})^{2} + (8 b^{2})^{2} + 2 (a^{2}) (8 b^{2}) - 2 (a^{2}) (8 b^{2}) \\ = [(a^{2})^{2} + 2 (a^{2}) (8 b^{2}) + (8 b^{2})^{2}] - 16 a^{2} b^{2} \\ = (a^{2} + 8 b^{2})^{2} - (4 a b)^{2} \end{matrix}$
Apply the difference of two squares identity.

Using $a^{2} - b^{2} = (a + b) (a - b)$ where the first term is $a^{2} + 8 b^{2}$ and the second term is $4 a b$ :
$\begin{matrix} = (a^{2} + 8 b^{2} + 4 a b) (a^{2} + 8 b^{2} - 4 a b) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = (a^{2} + 4 a b + 8 b^{2}) (a^{2} - 4 a b + 8 b^{2}) \end{matrix}$

Answer

$(a^{2} + 4 a b + 8 b^{2}) (a^{2} - 4 a b + 8 b^{2})$

Quick Solution

Q1Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Rearrange as $(x^{2})^{2} + 4^{2} + 4 x^{2}$ . Complete the square by adding and subtracting $2 (x^{2}) (4) = 8 x^{2}$ to form $(x^{2} + 4)^{2} - 4 x^{2} = (x^{2} + 4)^{2} - (2 x)^{2}$ , then factorize using $a^{2} - b^{2} = (a + b) (a - b)$ .

Solution

Rearrange and identify extreme square terms.

Rearrange terms to group the perfect square terms together:
$\begin{matrix} x^{4} + 4 x^{2} + 16 & = x^{4} + 16 + 4 x^{2} \\ = (x^{2})^{2} + 4^{2} + 4 x^{2} \end{matrix}$
Complete the square by adding and subtracting $2 (A) (B)$ .

For the extreme terms $A = x^{2}$ and $B = 4$ , the middle term needed to complete the square is:
$2 (A) (B) = 2 (x^{2}) (4) = 8 x^{2}$
Add and subtract $2 (x^{2}) (4)$ while retaining the original term $+ 4 x^{2}$ :
$\begin{matrix} x^{4} + 4 x^{2} + 16 & = (x^{2})^{2} + 4^{2} + 2 (x^{2}) (4) - 2 (x^{2}) (4) + 4 x^{2} \\ = [(x^{2})^{2} + 2 (x^{2}) (4) + 4^{2}] - 8 x^{2} + 4 x^{2} \\ = (x^{2} + 4)^{2} - 4 x^{2} \\ = (x^{2} + 4)^{2} - (2 x)^{2} \end{matrix}$
Factorize using the difference of two squares.

Using $a^{2} - b^{2} = (a + b) (a - b)$ :
$\begin{matrix} = (x^{2} + 4 + 2 x) (x^{2} + 4 - 2 x) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = (x^{2} + 2 x + 4) (x^{2} - 2 x + 4) \end{matrix}$

Answer

$(x^{2} + 2 x + 4) (x^{2} - 2 x + 4)$

Quick Solution

Q1Part (iv)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Rearrange as $(x^{2})^{2} + 1^{2} - 14 x^{2}$ . Add and subtract $2 (x^{2}) (1) = 2 x^{2}$ to form $(x^{2} + 1)^{2} - 16 x^{2} = (x^{2} + 1)^{2} - (4 x)^{2}$ , then apply $a^{2} - b^{2} = (a + b) (a - b)$ .

Solution

Rearrange and identify extreme square terms.

Rearrange terms to group the perfect square terms together:
$\begin{matrix} x^{4} - 14 x^{2} + 1 & = x^{4} + 1 - 14 x^{2} \\ = (x^{2})^{2} + 1^{2} - 14 x^{2} \end{matrix}$
Complete the square by adding and subtracting $2 (A) (B)$ .

For the extreme square terms $A = x^{2}$ and $B = 1$ , the middle term needed to complete the square is:
$2 (A) (B) = 2 (x^{2}) (1) = 2 x^{2}$
Add and subtract $2 (x^{2}) (1)$ while retaining the existing middle term $- 14 x^{2}$ :
$\begin{matrix} x^{4} - 14 x^{2} + 1 & = (x^{2})^{2} + 1^{2} + 2 (x^{2}) (1) - 2 (x^{2}) (1) - 14 x^{2} \\ = [(x^{2})^{2} + 2 (x^{2}) (1) + 1^{2}] - 2 x^{2} - 14 x^{2} \\ = (x^{2} + 1)^{2} - 16 x^{2} \\ = (x^{2} + 1)^{2} - (4 x)^{2} \end{matrix}$
Apply the difference of two squares identity.

