NCERT Solutions for Class 8 Maths Chapter 12: Factorisation — Free PDF

Factorisation is the reverse of expansion. While Chapter 8 taught you how to multiply and expand algebraic expressions, Chapter 12 teaches you how to break an expression into a product of simpler factors.

You will learn four methods of factorisation: taking out common factors, regrouping, using algebraic identities, and dividing polynomials. These skills are critical for simplifying expressions and solving equations in Classes 9 and 10.

This chapter has three exercises. Exercise 12.1 covers factorisation by common factors, regrouping, and identities. Exercise 12.2 focuses on division of algebraic expressions (monomial by monomial, polynomial by monomial, and polynomial by polynomial). Exercise 12.3 contains error-finding questions where you identify and correct mistakes in simplification — these are excellent for building deep understanding.

Factor: A factor of an algebraic expression is an expression that divides it exactly. For example, $2x$ and $(x + 3)$ are factors of $2x(x + 3) = 2x^2 + 6x$.

Factorisation: The process of writing an algebraic expression as a product of its factors. It is the reverse of expansion (multiplication of brackets).

Irreducible Factors: A factor that cannot be broken down further. For example, in $6x^2y = 2 \times 3 \times x \times x \times y$, the irreducible factors are $2$, $3$, $x$, $x$, and $y$.

HCF (Highest Common Factor): The largest factor common to all terms. This is key to the common factor method. For $12x^3y$ and $8x^2y^2$, the HCF is $4x^2y$.

Algebraic Identities used in factorisation:
- $a^2 + 2ab + b^2 = (a + b)^2$ — perfect square (sum)
- $a^2 - 2ab + b^2 = (a - b)^2$ — perfect square (difference)
- $a^2 - b^2 = (a + b)(a - b)$ — difference of squares
- $x^2 + (a + b)x + ab = (x + a)(x + b)$ — splitting the middle term

Recognising which identity applies is the most important skill in this chapter. With practice, the patterns become second nature.

Method 1: Common Factor Method

Find the HCF of all terms and take it out as a common factor.

Steps: (i) Find the HCF of all terms. (ii) Divide each term by the HCF. (iii) Write as HCF $\times$ (remaining expression).

Another example: $15a^3b^2 - 25a^2b^3 + 10ab = 5ab(3a^2b - 5ab^2 + 2)$.

Method 2: Regrouping

When no single factor is common to all terms, group terms that share a common factor, then factor again.

The key is choosing the right grouping. Sometimes the first grouping you try does not work, and you need to rearrange the terms.

Method 3: Using Identities (in Reverse)

Recognise the pattern and apply the appropriate identity.
- $x^2 + 6x + 9 = x^2 + 2(x)(3) + 3^2 = (x + 3)^2$
- $25m^2 - 49 = (5m)^2 - 7^2 = (5m + 7)(5m - 7)$

Method 4: Division of Polynomials

When dividing one polynomial by another, first factorise both, then cancel common factors.
- Monomial $\div$ Monomial: Divide coefficients, subtract exponents.
- Polynomial $\div$ Monomial: Divide each term separately.
- Polynomial $\div$ Polynomial: Factor the dividend, then cancel.

**Q1. Factorise: $15x^3y^2 + 10x^2y^3$.**

Solution:

Step 1: Find HCF. HCF of $15$ and $10$ is $5$. Common variable factors: $x^2y^2$.
HCF $= 5x^2y^2$.

Step 2: Divide each term by the HCF.
$\dfrac{15x^3y^2}{5x^2y^2} = 3x$ and $\dfrac{10x^2y^3}{5x^2y^2} = 2y$.

Step 3: Write the result.

---

**Q2. Factorise by regrouping: $x^2 + xy + 8x + 8y$.**

Solution:

Group the first two and last two terms:

Verification: $(x + 8)(x + y) = x^2 + xy + 8x + 8y$. Correct.

