Class 9-Online MCQ Tests-Chapter-02-Polynomials

Chapter 2: Polynomials

Class 9 Maths | 20 Questions (MCQ & Assertion-Reason)
1. A polynomial has a degree of 0. What is always true about it?
Solution A polynomial of degree 0 is a non-zero constant (e.g., \(P(x) = 5\)).
2. The polynomial \(p(x) = x^3 + ax + 1\) has no \(x^2\) term. What is the coefficient of \(x^2\)?
Solution If a term is missing, its coefficient is 0.
3. Which of these is a binomial?
Solution A binomial has exactly two terms. \(x^2 + 4x\) fits this description.
4. A student adds two polynomials of degrees 3 and 2. The result will be a polynomial of degree:
Solution When adding polynomials, the degree is the maximum of the degrees of the addends. Max(3, 2) = 3.
5. Which of these is a zero of the polynomial \(p(x) = x^2 - 4\)?
Solution \(x^2 - 4 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).
6. The number of terms in the polynomial \(x^2 + 2x - 3x + 7\) is:
Solution Simplify first: \(x^2 + (2x - 3x) + 7 = x^2 - x + 7\). This has 3 terms.
7. What is the degree of the polynomial \(3x^4 + 0x^3 + 2x + 5\)?
Solution The highest power with a non-zero coefficient is 4.
8. If \(x = -1\) is a zero of \(p(x) = x^2 + kx + 1\), what is the value of k?
Solution \(p(-1) = (-1)^2 + k(-1) + 1 = 0 \Rightarrow 1 - k + 1 = 0 \Rightarrow k = 2\).
9. Which polynomial has all zero coefficients except the constant term?
Solution This describes a constant polynomial, like 5.
10. A polynomial is said to be monic if the coefficient of its highest degree term is:
Solution Definition of a Monic Polynomial: Leading coefficient is 1.
DIRECTION: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
11. Assertion: The degree of the polynomial \(2x^3 + 3x^2 - 5x + 1\) is 3.
Reason: The degree of a polynomial is determined by the highest power of the variable.
Solution Both are true and Reason explains Assertion.
12. Assertion: The polynomial \(x^2 + 4x + 4\) can be factorized as \((x + 2)^2\).
Reason: The polynomial \(x^2 + 4x + 4\) is a perfect square trinomial.
Solution Both are true and Reason explains why it factorizes that way.
13. Assertion: The remainder when \(x^3 + 2x^2 - 7x - 12\) is divided by \(x + 2\) is 2.
Reason: According to the Remainder Theorem, the remainder is \(p(-2)\).
Solution Calculation: \((-2)^3 + 2(-2)^2 - 7(-2) - 12 = -8 + 8 + 14 - 12 = 2\). Assertion is True. Reason is True and correct explanation. (Note: Original image had a typo in Assertion value, corrected here for validity).
14. Assertion: The polynomial \(2x^2 + 5x - 3\) can be factored as \((2x - 1)(x + 3)\).
Reason: The factors of -6 that add up to 5 are 6 and -1.
Solution Splitting the middle term requires finding numbers that multiply to \(ac (-6)\) and add to \(b (5)\). The reason states this correctly.
15. Assertion: The degree of the polynomial 0 is undefined.
Reason: The polynomial 0 has no terms with a variable.
Solution Both are true and Reason explains why degree is undefined.
16. Assertion: The polynomial \(x^3 - 8\) can be factored as \((x - 2)(x^2 + 2x + 4)\).
Reason: \(x^3 - 8\) is a difference of cubes.
Solution The formula \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\) applies here.
17. Assertion: If \(x + 1\) is a factor of \(p(x)\), then \(p(-1) = 0\).
Reason: According to the Factor Theorem, if \(x - a\) is a factor of \(p(x)\), then \(p(a) = 0\).
Solution Both are true and Reason is the correct explanation.
18. Assertion: The polynomial \(x^2 + 1\) has no real roots.
Reason: The discriminant of \(x^2 + 1\) is negative.
Solution Discriminant \(D = b^2 - 4ac = 0 - 4(1)(1) = -4\). Since \(D < 0\), roots are not real.
19. Assertion: The polynomial \(3x^2 + 2x - 5\) is a quadratic polynomial.
Reason: The highest power of the variable in the polynomial is 2.
Solution Definition of quadratic polynomial is degree 2.
20. Assertion: If \(p(x) = x^3 + ax^2 + bx + c\), then \(p(0) = c\).
Reason: When \(x = 0\), all terms with x become 0.
Solution \(p(0) = 0 + 0 + 0 + c = c\). Reason is correct.
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