Using $a^{2} - b^{2} = (a + b) (a - b)$ :
$\begin{matrix} = (x^{2} + 1 + 4 x) (x^{2} + 1 - 4 x) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = (x^{2} + 4 x + 1) (x^{2} - 4 x + 1) \end{matrix}$

Answer

$(x^{2} + 4 x + 1) (x^{2} - 4 x + 1)$

Quick Solution

Q1Part (v)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Rearrange as $(x^{2})^{2} + (3 y^{2})^{2} - 30 x^{2} y^{2}$ . Add and subtract $2 (x^{2}) (3 y^{2}) = 6 x^{2} y^{2}$ to form $(x^{2} + 3 y^{2})^{2} - 36 x^{2} y^{2} = (x^{2} + 3 y^{2})^{2} - (6 x y)^{2}$ , then apply $a^{2} - b^{2} = (a + b) (a - b)$ .

Solution

Rearrange and identify extreme square terms.

Rearrange terms to group the perfect square monomials together:
$\begin{matrix} x^{4} - 30 x^{2} y^{2} + 9 y^{4} & = x^{4} + 9 y^{4} - 30 x^{2} y^{2} \\ = (x^{2})^{2} + (3 y^{2})^{2} - 30 x^{2} y^{2} \end{matrix}$
Complete the square by adding and subtracting $2 (A) (B)$ .

For $A = x^{2}$ and $B = 3 y^{2}$ , the middle term required to complete the square is:
$2 (A) (B) = 2 (x^{2}) (3 y^{2}) = 6 x^{2} y^{2}$
Add and subtract $2 (x^{2}) (3 y^{2})$ while retaining the existing term $- 30 x^{2} y^{2}$ :
$\begin{matrix} x^{4} - 30 x^{2} y^{2} + 9 y^{4} & = (x^{2})^{2} + (3 y^{2})^{2} + 2 (x^{2}) (3 y^{2}) - 2 (x^{2}) (3 y^{2}) - 30 x^{2} y^{2} \\ = [(x^{2})^{2} + 2 (x^{2}) (3 y^{2}) + (3 y^{2})^{2}] - 6 x^{2} y^{2} - 30 x^{2} y^{2} \\ = (x^{2} + 3 y^{2})^{2} - 36 x^{2} y^{2} \\ = (x^{2} + 3 y^{2})^{2} - (6 x y)^{2} \end{matrix}$
Factorize using the difference of two squares.

Using $a^{2} - b^{2} = (a + b) (a - b)$ :
$\begin{matrix} = (x^{2} + 3 y^{2} + 6 x y) (x^{2} + 3 y^{2} - 6 x y) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = (x^{2} + 6 x y + 3 y^{2}) (x^{2} - 6 x y + 3 y^{2}) \end{matrix}$

Answer

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

Quick Solution

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.

Quick Solution

Q2Part (i)

HardApproved by Miss Aniya • Sep 8, 2026

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

Hint

Pair the factors with equal constant sums $1 + 4 = 2 + 3 = 5$ as $[(x + 1) (x + 4)] [(x + 2) (x + 3)] + 1$ . Expand to $(x^{2} + 5 x + 4) (x^{2} + 5 x + 6) + 1$ , substitute $y = x^{2} + 5 x$ to obtain $(y + 5)^{2}$ , and back-substitute to get $(x^{2} + 5 x + 5)^{2}$ .

Solution

Pair linear factors with equal sums of constants.

Compare the constant terms: $1 + 4 = 5$ and $2 + 3 = 5$ . Regroup the four linear factors:
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 4) + 1 & = [(x + 1) (x + 4)] \cdot [(x + 2) (x + 3)] + 1 \end{matrix}$
Multiply each pair of linear factors.

Expand each pair to form quadratic expressions with identical first two terms:
$\begin{matrix} (x + 1) (x + 4) & = x^{2} + 4 x + x + 4 \\ = x^{2} + 5 x + 4 \\ (x + 2) (x + 3) & = x^{2} + 3 x + 2 x + 6 \\ = x^{2} + 5 x + 6 \end{matrix}$
Substitute these products back into the expression:
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 4) + 1 & = (x^{2} + 5 x + 4) (x^{2} + 5 x + 6) + 1 \end{matrix}$
Substitute $y = x^{2} + 5 x$ and factorize the quadratic.

