CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 2 Marks - Part 1
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CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 2 Marks - Part 1
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SECTION B — Very Short Answer Type Questions
[2 Marks Each]
Q1. Look at the graph of $y = p(x)$ below. Find the number of zeroes of $p(x)$ and justify your answer.
(Imagine a curve intersecting the x-axis at exactly three distinct points).
Answer: Number of zeroes: 3
Justification: The number of zeroes of a polynomial $p(x)$ is equal to the total number of points where the graph of $y = p(x)$ intersects the x-axis. Since the given curve intersects the x-axis at exactly three distinct points, the polynomial has 3 zeroes.
Q2. If the graph of a quadratic polynomial $p(x) = ax^2 + bx + c$ touches the x-axis at exactly one point, what can you say about the nature and number of its zeroes?
Answer: Number of zeroes: 1 (or 2 equal zeroes)
Justification: A quadratic polynomial can have at most two zeroes. When its graph touches the x-axis at exactly one point, the two roots (zeroes) are coincident (real and equal). Therefore, the polynomial has 1 distinct real zero, which corresponds to the x-coordinate of the point of contact.
Q3. Can a quadratic polynomial have exactly three zeroes? Explain briefly based on its degree.
Answer: No, a quadratic polynomial cannot have three zeroes.
Justification: A quadratic polynomial is a polynomial of degree 2 ($p(x) = ax^2 + bx + c$, where $a \neq 0$). By the Fundamental Theorem of Algebra, a polynomial of degree $n$ can have at most $n$ distinct zeroes. Therefore, a quadratic polynomial can have at most two zeroes.
Q4. If a polynomial $p(x)$ has no real zeroes, does its graph intersect the x-axis? Explain with a short reason.
Answer: No, its graph does not intersect the x-axis.
Justification: The zeroes of a polynomial $p(x)$ are the x-coordinates of the points where its graph intersects the x-axis. Since the polynomial has no real zeroes, the curve never meets or crosses the x-axis.
Q5. Find the number of zeroes of the polynomial $p(x)$ whose graph is a straight line parallel to the x-axis.
Answer: Number of zeroes: 0 (unless the line coincides with the x-axis, in which case it has infinitely many zeroes)
Justification: The zeroes of a polynomial are given by the points where its graph intersects the x-axis. A straight line that is parallel to the x-axis and distinct from it never intersects the x-axis, meaning there are no real zeroes. (If the line is the x-axis itself, every point is an intersection, leading to infinitely many zeroes).
Q6. The graph of a polynomial passes through the points $(-2, 0)$, $(0, 3)$, and $(4, 0)$. State the real zeroes of this polynomial.
Answer: Real zeroes: $-2$ and $4$
Justification: The real zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis (i.e., points where the y-coordinate is $0$). From the given points, the graph intersects the x-axis at $(-2, 0)$ and $(4, 0)$, making $-2$ and $4$ the real zeroes. (The point $(0, 3)$ represents the y-intercept, not a zero).
Q7. If the graph of $y = p(x)$ cuts the y-axis at $(0, -5)$ and the x-axis at $(2, 0)$ and $(-3, 0)$, how many zeroes does it definitely have in this range? State them.
Answer: Number of zeroes: 2 Zeroes: $2$ and $-3$
Justification: The zeroes of a polynomial $p(x)$ are the x-coordinates of the points where its graph intersects the x-axis. Since the graph intersects the x-axis at $(2, 0)$ and $(-3, 0)$, the polynomial has 2 definite real zeroes, which are $2$ and $-3$. (The point $(0, -5)$ is the y-intercept and does not represent a zero).
Q8. True or False: "The number of points where a graph intersects the y-axis gives the number of zeroes of the polynomial." Justify your stance.
Answer: False
Justification: The number of zeroes of a polynomial $p(x)$ is determined by the number of points where its graph intersects the **x-axis**, not the y-axis. The y-intercept simply gives the value of the polynomial when $x = 0$ (i.e., $p(0)$), which has no relation to the number of zeroes.
