Worksheet Solutions: Remainder and Factor Theorems | Mathematics Class 10 ICSE PDF Download

Very Short Questions 

Q1. Find the remainder when f(x) = x² + 3x – 5 is divided by x – 2.
Solution:
f(2) = (2)² + 3(2) – 5 = 4 + 6 – 5 = 5.
Answer: Remainder = 5

Q2. If f(x) = 2x³ – 7x + 4, find the remainder when divided by x + 1.
Solution:
x + 1 = 0 → x = –1.
f(–1) = 2(–1)³ – 7(–1) + 4 = –2 + 7 + 4 = 9.
Answer: Remainder = 9

Q3. Find the value of k if x – 3 is a factor of f(x) = x² + kx – 21.
Solution:
If x – 3 is a factor, f(3) = 0.
f(3) = (3)² + k(3) – 21 = 9 + 3k – 21 = 3k – 12.
3k – 12 = 0 → k = 4.
Answer: k = 4

Q4. Check whether x – 2 is a factor of f(x) = x³ – 3x² – 4x + 12.
Solution:
f(2) = (2)³ – 3(2)² – 4(2) + 12 = 8 – 12 – 8 + 12 = 0.
Since f(2) = 0, x – 2 is a factor.

Q5. Find the remainder when f(x) = 5x³ – 2x² + 7 is divided by x – 1.
Solution:
f(1) = 5(1)³ – 2(1)² + 7 = 5 – 2 + 7 = 10.
Answer: Remainder = 10

Short Answer Questions 

Q6. If f(x) = x³ + 2x² – kx – 10 and f(2) = 0, find the value of k.
Solution:
f(2) = (2)³ + 2(2)² – k(2) – 10
= 8 + 8 – 2k – 10 = 6 – 2k.
f(2) = 0 → 6 – 2k = 0 → k = 3.
Answer: k = 3

Q7. If f(x) = ax³ – 2x² + 3 and f(1) = 5, find a.
Solution:
f(1) = a(1)³ – 2(1)² + 3 = a – 2 + 3 = a + 1.
Given f(1) = 5 → a + 1 = 5 → a = 4.
Answer: a = 4

Q8. Check if x + 2 is a factor of f(x) = 2x³ + 5x² – 4x – 20.
Solution:
x + 2 = 0 → x = –2.
f(–2) = 2(–2)³ + 5(–2)² – 4(–2) – 20
= 2(–8) + 5(4) + 8 – 20
= –16 + 20 + 8 – 20 = –8.
Since f(–2) ≠ 0, x + 2 is not a factor.

Q9. Find the remainder when f(x) = 3x³ – x² + 2 is divided by 3x – 1.
Solution:
3x – 1 = 0 → x = 1/3.
f(1/3) = 3(1/27) – (1/9) + 2
= 1/9 – 1/9 + 2 = 2.
Answer: Remainder = 2

Q10. Find the value of m if (x – 1) is a factor of f(x) = x³ + mx² – 4x – 4.
Solution:
f(1) = (1)³ + m(1)² – 4(1) – 4
= 1 + m – 4 – 4 = m – 7.
Since x – 1 is a factor, f(1) = 0 → m – 7 = 0 → m = 7.
Answer: m = 7

Long Answer Questions

Q11. Show that x − 3 is a factor of f(x) = 2x³ − 3x² − 11x + 6 and factorise completely.

Solution (stepwise):

1. Check factor by remainder theorem.
   Substitute x = 3 into f(x):
   f(3) = 2(3)³ − 3(3)² − 11(3) + 6
   = 2(27) − 3(9) − 33 + 6
   = 54 − 27 − 33 + 6
   = 0.
   Since f(3) = 0, (x − 3) is a factor.

2. Divide polynomial by (x − 3).
   Do synthetic or long division:
   (2x³ − 3x² − 11x + 6) ÷ (x − 3) = 2x² + 3x − 2 (quotient).

3. Factorise the quadratic quotient.
   2x² + 3x − 2 = (2x − 1)(x + 2).

4. Write full factorisation.
   f(x) = (x − 3)(2x − 1)(x + 2).

Q12. If x – 4 is a factor of f(x) = x³ – px² + qx – 24 and f(2) = 0, find p and q.
Solution:
Step 1: Since x – 4 is a factor → f(4) = 0.
f(4) = (4)³ – p(16) + q(4) – 24 = 64 – 16p + 4q – 24 = 40 – 16p + 4q.
So, 40 – 16p + 4q = 0 …(i).

Step 2: f(2) = 0.
f(2) = (2)³ – p(4) + q(2) – 24 = 8 – 4p + 2q – 24 = –16 – 4p + 2q.
So, –16 – 4p + 2q = 0 → –4p + 2q = 16 …(ii).

Step 3: Solve equations.
From (ii): 2q = 4p + 16 → q = 2p + 8.
Substitute in (i): 40 – 16p + 4(2p + 8) = 0
40 – 16p + 8p + 32 = 0 → –8p + 72 = 0 → p = 9.
Then q = 2(9) + 8 = 26.

Answer: p = 9, q = 26

Q13. Factorise completely: f(x) = 2x³ + x² – 13x + 6.
Solution:
Step 1: Possible roots = ±1, ±2, ±3, ±6.
Try x = 2: f(2) = 2(8) + 4 – 26 + 6 = 0.
So, x – 2 is a factor.

Step 2: Divide by (x – 2).
(2x³ + x² – 13x + 6) ÷ (x – 2) = 2x² + 5x – 3.

Step 3: Factorise 2x² + 5x – 3.
= (2x – 1)(x + 3).

Final Answer:
2x³ + x² – 13x + 6 = (x – 2)(2x – 1)(x + 3)