Let $y = x^{2} + 5 x$ :
$\begin{matrix} (y + 4) (y + 6) + 1 & = y^{2} + 6 y + 4 y + 24 + 1 \\ = y^{2} + 10 y + 25 \\ = y^{2} + 2 (y) (5) + 5^{2} \\ = (y + 5)^{2} \end{matrix}$
Back-substitute $y = x^{2} + 5 x$ .

Replacing $y$ with $x^{2} + 5 x$ :
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 4) + 1 & = (y + 5)^{2} \\ = (x^{2} + 5 x + 5)^{2} \end{matrix}$
(Note: Since $Δ = 5^{2} - 4 (1) (5) = 5$ is not a perfect square, $x^{2} + 5 x + 5$ cannot be factored further over rational numbers.)

Answer

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

Quick Solution

Q2Part (ii)

HardApproved by Miss Aniya • Sep 8, 2026

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

Hint

Pair factors whose constants have equal sums $2 + (- 7) = (- 4) + (- 1) = - 5$ : $[(x + 2) (x - 7)] [(x - 4) (x - 1)] + 17$ . Let $y = x^{2} - 5 x$ , factorize $y^{2} - 10 y - 39 = (y - 13) (y + 3)$ , and back-substitute to obtain $(x^{2} - 5 x - 13) (x^{2} - 5 x + 3)$ .

Solution

Pair linear factors with equal constant sums.

Compare the constant sums: $2 + (- 7) = - 5$ and $(- 4) + (- 1) = - 5$ . Regroup the factors:
$\begin{matrix} (x + 2) (x - 7) (x - 4) (x - 1) + 17 & = [(x + 2) (x - 7)] \cdot [(x - 4) (x - 1)] + 17 \end{matrix}$
Expand each pair of linear factors.
$\begin{matrix} (x + 2) (x - 7) & = x^{2} - 7 x + 2 x - 14 \\ = x^{2} - 5 x - 14 \\ (x - 4) (x - 1) & = x^{2} - x - 4 x + 4 \\ = x^{2} - 5 x + 4 \end{matrix}$
Substitute these products into the expression:
$\begin{matrix} (x + 2) (x - 7) (x - 4) (x - 1) + 17 & = (x^{2} - 5 x - 14) (x^{2} - 5 x + 4) + 17 \end{matrix}$
Substitute $y = x^{2} - 5 x$ and factorize the quadratic.

Let $y = x^{2} - 5 x$ :
$\begin{matrix} (y - 14) (y + 4) + 17 & = y^{2} + 4 y - 14 y - 56 + 17 \\ = y^{2} - 10 y - 39 \\ = y^{2} - 13 y + 3 y - 39 \\ = y (y - 13) + 3 (y - 13) \\ = (y - 13) (y + 3) \end{matrix}$
Back-substitute $y = x^{2} - 5 x$ .

Replacing $y$ with $x^{2} - 5 x$ :
$\begin{matrix} (x + 2) (x - 7) (x - 4) (x - 1) + 17 & = (y - 13) (y + 3) \\ = (x^{2} - 5 x - 13) (x^{2} - 5 x + 3) \end{matrix}$
(Note: Neither $x^{2} - 5 x - 13$ nor $x^{2} - 5 x + 3$ has a square discriminant, so both factors are irreducible over rational numbers.)

Answer

$(x^{2} - 5 x - 13) (x^{2} - 5 x + 3)$

Quick Solution

Q2Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Substitute $y = 2 x^{2} + 7 x$ into the expression to obtain $(y + 3) (y + 5) + 1 = y^{2} + 8 y + 16 = (y + 4)^{2}$ , then replace $y$ with $2 x^{2} + 7 x$ to get $(2 x^{2} + 7 x + 4)^{2}$ .

Solution

Identify the common quadratic expression.

Notice that both factors share the identical quadratic expression $2 x^{2} + 7 x$ .
Substitute $y = 2 x^{2} + 7 x$ and simplify.

