Class 10 Mathematics Ch 2 MCQ with Answers: Polynomials

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20 multiple-choice (MCQ / objective) questions from this chapter, in the common exam-style shapes. Every question comes with an explanation along with the answer, so you remember why the option is correct, not just which one it is.

1 Direct Questions

Direct Questions

1 markWhich of the following is NOT a polynomial?
(A) $x^3$ (B) $\frac{1}{x – 1}$ (C) $y + \sqrt{2}$ (D) $2 – x^2 + \sqrt{3}x$
1 markThe zeroes of $x^2 – 7x + 12$ are:
(A) $-3$ and $-4$ (B) $2$ and $6$ (C) $1$ and $12$ (D) $3$ and $4$
1 markThe sum of the zeroes of $3x^2 – 6x + 5$ is:
(A) $2$ (B) $-2$ (C) $\frac{5}{3}$ (D) $6$
1 markThe product of the zeroes of $x^2 + 7x + 10$ is:
(A) $-10$ (B) $7$ (C) $-7$ (D) $10$
1 markA quadratic polynomial whose zeroes are $2$ and $-3$ is:
(A) $x^2 – x – 6$ (B) $x^2 + x – 6$ (C) $x^2 + x + 6$ (D) $x^2 – x + 6$
1 markThe graph of $y = p(x)$ is a straight line parallel to the x-axis and above it. The number of zeroes of $p(x)$ is:
(A) 1(B) 2(C) 0(D) infinitely many
1 markThe maximum number of zeroes a polynomial of degree 5 can have is:
(A) 4(B) 6(C) 10(D) 5
1 markThe zeroes of $x^3 – x$ are:
(A) $0$ only(B) $-1$ , $0$ and $1$ (C) $1$ and $-1$ only(D) $0$ and $1$ only
1 markIf $\alpha, \beta$ are the zeroes of $x^2 – 4x + 3$ , then $\frac{1}{\alpha} + \frac{1}{\beta}$ is:
(A) $\frac{3}{4}$ (B) $-\frac{4}{3}$ (C) $\frac{4}{3}$ (D) $\frac{1}{3}$
1 markIf one zero of $3x^2 + 8x + k$ is the reciprocal of the other, then $k$ is:
(A) $-3$ (B) $8$ (C) $\frac{1}{3}$ (D) $3$
1 markIf the zeroes of $x^2 + px + q$ are equal in size but opposite in sign, then:
(A) $p = 0$ (B) $q = 0$ (C) $p = q$ (D) $p = 1$
1 markThe zeroes of $t^2 – 15$ are:
(A) $15$ and $-15$ (B) $\sqrt{15}$ only(C) $\sqrt{15}$ and $-\sqrt{15}$ (D) $0$ and $15$

2 Assertion-Reason

Assertion-Reason

1 markAssertion (A): The graph of $y = -2x^2 + 3x – 1$ is a parabola that opens downwards.
Reason (R): The graph of $y = ax^2 + bx + c$ opens downwards when $a \lt 0$ .
(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
1 markAssertion (A): Every quadratic polynomial has exactly two real zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
(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

3 Match the Following

Match the Following

1 markMatch each polynomial in Column A with its type in Column B.
Column A: (i) $2x – 3$ (ii) $y^2 – 2$ (iii) $3 – x^2 + x^3$ (iv) $7u^6 + u – 8$
Column B: (a) cubic (b) degree 6 (c) linear (d) quadratic
(A) i-c, ii-d, iii-a, iv-b(B) i-d, ii-c, iii-a, iv-b(C) i-c, ii-a, iii-d, iv-b(D) i-c, ii-d, iii-b, iv-a
1 markMatch the way a parabola meets the x-axis (Column A) with the number of zeroes (Column B).
Column A: (i) cuts it at two distinct points (ii) touches it at one point (iii) lies completely above it
Column B: (a) no zero (b) two zeroes (c) one zero
(A) i-c, ii-b, iii-a(B) i-a, ii-c, iii-b(C) i-b, ii-c, iii-a(D) i-b, ii-a, iii-c

