Polynomials MCQ: Answers and Explanation

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Jasmine Grover

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The name "polynomial" comes from the words "poly" (meaning many) and "nomial" (meaning phrase), so it means "many terms."

  • A polynomial is made up of solely added, subtracted, and multiplied terms.
  • ax² + bx + c is the form of a quadratic polynomial in x with real coefficients, where a, b, and c are real numbers with a = 0.
  • The degree of a polynomial refers to the variable's highest exponent in the polynomial. 2x² + 4 is an example where the degree is 2.
  • Linear, quadratic, and cubic polynomials have the degrees 1, 2, and 3 polynomials, respectively.
  • A polynomial can have terms with constants such as 5, -2, and others, as well as variables like x and y and exponents such as 2 in y².
  • Addition, subtraction, and multiplication can be used to combine these, but not division.
  • The x-coordinates of the locations where the graph of y = p(x) contacts the x-axis are the zeroes of polynomial p(x).
  • If α and β are the zeroes of the quadratic polynomial ax² + bx + c, then:

Polynomial formula

  • If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d = 0, then:

Polynomial formula

  • Zeroes, also known as α, β, γ, follow the rules of algebraic identities, i.e.,

(α + β)² = α² + β² + 2αβ

∴(α² + β²) = (α + β)² – 2αβ

  • If p(x) and g(x) are any two polynomials with g(x) = 0, then p(x) = g(x) * q + r is the division algorithm:

Dividend = Divisor x Quotient + Remainder

The video below explains this:

Polynomials Detailed Video Explanation:


Question 1: If the zeroes of the quadratic polynomial ay² + by + c, c ≠ 0 are equal, then

  1. c and b have opposite signs
  2. c and a have opposite signs
  3. c and b have the same signs
  4. c and a have the same signs

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Answer: (d) c and a have the same signs

Explanation: Discriminant will be equal to zero for equal roots:

b² - 4ac = 0

b² = 4ac

ac = b²/4

ac > 0 (square of any number cannot be negative)

Question 2: If the sum of zeroes of the quadratic polynomial 3x² – kx + 6 is 3, then the value of k is:

  1. 4
  2. 5
  3. 7
  4. 9

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Answer: (d) 9

Explanation: Here a = 3, b = -k, c = 6

Sum of the zeroes, (α + β) = -b/a = 3 …..(given)

⇒ −(−k)/3 = 3

⇒ k = 9

Question 3: The degree of the polynomial, x5 – 2x2 + 2 is:

  1. 2
  2. 4
  3. 1
  4. 5

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Answer: (d) 5

Explanation: In every polynomial, the highest power of the variable is called a degree.

Question 4: If p(x) is a polynomial of degree one and p(y) = 0, then y is said to be:

  1. Zero of p(x)
  2. Value of p(x)
  3. Constant of p(x)
  4. None of the above

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Answer: (a) Zero of p(x)

Explanation: Let p(x) = mx + n

Put x = y

p(y) = my + n = 0

So, y is zero of p(x).

Question 5: A polynomial's zeros can be represented graphically. The number of polynomial zeros equals the number of points on the graph of the polynomial:

  1. Intersects y-axis
  2. Intersects x-axis
  3. Intersects y-axis or x-axis
  4. None of the above

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Answer: (b) Intersects x-axis

Explanation: The number of zeroes of a polynomial is equal to the number of points where the graph of polynomial intersects the x-axis.

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Question 6: A polynomial of degree p has:

  1. Only one zero
  2. At least p zeroes
  3. More than p zeroes
  4. At most p zeroes

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Answer: (d)

Explanation: A polynomial's maximum number of zeroes equals the polynomial's degree.

Question 7: If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then the polynomial is:

  1. x² + 6x + 5 = 0
  2. x² + 6x - 5 = 0
  3. x² - 6x + 5 = 0
  4. x² - 6x - 5 = 0

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Answer: (a) x² + 6x + 5 = 0

Explanation: Quadratic polynomial is x² – Sx + P = 0, where S is the sum and P is the product

⇒ x² – (-6)x + 5 = 0

⇒ x² + 6x + 5 = 0

Question 8: Zeros of p(x) = x² - 27 are:

  1. ± 3√3
  2. ± 9√3
  3. ± 7√3
  4. None of the above

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Answer: (a)

Explanation: x² - 27 = 0

x² = 27

x = √27

x = ±3√3

Question 9: The quadratic polynomial whose zeroes are 3 + √2 and 3 – √2 is:

  1. x² – 6x - 7
  2. x² + 6x + 7
  3. x² – 6x + 7
  4. x² + 6x - 7

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Answer: (c) x² – 6x + 7

Explanation: S is the sum of zeroes and P is the product of zeroes: 

S = (3 + √2) + (3 – √2) = 6

P = (3 + √2) x (3 – √2) = (3)² – (√2)² = 9 – 2 = 7

So, Quadratic polynomial = x² – Sx + P = x² – 6x + 7

Question 10: If a quadratic polynomial's discriminant, D, is greater than zero, the polynomial has:

  1. two real and equal roots
  2. two real and unequal roots
  3. imaginary roots
  4. no roots

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Answer: (b) two real and unequal roots

Explanation: If the discriminant of a quadratic polynomial, D > 0, then the polynomial has two real and unequal roots.


Also Read:

CBSE X Related Questions

  • 1.
    If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

      • 3
      • –3
      • –4
      • \(\pm 3\)

    • 2.
      If the pair of linear equations : \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) is consistent and dependent, then

        • \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
        • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
        • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
        • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)

      • 3.
        In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.


          • 4.
            If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).


              • 5.
                \(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

                  • \(\frac{21}{4} \text{ cm}\)
                  • \(\frac{28}{3} \text{ cm}\)
                  • \(\frac{12}{7} \text{ cm}\)
                  • \(5.5 \text{ cm}\)

                • 6.
                  PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

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