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

Replacing $y$ with $2 x^{2} + 7 x$ :
$\begin{matrix} (2 x^{2} + 7 x + 3) (2 x^{2} + 7 x + 5) + 1 & = (y + 4)^{2} \\ = (2 x^{2} + 7 x + 4)^{2} \end{matrix}$
(Note: The discriminant of $2 x^{2} + 7 x + 4$ is $Δ = 7^{2} - 4 (2) (4) = 49 - 32 = 17$ , which is not a square, so it is irreducible over rational numbers.)

Answer

$(2 x^{2} + 7 x + 4)^{2}$

Quick Solution

Q2Part (iv)

MediumApproved by Miss Aniya • Sep 10, 2026

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

Hint

Let $y = 3 x^{2} + 5 x$ to get $(y + 3) (y + 5) - 3 = y^{2} + 8 y + 12 = (y + 6) (y + 2)$ . Back-substitute to obtain $(3 x^{2} + 5 x + 6) (3 x^{2} + 5 x + 2)$ .

Solution

Identify the common quadratic expression.

Both factors share the common expression $3 x^{2} + 5 x$ .
Substitute $y = 3 x^{2} + 5 x$ and simplify.

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

Replacing $y$ with $3 x^{2} + 5 x$ :
$\begin{matrix} (y + 6) (y + 2) & = (3 x^{2} + 5 x + 6) (3 x^{2} + 5 x + 2) \end{matrix}$
Therefore, the factored form is:
$\begin{matrix} (3 x^{2} + 5 x + 3) (3 x^{2} + 5 x + 5) - 3 & = (3 x^{2} + 5 x + 6) (3 x^{2} + 5 x + 2) \end{matrix}$

Answer

$(3 x^{2} + 5 x + 6) (3 x^{2} + 5 x + 2)$

Quick Solution

Q2Part (v)

HardApproved by Miss Aniya • Sep 8, 2026

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

Hint

Pair factors having equal constant products $1 \times 6 = 2 \times 3 = 6$ : $[(x + 1) (x + 6)] [(x + 2) (x + 3)] - 3 x^{2}$ . Group as $[(x^{2} + 6) + 7 x] [(x^{2} + 6) + 5 x] - 3 x^{2}$ , substitute $y = x^{2} + 6$ , factorize $y^{2} + 12 x y + 32 x^{2} = (y + 8 x) (y + 4 x)$ , and back-substitute to obtain $(x^{2} + 8 x + 6) (x^{2} + 4 x + 6)$ .

Solution

Pair linear factors with equal products of constant terms.

Notice the constant terms: $1 \times 6 = 6$ and $2 \times 3 = 6$ . Regroup the linear factors:
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2} & = [(x + 1) (x + 6)] \cdot [(x + 2) (x + 3)] - 3 x^{2} \end{matrix}$
Multiply each pair of factors.
$\begin{matrix} (x + 1) (x + 6) & = x^{2} + 6 x + x + 6 \\ = x^{2} + 7 x + 6 \\ = (x^{2} + 6) + 7 x \\ (x + 2) (x + 3) & = x^{2} + 3 x + 2 x + 6 \\ = x^{2} + 5 x + 6 \\ = (x^{2} + 6) + 5 x \end{matrix}$
Substitute these products into the expression:
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2} & = [(x^{2} + 6) + 7 x] [(x^{2} + 6) + 5 x] - 3 x^{2} \end{matrix}$
Substitute $y = x^{2} + 6$ and factorize.

Let $y = x^{2} + 6$ :
$\begin{matrix} (y + 7 x) (y + 5 x) - 3 x^{2} & = y^{2} + 5 x y + 7 x y + 35 x^{2} - 3 x^{2} \\ = y^{2} + 12 x y + 32 x^{2} \\ = y^{2} + 8 x y + 4 x y + 32 x^{2} \\ = y (y + 8 x) + 4 x (y + 8 x) \\ = (y + 8 x) (y + 4 x) \end{matrix}$
Back-substitute $y = x^{2} + 6$ .

Replacing $y$ with $x^{2} + 6$ :
$\begin{matrix} (x + 1) (x + 2) (x + 3) (x + 6) - 3 x^{2} & = (y + 8 x) (y + 4 x) \\ = (x^{2} + 6 + 8 x) (x^{2} + 6 + 4 x) \\ = (x^{2} + 8 x + 6) (x^{2} + 4 x + 6) \end{matrix}$
(Note: Neither $x^{2} + 8 x + 6$ nor $x^{2} + 4 x + 6$ can be factored further over rational numbers.)

