CBSE Class 8 Mathematics Factorisation Practice Paper Worksheet Set 04 with Answers (Old NCERT Syllabus)

This Class 8 Maths Factorisation practice paper carries 40 marks and has 17 questions in five sections (A to E), with a worked step-by-step answer for every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it is useful for state-board students and for revising this chapter. The paper tests taking out common factors, factorising by regrouping, splitting the middle term of x² + px + q, using identities such as a² − b² and perfect squares, and dividing a polynomial by a factor.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Factorise: 6ab − b² + 12ac − 2bc
Answer:
Rearrange the terms: 6ab + 12ac − b² − 2bc.
= 6a(b + 2c) − b(b + 2c).
= (b + 2c)(6a − b).
In simple words: Put together the terms with 6a and the terms with b; both groups share (b + 2c).

Teacher's Note:
1. Look for pairs of terms that leave the same bracket.
2. Check: (b + 2c)(6a − b) = 6ab − b² + 12ac − 2bc.

Question 8. Factorise: ab(x² + y²) − xy(a² + b²)
Answer:
Expand: abx² + aby² − a²xy − b²xy.
Regroup: (abx² − a²xy) + (aby² − b²xy).
= ax(bx − ay) + by(ay − bx) = ax(bx − ay) − by(bx − ay).
= (bx − ay)(ax − by).
In simple words: Open the brackets, pair terms with common factors ax and by, and take out (bx − ay).

Teacher's Note:
1. (ay − bx) = −(bx − ay); changing the sign this way makes the brackets match.
2. Check by expanding: (bx − ay)(ax − by) = abx² − b²xy − a²xy + aby².

Question 9. Factorise: 100 − (x − 5)²
Answer:
Write 100 = 10², so the expression is 10² − (x − 5)².
Use a² − b² = (a + b)(a − b): {10 + (x − 5)}{10 − (x − 5)}.
= (10 + x − 5)(10 − x + 5).
= (x + 5)(15 − x).
In simple words: It is a difference of two squares, 10² minus (x − 5)², which splits into (x + 5)(15 − x).

Teacher's Note:
1. Keep (x − 5) in a bracket when subtracting so that −(−5) becomes +5.
2. Check with x = 5: 100 − 0 = 100 and (10)(10) = 100.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Factorise: (i) y² + 10y + 24 (ii) x² + 5x + 6
Answer:
(i) Numbers with product 24 and sum 10: 6 and 4.
y² + 10y + 24 = y² + 6y + 4y + 24 = y(y + 6) + 4(y + 6) = (y + 6)(y + 4).
(ii) Numbers with product 6 and sum 5: 3 and 2.
x² + 5x + 6 = x² + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2).
In simple words: Split the middle term using two numbers that multiply to the constant and add to the middle coefficient.

Teacher's Note:
1. List factor pairs of the constant and pick the pair with the right sum.
2. Both signs are + here, so both numbers are positive.

Question 11. Factorise: (i) p² + 6p − 16 (ii) x² − 10x + 24
Answer:
(i) Numbers with product −16 and sum 6: 8 and −2.
p² + 6p − 16 = p² + 8p − 2p − 16 = p(p + 8) − 2(p + 8) = (p + 8)(p − 2).
(ii) Numbers with product 24 and sum −10: −6 and −4.
x² − 10x + 24 = x² − 6x − 4x + 24 = x(x − 6) − 4(x − 6) = (x − 6)(x − 4).
In simple words: Find the right pair of numbers for each expression, split the middle term, and group.

Teacher's Note:
1. A negative constant means one number is positive and one negative.
2. Check (i): 8 × (−2) = −16 and 8 + (−2) = 6.

Question 12. Factorise: (i) 75a³b² − 108ab⁴ (ii) 256x³ − 81x
Answer:
(i) Common factor 3ab²: 75a³b² − 108ab⁴ = 3ab²(25a² − 36b²).
= 3ab²[(5a)² − (6b)²] = 3ab²(5a + 6b)(5a − 6b).
(ii) Common factor x: 256x³ − 81x = x(256x² − 81).
= x[(16x)² − 9²] = x(16x + 9)(16x − 9).
In simple words: Take out the common factor first, and what remains is a difference of two squares.

Teacher's Note:
1. HCF of 75 and 108 is 3; the lowest powers are a and b².
2. Use a² − b² = (a + b)(a − b) after taking out the common factor.

Question 13. Factorise: 49(a − b)² − 25(a + b)²
Answer:
Write it as [7(a − b)]² − [5(a + b)]².
Use x² − y² = (x + y)(x − y): [7(a − b) + 5(a + b)][7(a − b) − 5(a + b)].
= (7a − 7b + 5a + 5b)(7a − 7b − 5a − 5b) = (12a − 2b)(2a − 12b).
= 2(6a − b) × 2(a − 6b) = 4(6a − b)(a − 6b).
In simple words: Both parts are perfect squares, so use the difference of squares, simplify the brackets and take out the 2s.