---

**Q3. Factorise: $x^2 - 16$.**

Solution:

This is a difference of squares: $a^2 - b^2 = (a+b)(a-b)$ with $a = x, b = 4$:

---

**Q4. Factorise: $4x^2 + 12x + 9$.**

Solution:

Check if this is a perfect square trinomial.
$4x^2 = (2x)^2$, $9 = 3^2$, and $12x = 2 \times 2x \times 3$.

Yes! This matches $a^2 + 2ab + b^2 = (a+b)^2$ with $a = 2x, b = 3$:

---

**Q5. Factorise: $x^2 + 5x + 6$.**

Solution:

We need two numbers that add to $5$ and multiply to $6$. Those numbers are $2$ and $3$.

---

**Q6. Factorise: $9p^2 - 24pq + 16q^2$.**

Solution:

$9p^2 = (3p)^2$, $16q^2 = (4q)^2$, $24pq = 2 \times 3p \times 4q$.

This matches $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3p, b = 4q$:

**Q1. Divide: $14x^3y^2 \div 2x^2y$.**

Solution:

Divide the coefficients and subtract exponents:

---

**Q2. Divide: $6x^3 + 9x^2 + 3x$ by $3x$.**

Solution:

Divide each term of the polynomial by the monomial:

---

**Q3. Divide $x^2 + 7x + 12$ by $(x + 3)$.**

Solution:

First factorise the numerator:

Now divide:

---

**Q4. Divide $a^2 - b^2$ by $(a + b)$.**

Solution:

---

**Q5. Find the error: $\dfrac{3x + 1}{3} = x + 1$.**

Solution:

This is incorrect. You cannot cancel the $3$ with only the coefficient of $x$. Every term in the numerator must be divided by the denominator:

The mistake is treating the $3$ as if it were a factor of the entire numerator, but $3$ is not a factor of $(3x + 1)$.

---

**Q6. Find the error: $\dfrac{4x + 5}{4x} = 5$.**

Solution:

This is incorrect. You cannot cancel $4x$ with the $4x$ in the numerator because $4x$ is a term, not a factor of the entire numerator.

Correct simplification: $\dfrac{4x + 5}{4x} = \dfrac{4x}{4x} + \dfrac{5}{4x} = 1 + \dfrac{5}{4x}$.

Alternatively, since $(4x + 5)$ cannot be factored further, the expression is already in simplest form as a fraction.

**Example 1. Factorise: $12m^2n - 18mn^2 + 24mn$.**

Solution:

HCF of $12, 18, 24$ is $6$. Common variables: $mn$. HCF $= 6mn$.

Verification: $6mn \times 2m = 12m^2n$, $6mn \times (-3n) = -18mn^2$, $6mn \times 4 = 24mn$. Correct.

---

**Example 2. Factorise: $x^2 - 11x + 30$.**

Solution:

We need two numbers that multiply to $30$ and add to $-11$. Those are $-5$ and $-6$.

---

**Example 3. Factorise: $49a^2 - 64b^2$.**

Solution:

$(7a)^2 - (8b)^2 = (7a + 8b)(7a - 8b)$.

---

**Example 4. Factorise by regrouping: $pq - pr + sq - sr$.**

Solution:

Group: $(pq - pr) + (sq - sr) = p(q - r) + s(q - r) = (p + s)(q - r)$.

---

**Example 5. Divide: $(x^2 - 9x + 20) \div (x - 5)$.**

Solution:

Factorise: $x^2 - 9x + 20 = (x - 4)(x - 5)$.

Mistake 1: Cancelling terms instead of factors.
This is the single most important error to avoid. You can only cancel complete factors, never individual terms.

Wrong: $\dfrac{x^2 + 3x}{x} = x + 3x = 4x$.
Correct: $\dfrac{x^2 + 3x}{x} = \dfrac{x(x + 3)}{x} = x + 3$.

Wrong: $\dfrac{5x + 10}{5} = x + 10$.
Correct: $\dfrac{5x + 10}{5} = \dfrac{5(x + 2)}{5} = x + 2$.