Answer

$(x^{2} + 8 x + 6) (x^{2} + 4 x + 6)$

Quick Solution

Q2Part (vi)

HardUnder Review

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

Hint

Multiply the conjugate pairs: $(x^{2} - 1) (x^{2} - 4) + 13 x^{2} = x^{4} + 8 x^{2} + 4$ . Adding and subtracting $4 x^{2}$ gives $(x^{2} + 2)^{2} + (2 x)^{2}$ , a sum of squares, so it cannot be factorized. With the intended $+ 5 x^{2}$ you get $x^{4} + 4 = (x^{2} + 2)^{2} - (2 x)^{2} = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$ .

Solution

Multiply the conjugate binomial pairs.

Use $(a + b) (a - b) = a^{2} - b^{2}$ on each pair:
$\begin{matrix} (x + 1) (x - 1) & = x^{2} - 1 \\ (x + 2) (x - 2) & = x^{2} - 4 \end{matrix}$
Expand and simplify.
$\begin{matrix} (x + 1) (x - 1) (x + 2) (x - 2) + 13 x^{2} & = (x^{2} - 1) (x^{2} - 4) + 13 x^{2} \\ = x^{4} - 4 x^{2} - x^{2} + 4 + 13 x^{2} \\ = x^{4} + 8 x^{2} + 4 \end{matrix}$
Try to make a difference of two squares.

The outer terms are $(x^{2})^{2}$ and $(2)^{2}$ , so the middle term needed for a perfect square is $2 (x^{2}) (2) = 4 x^{2}$ :
$\begin{matrix} x^{4} + 8 x^{2} + 4 & = (x^{4} + 4 x^{2} + 4) + 4 x^{2} \\ = (x^{2} + 2)^{2} + (2 x)^{2} \end{matrix}$
Using $(x^{2} - 2)^{2} = x^{4} - 4 x^{2} + 4$ instead:
$\begin{matrix} x^{4} + 8 x^{2} + 4 & = (x^{2} - 2)^{2} + 12 x^{2} \end{matrix}$
Both ways give a sum of squares, not a difference, so $a^{2} - b^{2} = (a + b) (a - b)$ cannot be used. $x^{4} + 8 x^{2} + 4$ has no factors with integer coefficients.

[!NOTE] Misprint in the textbook. As printed ( $+ 13 x^{2}$ ), the expression only simplifies to $x^{4} + 8 x^{2} + 4$ and cannot be factorized. The intended term is $+ 5 x^{2}$ , which factorizes neatly:
$\begin{matrix} (x + 1) (x - 1) (x + 2) (x - 2) + 5 x^{2} & = (x^{2} - 1) (x^{2} - 4) + 5 x^{2} \\ = x^{4} - 5 x^{2} + 4 + 5 x^{2} \\ = x^{4} + 4 \\ = (x^{2})^{2} + (2)^{2} + 2 (x^{2}) (2) - 2 (x^{2}) (2) \\ = (x^{2} + 2)^{2} - 4 x^{2} \\ = (x^{2} + 2)^{2} - (2 x)^{2} \\ = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2) \end{matrix}$

Answer

As printed, it simplifies to $x^{4} + 8 x^{2} + 4$ , which cannot be factorized (textbook misprint). With the intended $+ 5 x^{2}$ : $(x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$

Quick Solution

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.

Quick Solution

Q3Part (i)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Express the first and last terms as perfect cubes $(2 x)^{3}$ and $(1)^{3}$ , verify the middle terms match $3 (2 x)^{2} (1) = 12 x^{2}$ and $3 (2 x) (1)^{2} = 6 x$ , and apply the identity $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = (a + b)^{3}$ to obtain $(2 x + 1)^{3}$ .

Solution

Identify the cube terms and base values.

Recall the algebraic identity for the cube of a binomial sum:
$(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
Express the first and last terms as perfect cubes:
$\begin{matrix} 8 x^{3} & = (2 x)^{3} ⟹ a = 2 x \\ 1 & = (1)^{3} ⟹ b = 1 \end{matrix}$
Express middle terms in standard identity form and verify.

Check that the middle terms match the identity with $a = 2 x$ and $b = 1$ :
$\begin{matrix} 3 a^{2} b & = 3 (2 x)^{2} (1) \\ = 3 (4 x^{2}) (1) \\ = 12 x^{2} \\ 3 a b^{2} & = 3 (2 x) (1)^{2} \\ = 3 (2 x) (1) \\ = 6 x \end{matrix}$
Both terms match the given expression exactly.
Apply the identity.

