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About Polynomials — Class 9 Olympiad

Factor polynomials, apply remainder and factor theorems, use algebraic identities in competition problems. This topic is part of the Olympiad Class 9 mathematics syllabus (chapter: Module 2). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Polynomials — solved examples for Class 9 Olympiad

Example 1easy

Which of the following expressions is NOT a polynomial?
  1. A)A) 3x² - √2x + 5
  2. B)B) y³ + 4y
  3. C)C) (x+1)(x-2)
  4. D)D) x + 7√x

Step-by-step solution

  1. A polynomial is an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
  2. Option A, B, and C all have variables with non-negative integer exponents (x², x¹, y³, y¹, x² from expansion).
  3. Option D contains √x, which can be written as x^(1/2). The exponent 1/2 is not a non-negative integer, so it is not a polynomial.

Answer: D) x + 7√x

Example 2medium

A polynomial P(x) leaves a remainder 5 when divided by (x-2) and a remainder 3 when divided by (x-3). What is the remainder when P(x) is divided by (x-2)(x-3)?
  1. A)A. 2x+1
  2. B)B. -2x+9
  3. C)C. x+3
  4. D)D. 3x-1

Step-by-step solution

  1. By the Remainder Theorem, P(2) = 5 and P(3) = 3.
  2. When P(x) is divided by (x-2)(x-3), the remainder is a linear polynomial, say Ax+B. So, P(x) = Q(x)(x-2)(x-3) + (Ax+B).
  3. Substitute x=2: P(2) = A(2)+B ⇒ 2A+B = 5.
  4. Substitute x=3: P(3) = A(3)+B ⇒ 3A+B = 3.
  5. Subtracting the first equation from the second: (3A+B) - (2A+B) = 3 - 5 ⇒ A = -2.
  6. Substitute A=-2 into 2A+B=5: 2(-2)+B=5 ⇒ -4+B=5 ⇒ B=9.
  7. The remainder is -2x+9.

Answer: B. -2x+9

Example 3hard

A polynomial P(x) = x³ + ax² + bx + c has factors (x-1) and (x-2). When P(x) is divided by (x+1), the remainder is 6. Find the value of a+b+c.
  1. A)-2
  2. B)0
  3. C)1
  4. D)-1

Step-by-step solution

  1. Since (x-1) and (x-2) are factors, P(1)=0 and P(2)=0. Applying these conditions:
    P(1)=1+a+b+c=0=>a+b+c=−1(Eq1)P(2)=8+4a+2b+c=0=>4a+2b+c=−8(Eq2)P(1) = 1 + a + b + c = 0 => a + b + c = -1 (Eq 1) P(2) = 8 + 4a + 2b + c = 0 => 4a + 2b + c = -8 (Eq 2)
  2. Using the Remainder Theorem, P(-1) = 6:
    P(−1)=(−1)3+a(−1)2+b(−1)+c=6=>−1+a−b+c=6=>a−b+c=7(Eq3)P(-1) = (-1)³ + a(-1)² + b(-1) + c = 6 => -1 + a - b + c = 6 => a - b + c = 7 (Eq 3)
  3. From (Eq 1), we already have a+b+c = -1. This is the value we need to find.
    Therequiredvalueisa+b+c=−1.The required value is a+b+c = -1.

Answer: -1

Practice questions on Polynomials

  1. Q1.easy

    If x + 1/x = 5, what is the value of x² + 1/x²?
    1. A)A) 23
    2. B)B) 25
    3. C)C) 27
    4. D)D) 29
    Show answer

    Answer: A) 23

    Hint: Consider squaring the given expression (x + 1/x) and observe how it relates to the desired value.

  2. Q2.easy

    If x=-1 is a zero of the polynomial p(x) = x³ - 2x² + kx + 6, what is the value of k?
    1. A)A) 1
    2. B)B) 3
    3. C)C) 5
    4. D)D) 7
    Show answer

    Answer: B) 3

    Hint: If 'a' is a zero of a polynomial p(x), then p(a) must be equal to zero.

  3. Q3.easy

    What is the remainder when x³ - 2x² + 3x + 5 is divided by (x+1)?
    1. A)A) -5
    2. B)B) -3
    3. C)C) -1
    4. D)D) 1
    Show answer

    Answer: C) -1

    Hint: Apply the Remainder Theorem, which states that if a polynomial p(x) is divided by (x-a), the remainder is p(a).

  4. Q4.medium

    If (x-1) and (x+1) are factors of the polynomial P(x) = x⁴ + ax³ + bx² + cx + d, and the remainder when P(x) is divided by (x-2) is 30, find the value of a + b + c + d.
    1. A)A. 20
    2. B)B. 24
    3. C)C. -1
    4. D)D. 30
    Show answer

    Answer: C. -1

    Hint: Use the Factor Theorem for (x-1). The sum of coefficients of P(x) is P(1).

  5. Q5.medium

    Factor the expression: (a²-b²)³ + (b²-c²)³ + (c²-a²)³.
    1. A)A. 3(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)
    2. B)B. (a²-b²)(b²-c²)(c²-a²)
    3. C)C. 3(a²+b²)(b²+c²)(c²+a²)
    4. D)D. 3(a-b)(b-c)(c-a)
    Show answer

    Answer: A. 3(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)

    Hint: Recall the identity: If x+y+z = 0, then x³+y³+z³ = 3xyz.

  6. Q6.medium

    Let P(x) = x² - 5x + 7. If 'a' is a root of the equation P(x) = 0, find the value of a³ - 18a + 37.
    1. A)A. -1
    2. B)B. 0
    3. C)C. 1
    4. D)D. 2
    Show answer

    Answer: D. 2

    Hint: Since 'a' is a root, P(a)=0. Use this relationship to simplify higher powers of 'a' in the target expression.

  7. Q7.hard

    If x + 1/x = 3, what is the value of x⁵ + 1/x⁵?
    1. A)118
    2. B)123
    3. C)125
    4. D)130
    Show answer

    Answer: 123

    Hint: Calculate x² + 1/x² and x³ + 1/x³ first, then find a way to combine them to get x⁵ + 1/x⁵.

  8. Q8.hard

    When the polynomial x¹⁰⁰ is divided by x²-3x+2, what is the remainder?
    1. A)(2¹⁰⁰ - 2)x + (1 - 2¹⁰⁰)
    2. B)(2¹⁰⁰ - 1)x + (2 - 2¹⁰⁰)
    3. C)(2¹⁰⁰ - 1)x + (2¹⁰⁰ - 2)
    4. D)(2¹⁰⁰ + 1)x + (2 - 2¹⁰⁰)
    Show answer

    Answer: (2¹⁰⁰ - 1)x + (2 - 2¹⁰⁰)

    Hint: Factorize the divisor and use the Remainder Theorem. The remainder will be a linear polynomial, ax+b.

  9. Q9.hard

    Factorize the expression (x²-x+1)(x²-x+2) - 12 completely.
    1. A)(x²-x+5)(x-2)(x+1)
    2. B)(x²-x-5)(x-2)(x+1)
    3. C)(x²-x+5)(x-1)(x+2)
    4. D)(x²-x-5)(x-1)(x+2)
    Show answer

    Answer: (x²-x+5)(x-2)(x+1)

    Hint: Look for a common expression that can be substituted to simplify the problem into a quadratic form.

These are 9 of the 60 questions available for Polynomials. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.