4 Statement-Based Questions

Statement-Based Questions

1 markStatement I: A real number $k$ is a zero of $p(x)$ if $p(k) = 0$ .
Statement II: $-1$ and $4$ are the zeroes of $x^2 – 3x – 4$ .
(A) Only Statement I is correct(B) Only Statement II is correct(C) Both Statements are correct(D) Neither Statement is correct
1 markStatement I: A quadratic polynomial always has two distinct zeroes.
Statement II: The zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis.
(A) Only Statement I is correct(B) Only Statement II is correct(C) Both Statements are correct(D) Neither Statement is correct

5 Fill in the Blanks

Fill in the Blanks

1 markThe word ‘quadratic’ comes from the word ‘quadrate’, which means ______.
(A) four(B) square(C) cube(D) equal

6 Odd One Out

Odd One Out

1 markChoose the odd one out (by type of polynomial).
(A) $2x + 5$ (B) $x^3 – 4x$ (C) $3z – 1$ (D) $\frac{2}{3}u + 1$

Answer Key and Explanations

Direct Questions

(B) $\frac{1}{x-1}$ has the variable in the denominator, so it is not a polynomial. The book lists it as an example of a non-polynomial.
(D) $x^2 – 7x + 12 = (x – 3)(x – 4)$ , so the zeroes are 3 and 4. Check: sum $7 = \frac{-(-7)}{1}$ , product $12 = \frac{12}{1}$ .
(A) Sum $= \frac{-b}{a} = \frac{-(-6)}{3} = 2$ .
(D) Product $= \frac{c}{a} = \frac{10}{1} = 10$ . The zeroes are $-2$ and $-5$ , and $(-2)(-5) = 10$ .
(B) Sum $= 2 + (-3) = -1$ , product $= -6$ . So $x^2 – (-1)x + (-6) = x^2 + x – 6$ .
(C) The line never meets the x-axis, so $p(x)$ has no zero. This is Exercise 2.1 (i).
(D) A polynomial of degree $n$ has at most $n$ zeroes, because its graph meets the x-axis at most $n$ times.
(B) $x^3 – x = x(x^2 – 1) = x(x – 1)(x + 1)$ , so the zeroes are $0$ , $1$ and $-1$ .
(C) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{4}{3}$ .
(D) If the zeroes are $\alpha$ and $\frac{1}{\alpha}$ , their product is 1. So $\frac{k}{3} = 1$ , giving $k = 3$ .
(A) Zeroes $\alpha$ and $-\alpha$ have sum 0. So $-\frac{p}{1} = 0$ , giving $p = 0$ .
(C) $t^2 – 15 = (t – \sqrt{15})(t + \sqrt{15})$ , so the zeroes are $\pm\sqrt{15}$ .

Assertion-Reason

(A) Both A and R are true: here $a = -2 \lt 0$ , and R is exactly the reason the parabola opens downwards.
(D) A is false: a quadratic can have two, one or no zero (for example $x^2 + 2$ has none). R is true: a polynomial of degree 2 has at most two zeroes.

Match the Following

(A) (i) $2x – 3$ has degree 1, linear (c); (ii) $y^2 – 2$ degree 2, quadratic (d); (iii) $3 – x^2 + x^3$ degree 3, cubic (a); (iv) $7u^6 + u – 8$ degree 6 (b).
(C) Case (i): two distinct points give two zeroes (b). Case (ii): two coincident points give one zero (c). Case (iii): no meeting point gives no zero (a).

Statement-Based Questions

(C) Statement I is true: it is the definition of a zero. Statement II is true: $p(-1) = 1 + 3 – 4 = 0$ and $p(4) = 16 – 12 – 4 = 0$ .
(B) Statement I is false: a quadratic can have two, one or no zero. Statement II is true: it is point 3 of the book’s summary.

Fill in the Blanks

(B) The book: the name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’.

Odd One Out

(B) $x^3 – 4x$ is cubic (degree 3); the other three are all linear (degree 1).

More for this chapter

Notes Important Questions NCERT Solutions Short Notes