**Mistake 2: Sign errors with $(a - b)^2$.**
$(a - b)^2 = a^2 - 2ab + b^2$, NOT $a^2 - b^2$. The middle term $-2ab$ is crucial.

Mistake 3: Incomplete factorisation.
Always check if your factors can be factored further. For example, $2(x^2 - 4)$ is not fully factorised — it should be $2(x + 2)(x - 2)$.

Mistake 4: Wrong identification of the identity.
$x^2 + 9$ is NOT $(x + 3)^2$. The identity $(a + b)^2$ requires a middle term: $(x + 3)^2 = x^2 + 6x + 9$. The expression $x^2 + 9$ cannot be factorised using real numbers at this level.

Mistake 5: Errors in finding the splitting numbers.
When factorising $x^2 + bx + c$, you need two numbers that ADD to $b$ and MULTIPLY to $c$. Getting these mixed up (adding to $c$, multiplying to $b$) leads to wrong factors.

Here is a decision flowchart for choosing the right factorisation method:

Step 1: Is there a common factor in all terms?
If yes, take it out first. Example: $6x^2 + 12x = 6x(x + 2)$.

Step 2: How many terms are there after taking out common factors?
- 2 terms: Check for difference of squares ($a^2 - b^2$). Example: $x^2 - 25 = (x+5)(x-5)$. Note: sum of squares ($a^2 + b^2$) cannot be factorised.
- 3 terms (trinomial): Check for perfect square ($a^2 \pm 2ab + b^2$). If not, try splitting the middle term. Example: $x^2 + 7x + 12 = (x+3)(x+4)$.
- 4 terms: Try regrouping. Example: $ax + ay + bx + by = (a+b)(x+y)$.

**Step 3: For splitting the middle term of $x^2 + bx + c$:**
Find two numbers $p$ and $q$ such that $p + q = b$ and $p \times q = c$.
- If $c > 0$ and $b > 0$: both numbers are positive.
- If $c > 0$ and $b < 0$: both numbers are negative.
- If $c < 0$: one number is positive and the other is negative.

Q1. Factorise: $8a^3b - 12a^2b^2 + 4ab$.

Answer: HCF $= 4ab$. $8a^3b - 12a^2b^2 + 4ab = 4ab(2a^2 - 3ab + 1)$.

---

Q2. Factorise: $x^2 - 81$.

Answer: $(x + 9)(x - 9)$.

---

Q3. Factorise: $x^2 + 14x + 49$.

Answer: $(x + 7)^2$. (Since $49 = 7^2$ and $14x = 2 \times x \times 7$.)

---

Q4. Factorise: $x^2 - 3x - 10$.

Answer: We need numbers that multiply to $-10$ and add to $-3$: those are $-5$ and $2$.
$x^2 - 5x + 2x - 10 = x(x - 5) + 2(x - 5) = (x + 2)(x - 5)$.

---

Q5. Divide: $(x^2 + 5x + 6) \div (x + 2)$.

Answer: $x^2 + 5x + 6 = (x + 2)(x + 3)$. So the answer is $(x + 3)$.

---

Q6. Find and correct the error: $(x + 2)^2 = x^2 + 4$.

Answer: Wrong! The correct expansion is $(x + 2)^2 = x^2 + 4x + 4$. The middle term $4x$ was missing.

• Factorisation is the reverse of expansion: expressing an expression as a product of factors.
  - Four methods: common factors, regrouping, identities, and division.
  - The algebraic identities from Chapter 8 are used in reverse here.
  - When dividing polynomials, factorise first and then cancel common factors.
  - Never cancel parts of terms — only complete factors can be cancelled.
  - Always check your work by expanding back to the original expression.
  - Factorisation is a foundation skill for solving quadratic equations in Class 10 and beyond.

Practise factorisation on SparkEd for adaptive questions that build your skill step by step!