Teacher's Note:
1. Open the brackets carefully: −5(a + b) = −5a − 5b.
2. Take out the common factor 2 from each bracket at the end.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Factorise the expressions and divide them as directed.
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² − 14m − 32) ÷ (m + 2)
Answer:
(i) y² + 7y + 10 = y² + 2y + 5y + 10 = y(y + 2) + 5(y + 2) = (y + 2)(y + 5).
(y² + 7y + 10) ÷ (y + 5) = \( \frac{(y + 2)(y + 5)}{y + 5} \) = y + 2.
(ii) m² − 14m − 32 = m² + 2m − 16m − 32 = m(m + 2) − 16(m + 2) = (m − 16)(m + 2).
(m² − 14m − 32) ÷ (m + 2) = \( \frac{(m - 16)(m + 2)}{m + 2} \) = m − 16.
In simple words: Factorise the top, then cancel the bracket that matches the divisor.

Teacher's Note:
1. The divisor is a hint: one factor of the expression must be that bracket.
2. Check (ii): (m − 16)(m + 2) = m² − 14m − 32.

Question 15. Factorise. (i) a⁴ − b⁴ (ii) p⁴ − 81
Answer:
(i) a⁴ − b⁴ = (a²)² − (b²)² = (a² − b²)(a² + b²).
a² − b² = (a − b)(a + b), so a⁴ − b⁴ = (a − b)(a + b)(a² + b²).
(ii) p⁴ − 81 = (p²)² − 9² = (p² − 9)(p² + 9).
p² − 9 = p² − 3² = (p − 3)(p + 3), so p⁴ − 81 = (p − 3)(p + 3)(p² + 9).
In simple words: Use the difference of squares twice; the sum of squares bracket cannot be split further.

Teacher's Note:
1. After the first step, check whether any bracket is again a difference of squares.
2. a² + b² and p² + 9 do not factorise further using these identities.

Question 16. Factorise the following expressions.
(i) a² + 8a + 16 (ii) p² − 10p + 25 (iii) 25m² + 30m + 9 (iv) 49y² + 84yz + 36z²
Answer:
(i) a² + 8a + 16 = a² + 2 × a × 4 + 4² = (a + 4)², using (x + y)² = x² + 2xy + y².
(ii) p² − 10p + 25 = p² − 2 × p × 5 + 5² = (p − 5)², using (x − y)² = x² − 2xy + y².
(iii) 25m² + 30m + 9 = (5m)² + 2 × 5m × 3 + 3² = (5m + 3)².
(iv) 49y² + 84yz + 36z² = (7y)² + 2 × 7y × 6z + (6z)² = (7y + 6z)².
In simple words: Each one is a perfect square: spot the two square terms and check that the middle term is twice their product.

Teacher's Note:
1. Test: is the middle term exactly 2 × (first root) × (second root)?
2. The sign of the middle term gives the sign inside the bracket.

Section E: Case Study Based Question (4 Marks)

Question 17. One day in Class VIII-B, the teacher was explaining factorisation by regrouping terms and factorisation using algebraic identities. In some algebraic expressions, not every term has a common factor. To factorise such expressions, terms having common factors are grouped together. Algebraic identities can also be used for factorisation.
Answer the following questions based on the above information:
(a) Factorise: 23xy − 46x + 54y − 108 (1)
(b) Find the factors of x² − 4. (1)
(c) Factorise: xa² + xb² − ya² − yb² (1)
(d) Factorise: (x + 2y)² − 4(2x − y)² (1)
Answer:
(a) 23xy − 46x + 54y − 108 = 23x(y − 2) + 54(y − 2) = (23x + 54)(y − 2).
(b) x² − 4 = x² − 2² = (x + 2)(x − 2).
(c) xa² + xb² − ya² − yb² = x(a² + b²) − y(a² + b²) = (x − y)(a² + b²).
(d) 4(2x − y)² = [2(2x − y)]² = (4x − 2y)². So the expression is (x + 2y)² − (4x − 2y)².
= [(x + 2y) + (4x − 2y)][(x + 2y) − (4x − 2y)] = (5x)(4y − 3x) = 5x(4y − 3x).
In simple words: Group terms in (a) and (c) to get a common bracket; (b) and (d) are differences of two squares.

Teacher's Note:
1. In (d), bring the 4 inside the square as 2² before using a² − b².
2. Check (a): 46 = 2 × 23 and 108 = 2 × 54, so (y − 2) is common.