Rewrite the expression in standard identity form and factorize:
$\begin{matrix} 8 x^{3} + 12 x^{2} + 6 x + 1 & = (2 x)^{3} + 3 (2 x)^{2} (1) + 3 (2 x) (1)^{2} + (1)^{3} \\ = (2 x + 1)^{3} \end{matrix}$

Answer

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

Quick Solution

Q3Part (ii)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Express the first and last terms as perfect cubes $(3 a)^{3}$ and $(4 b)^{3}$ , verify $3 (3 a)^{2} (4 b) = 108 a^{2} b$ and $3 (3 a) (4 b)^{2} = 144 a b^{2}$ , and apply the identity $A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3} = (A + B)^{3}$ to obtain $(3 a + 4 b)^{3}$ .

Solution

Identify the cube terms and base values.

Recall the algebraic identity for the cube of a binomial sum:
$(A + B)^{3} = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$
Express the first and last terms as perfect cubes:
$\begin{matrix} 27 a^{3} & = (3 a)^{3} ⟹ A = 3 a \\ 64 b^{3} & = (4 b)^{3} ⟹ B = 4 b \end{matrix}$
Express middle terms in standard identity form and verify.

Check that the middle terms match the identity with $A = 3 a$ and $B = 4 b$ :
$\begin{matrix} 3 A^{2} B & = 3 (3 a)^{2} (4 b) \\ = 3 (9 a^{2}) (4 b) \\ = 108 a^{2} b \\ 3 A B^{2} & = 3 (3 a) (4 b)^{2} \\ = 3 (3 a) (16 b^{2}) \\ = 144 a b^{2} \end{matrix}$
Both terms match the given expression exactly.
Apply the identity.

Rewrite the expression in standard identity form and factorize:
$\begin{matrix} 27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3} & = (3 a)^{3} + 3 (3 a)^{2} (4 b) + 3 (3 a) (4 b)^{2} + (4 b)^{3} \\ = (3 a + 4 b)^{3} \end{matrix}$

Answer

$(3 a + 4 b)^{3}$

Quick Solution

Q3Part (iii)

MediumUnder Review

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

Hint

The book's $48 x^{2} y$ is a misprint for $18 x^{2} y$ . Write $x^{3} = (x)^{3}$ and $216 y^{3} = (6 y)^{3}$ , verify $3 (x)^{2} (6 y) = 18 x^{2} y$ and $3 (x) (6 y)^{2} = 108 x y^{2}$ , and apply $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = (a + b)^{3}$ to obtain $(x + 6 y)^{3}$ .

Solution

[!NOTE] Misprint in the textbook. The book prints $48 x^{2} y$ as the second term. With $a = x$ and $b = 6 y$ , the identity needs $3 a^{2} b = 3 (x)^{2} (6 y) = 18 x^{2} y$ , so the expression as printed is not a perfect cube and has no simple factors. The intended term is $18 x^{2} y$ ; the solution below factorizes $x^{3} + 18 x^{2} y + 108 x y^{2} + 216 y^{3}$ .

Identify the cube terms and base values.

Recall the algebraic identity for the cube of a binomial sum:
$(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
Express the first and last terms as perfect cubes:
$\begin{matrix} x^{3} & = (x)^{3} ⟹ a = x \\ 216 y^{3} & = (6 y)^{3} ⟹ b = 6 y \end{matrix}$
Express middle terms in standard identity form and verify.

Check that the middle terms match the identity with $a = x$ and $b = 6 y$ :
$\begin{matrix} 3 a^{2} b & = 3 (x)^{2} (6 y) \\ = 18 x^{2} y \\ 3 a b^{2} & = 3 (x) (6 y)^{2} \\ = 3 (x) (36 y^{2}) \\ = 108 x y^{2} \end{matrix}$
Both terms match the corrected expression exactly.
Apply the identity.

Rewrite the expression in standard identity form and factorize:
$\begin{matrix} x^{3} + 18 x^{2} y + 108 x y^{2} + 216 y^{3} & = (x)^{3} + 3 (x)^{2} (6 y) + 3 (x) (6 y)^{2} + (6 y)^{3} \\ = (x + 6 y)^{3} \end{matrix}$

Answer

$(x + 6 y)^{3}$ (with the intended term $18 x^{2} y$ ; as printed with $48 x^{2} y$ it is not a perfect cube)

Quick Solution

Q3Part (iv)

MediumApproved by Miss Aniya • Sep 8, 2026

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

Hint

Rearrange into standard descending order $8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3}$ , identify $a = 2 x$ and $b = 5 y$ , and apply the identity $a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} = (a - b)^{3}$ to obtain $(2 x - 5 y)^{3}$ .