Q9. Draw a rough sketch of the graph of the polynomial $p(x) = 3x - 6$ and mark its zero.
Answer: Zero of the polynomial: $2$
Justification & Sketch Description:
To find the zero, set $p(x) = 0 \implies 3x - 6 = 0 \implies x = 2$.
The graph of $p(x) = 3x - 6$ is a straight line.
* **x-intercept:** $(2, 0)$ — this is where the line crosses the x-axis, representing the zero.
* **y-intercept:** $(0, -6)$ — obtained by putting $x = 0$.
Rough Sketch Details: Plot the points $(2, 0)$ on the x-axis and $(0, -6)$ on the y-axis, then draw a straight line passing through them. The point where the line intersects the x-axis, i.e., $(2, 0)$, is marked as the zero.
Q10. If a cubic polynomial $p(x) = x^3 - x$ is plotted, at how many points will its graph intersect the x-axis?
Answer: Number of intersection points: 3
Justification: The number of points where the graph of a polynomial intersects the x-axis is equal to the number of its distinct real zeroes. For the cubic polynomial $p(x) = x^3 - x$, we can factor it as $p(x) = x(x^2 - 1) = x(x - 1)(x + 1)$. Setting $p(x) = 0$ gives three distinct real zeroes: $x = 0$, $x = 1$, and $x = -1$. Therefore, its graph intersects the x-axis at exactly 3 points.
Q11. State the maximum number of zeroes that a polynomial of degree $n$ can have.
Answer: Maximum number of zeroes: $n$
Justification: By the fundamental theorem of algebra (or properties of polynomial equations), a non-zero polynomial of degree $n$ can have at most $n$ distinct real roots (or zeroes). Consequently, the graph of a polynomial of degree $n$ can intersect the x-axis at most $n$ times.
Q12. If the graph of a quadratic polynomial is a parabola opening downwards, what is the sign of the coefficient of $x^2$?
Answer: Sign of the coefficient of $x^2$: Negative ($a < 0$)
Justification: For a general quadratic polynomial $p(x) = ax^2 + bx + c$ (where $a \neq 0$), the graph is a parabola. If $a > 0$, the parabola opens upwards (concave up), and if $a < 0$, the parabola opens downwards (concave down). Therefore, since the parabola opens downwards, the coefficient $a$ must be negative.
Q13. Find the zeroes of the quadratic polynomial $p(x) = x^2 - 2x - 8$.
Answer: Zeroes of the polynomial: $4$ and $-2$
Justification & Calculation:
To find the zeroes, set $p(x) = 0$:
$$x^2 - 2x - 8 = 0$$
By splitting the middle term:
$$x^2 - 4x + 2x - 8 = 0$$
$$x(x - 4) + 2(x - 4) = 0$$
$$(x - 4)(x + 2) = 0$$
Setting each factor to zero gives $x = 4$ or $x = -2$. Therefore, the zeroes are $4$ and $-2$.
Q14. Calculate the zeroes of the polynomial $g(x) = 4x^2 - 4x + 1$.
Answer: Zeroes of the polynomial: $\frac{1}{2}$ and $\frac{1}{2}$ (equal zeroes)
Justification & Calculation:
To find the zeroes, set $g(x) = 0$:
$$4x^2 - 4x + 1 = 0$$
Using the splitting the middle term method (or recognizing it as a perfect square, $(2x - 1)^2 = 0$):
$$4x^2 - 2x - 2x + 1 = 0$$
$$2x(2x - 1) - 1(2x - 1) = 0$$
$$(2x - 1)(2x - 1) = 0$$
Setting each factor to zero gives $2x - 1 = 0 \implies x = \frac{1}{2}$, twice. Therefore, the polynomial has two coincident (equal) real zeroes: $\frac{1}{2}$ and $\frac{1}{2}$.
Q15. Find the zeroes of the quadratic polynomial $f(x) = x^2 - 3$.