Solution

Rearrange terms in standard descending order.

Rearrange the terms according to descending powers of $x$ :
$\begin{matrix} 8 x^{3} - 125 y^{3} + 150 x y^{2} - 60 x^{2} y & = 8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3} \end{matrix}$
Identify the cube terms and base values.

Recall the algebraic identity for the cube of a binomial difference:
$(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$
Express the first and last terms as perfect cubes:
$\begin{matrix} 8 x^{3} & = (2 x)^{3} ⟹ a = 2 x \\ 125 y^{3} & = (5 y)^{3} ⟹ b = 5 y \end{matrix}$
Express middle terms in standard identity form and verify.

Check the middle terms with $a = 2 x$ and $b = 5 y$ :
$\begin{matrix} 3 a^{2} b & = 3 (2 x)^{2} (5 y) \\ = 3 (4 x^{2}) (5 y) \\ = 60 x^{2} y \\ 3 a b^{2} & = 3 (2 x) (5 y)^{2} \\ = 3 (2 x) (25 y^{2}) \\ = 150 x y^{2} \end{matrix}$
Substituting these into $a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ yields $8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3}$ , which matches the rearranged expression exactly.
Apply the identity.

Rewrite the expression in standard identity form and factorize:
$\begin{matrix} 8 x^{3} - 60 x^{2} y + 150 x y^{2} - 125 y^{3} & = (2 x)^{3} - 3 (2 x)^{2} (5 y) + 3 (2 x) (5 y)^{2} - (5 y)^{3} \\ = (2 x - 5 y)^{3} \end{matrix}$

Answer

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

Quick Solution

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.

Quick Solution

Q4Part (i)

EasyApproved by Miss Aniya • Sep 8, 2026

$125 a^{3} - 1$

Hint

Express the expression as $(5 a)^{3} - (1)^{3}$ and apply the difference of cubes formula $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ to obtain $(5 a - 1) (25 a^{2} + 5 a + 1)$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the difference of two cubes:
a3−b3=(a−b)(a2+ab+b2)
Express each term as a perfect cube:
- First term: $125 a^{3} = (5 a)^{3}$
- Second term: $1 = (1)^{3}$
Here, the base values are $a = 5 a$ and $b = 1$ .
Apply the difference of cubes formula.

Substitute the base values $a = 5 a$ and $b = 1$ into the identity:
$\begin{matrix} 125 a^{3} - 1 & = (5 a)^{3} - (1)^{3} \\ = (5 a - 1) [(5 a)^{2} + (5 a) (1) + (1)^{2}] \\ = (5 a - 1) (25 a^{2} + 5 a + 1) \end{matrix}$

Answer

$(5 a - 1) (25 a^{2} + 5 a + 1)$

Quick Solution

Q4Part (ii)

EasyApproved by Miss Aniya • Sep 8, 2026

$64 x^{3} + 125$

Hint

Express as $(4 x)^{3} + (5)^{3}$ and apply the sum of cubes formula $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ to obtain $(4 x + 5) (16 x^{2} - 20 x + 25)$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the sum of two cubes:
a3+b3=(a+b)(a2−ab+b2)
Express each term as a perfect cube:
- First term: $64 x^{3} = (4 x)^{3}$
- Second term: $125 = (5)^{3}$
Here, the base values are $a = 4 x$ and $b = 5$ .
Apply the sum of cubes formula.

Substitute the base values $a = 4 x$ and $b = 5$ into the identity:
$\begin{matrix} 64 x^{3} + 125 & = (4 x)^{3} + (5)^{3} \\ = (4 x + 5) [(4 x)^{2} - (4 x) (5) + (5)^{2}] \\ = (4 x + 5) (16 x^{2} - 20 x + 25) \end{matrix}$

Answer

$(4 x + 5) (16 x^{2} - 20 x + 25)$

Quick Solution

Q4Part (iii)

MediumApproved by Miss Aniya • Sep 8, 2026

$x^{6} - 27$

Hint

Express the first term as $(x^{2})^{3}$ and the second term as $(3)^{3}$ , then use the formula $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ to factorize into $(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the difference of two cubes:
a3−b3=(a−b)(a2+ab+b2)
Express each term as a perfect cube:
- First term: $x^{6} = (x^{2})^{3}$
- Second term: $27 = (3)^{3}$
Here, the base values are $a = x^{2}$ and $b = 3$ .
Apply the difference of cubes formula.

Substitute the base values $a = x^{2}$ and $b = 3$ into the identity:
$\begin{matrix} x^{6} - 27 & = (x^{2})^{3} - (3)^{3} \\ = (x^{2} - 3) [(x^{2})^{2} + (x^{2}) (3) + (3)^{2}] \\ = (x^{2} - 3) (x^{4} + 3 x^{2} + 9) \end{matrix}$

Answer

$(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$

Quick Solution

Q4Part (iv)

EasyApproved by Miss Aniya • Sep 8, 2026

$1000 a^{3} + 1$

Hint

Write $1000 a^{3}$ as $(10 a)^{3}$ and $1$ as $(1)^{3}$ , then apply $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ to get $(10 a + 1) (100 a^{2} - 10 a + 1)$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the sum of two cubes:
a3+b3=(a+b)(a2−ab+b2)
Express each term as a perfect cube:
- First term: $1000 a^{3} = (10 a)^{3}$
- Second term: $1 = (1)^{3}$
Here, the base values are $a = 10 a$ and $b = 1$ .
Apply the sum of cubes formula.

Substitute the base values $a = 10 a$ and $b = 1$ into the identity:
$\begin{matrix} 1000 a^{3} + 1 & = (10 a)^{3} + (1)^{3} \\ = (10 a + 1) [(10 a)^{2} - (10 a) (1) + (1)^{2}] \\ = (10 a + 1) (100 a^{2} - 10 a + 1) \end{matrix}$

Answer

$(10 a + 1) (100 a^{2} - 10 a + 1)$

Quick Solution

Q4Part (v)

MediumApproved by Miss Aniya • Sep 8, 2026

$343 x^{3} + 216$

Hint

Notice that $343 = 7^{3}$ and $216 = 6^{3}$ . Use the sum of cubes formula $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ to get $(7 x + 6) (49 x^{2} - 42 x + 36)$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the sum of two cubes:
a3+b3=(a+b)(a2−ab+b2)
Express each term as a perfect cube:
- First term: $343 x^{3} = (7 x)^{3}$ since $7^{3} = 343$
- Second term: $216 = (6)^{3}$ since $6^{3} = 216$
Here, the base values are $a = 7 x$ and $b = 6$ .
Apply the sum of cubes formula.

Substitute the base values $a = 7 x$ and $b = 6$ into the identity:
$\begin{matrix} 343 x^{3} + 216 & = (7 x)^{3} + (6)^{3} \\ = (7 x + 6) [(7 x)^{2} - (7 x) (6) + (6)^{2}] \\ = (7 x + 6) (49 x^{2} - 42 x + 36) \end{matrix}$

Answer

$(7 x + 6) (49 x^{2} - 42 x + 36)$

Quick Solution

Q4Part (vi)

MediumApproved by Miss Aniya • Sep 8, 2026

$27 - 512 y^{3}$

Hint

Notice that $27 = 3^{3}$ and $512 = 8^{3}$ . Use the difference of cubes identity $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ to obtain $(3 - 8 y) (9 + 24 y + 64 y^{2})$ .

Solution

Identify the terms as perfect cubes.

Recall the formula for the difference of two cubes:
a3−b3=(a−b)(a2+ab+b2)
Express each term as a perfect cube:
- First term: $27 = (3)^{3}$
- Second term: $512 y^{3} = (8 y)^{3}$ since $8^{3} = 512$
Here, the base values are $a = 3$ and $b = 8 y$ .
Apply the difference of cubes formula.

Substitute the base values $a = 3$ and $b = 8 y$ into the identity:
$\begin{matrix} 27 - 512 y^{3} & = (3)^{3} - (8 y)^{3} \\ = (3 - 8 y) [(3)^{2} + (3) (8 y) + (8 y)^{2}] \\ = (3 - 8 y) (9 + 24 y + 64 y^{2}) \end{matrix}$

Answer

$(3 - 8 y) (9 + 24 y + 64 y^{2})$

Quick Solution

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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