Answer: Zeroes of the polynomial: $\sqrt{3}$ and $-\sqrt{3}$
Justification & Calculation:
To find the zeroes, set $f(x) = 0$:
$$x^2 - 3 = 0$$
$$x^2 = 3$$
Taking the square root on both sides:
$$x = \pm\sqrt{3}$$
Therefore, the two zeroes of the polynomial are $\sqrt{3}$ and $-\sqrt{3}$.
Q16. Find the zeroes of the polynomial $p(u) = 4u^2 + 8u$.
Answer: Zeroes of the polynomial: $0$ and $-2$
Justification & Calculation:
To find the zeroes, set $p(u) = 0$:
$$4u^2 + 8u = 0$$
Factor out the common terms ($4u$):
$$4u(u + 2) = 0$$
Setting each factor to zero:
$$4u = 0 \implies u = 0$$
$$u + 2 = 0 \implies u = -2$$
Therefore, the zeroes of the polynomial are $0$ and $-2$.
Q17. Find the zeroes of the quadratic polynomial $t^2 - 15$.
Answer: Zeroes of the polynomial: $\sqrt{15}$ and $-\sqrt{15}$
Justification & Calculation:
To find the zeroes, set the polynomial equal to zero:
$$t^2 - 15 = 0$$
$$t^2 = 15$$
Taking the square root on both sides:
$$t = \pm\sqrt{15}$$
Therefore, the two zeroes of the polynomial are $\sqrt{15}$ and $-\sqrt{15}$.
Q18. Obtain the zeroes of the polynomial $h(x) = 3x^2 - x - 4$.
Answer: Zeroes of the polynomial: $\frac{4}{3}$ and $-1$
Justification & Calculation:
To find the zeroes, set $h(x) = 0$:
$$3x^2 - x - 4 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $3 \times (-4) = -12$ and add up to $-1$, which are $-4$ and $3$):
$$3x^2 - 4x + 3x - 4 = 0$$
$$x(3x - 4) + 1(3x - 4) = 0$$
$$(3x - 4)(x + 1) = 0$$
Setting each factor to zero:
$$3x - 4 = 0 \implies x = \frac{4}{3}$$
$$x + 1 = 0 \implies x = -1$$
Therefore, the zeroes of the polynomial are $\frac{4}{3}$ and $-1$.
Q19. Find the zeroes of the polynomial $p(x) = 2x^2 - 7x + 3$.
Answer: Zeroes of the polynomial: $3$ and $\frac{1}{2}$
Justification & Calculation:
To find the zeroes, set $p(x) = 0$:
$$2x^2 - 7x + 3 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $2 \times 3 = 6$ and add up to $-7$, which are $-6$ and $-1$):
$$2x^2 - 6x - x + 3 = 0$$
$$2x(x - 3) - 1(x - 3) = 0$$
$$(2x - 1)(x - 3) = 0$$
Setting each factor to zero:
$$2x - 1 = 0 \implies x = \frac{1}{2}$$
$$x - 3 = 0 \implies x = 3$$
Therefore, the zeroes of the polynomial are $3$ and $\frac{1}{2}$.
Q20. Determine the zeroes of the polynomial $f(x) = x^2 + 7x + 10$.
Answer: Zeroes of the polynomial: $-2$ and $-5$
Justification & Calculation:
To find the zeroes, set $f(x) = 0$:
$$x^2 + 7x + 10 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $10$ and add up to $7$, which are $5$ and $2$):
$$x^2 + 5x + 2x + 10 = 0$$
$$x(x + 5) + 2(x + 5) = 0$$
$$(x + 5)(x + 2) = 0$$
Setting each factor to zero:
$$x + 5 = 0 \implies x = -5$$
$$x + 2 = 0 \implies x = -2$$
Therefore, the zeroes of the polynomial are $-2$ and $-5$.
Q21. Find the zeroes of the quadratic polynomial $6x^2 - 3 - 7x$ by first rewriting it in standard form.
Answer: Zeroes of the polynomial: $\frac{3}{2}$ and $-\frac{1}{3}$
Justification & Calculation:
First, rewrite the polynomial in standard form ($ax^2 + bx + c$):
$$p(x) = 6x^2 - 7x - 3$$
To find the zeroes, set $p(x) = 0$:
$$6x^2 - 7x - 3 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $6 \times (-3) = -18$ and add up to $-7$, which are $-9$ and $2$):
$$6x^2 - 9x + 2x - 3 = 0$$
$$3x(2x - 3) + 1(2x - 3) = 0$$
$$(3x + 1)(2x - 3) = 0$$
Setting each factor to zero:
$$3x + 1 = 0 \implies x = -\frac{1}{3}$$
$$2x - 3 = 0 \implies x = \frac{3}{2}$$
Therefore, the zeroes of the polynomial are $\frac{3}{2}$ and $-\frac{1}{3}$.
Q22. Find the zeroes of the polynomial $p(x) = \sqrt{3}x^2 + 10x + 7\sqrt{3}$.
Answer: Zeroes of the polynomial: $-\sqrt{3}$ and $-\frac{7}{\sqrt{3}}$ (or $-\frac{7\sqrt{3}}{3}$)
Justification & Calculation:
To find the zeroes, set $p(x) = 0$:
$$\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$$
Using the splitting the middle term method (find two numbers that multiply to $\sqrt{3} \times 7\sqrt{3} = 21$ and add up to $10$, which are $7$ and $3$):
$$\sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0$$
Factor out $\sqrt{3}x$ from the first two terms and $7$ from the last two terms:
$$\sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0$$
$$(x + \sqrt{3})(\sqrt{3}x + 7) = 0$$
Setting each factor to zero:
$$x + \sqrt{3} = 0 \implies x = -\sqrt{3}$$
$$\sqrt{3}x + 7 = 0 \implies x = -\frac{7}{\sqrt{3}}$$
Therefore, the zeroes of the polynomial are $-\sqrt{3}$ and $-\frac{7}{\sqrt{3}}$.
Q23. Find the zeroes of the quadratic polynomial $x^2 - 5x$.
Answer: Zeroes of the polynomial: $0$ and $5$
Justification & Calculation:
To find the zeroes, set the polynomial equal to zero:
$$x^2 - 5x = 0$$
Factor out the common term $x$:
$$x(x - 5) = 0$$
Setting each factor to zero:
$$x = 0$$
$$x - 5 = 0 \implies x = 5$$
Therefore, the zeroes of the polynomial are $0$ and $5$.
Q24. Calculate the zeroes of the polynomial $p(x) = 4x^2 - 9$
Answer: Zeroes of the polynomial: $\frac{3}{2}$ and $-\frac{3}{2}$
Justification & Calculation:
To find the zeroes, set $p(x) = 0$:
$$4x^2 - 9 = 0$$
This can be written in the form of a difference of squares ($a^2 - b^2 = (a-b)(a+b)$):
$$(2x)^2 - (3)^2 = 0$$
$$(2x - 3)(2x + 3) = 0$$
Setting each factor to zero:
$$2x - 3 = 0 \implies x = \frac{3}{2}$$
$$2x + 3 = 0 \implies x = -\frac{3}{2}$$
Therefore, the zeroes of the polynomial are $\frac{3}{2}$ and $-\frac{3}{2}$.
Q25. Find the zeroes of the polynomial $g(t) = 2t^2 - 9t - 5$.
Answer: Zeroes of the polynomial: $5$ and $-\frac{1}{2}$
Justification & Calculation:
To find the zeroes, set $g(t) = 0$:
$$2t^2 - 9t - 5 = 0$$
Using the splitting the middle term method (find two numbers that multiply to $2 \times (-5) = -10$ and add up to $-9$, which are $-10$ and $1$):
$$2t^2 - 10t + t - 5 = 0$$
$$2t(t - 5) + 1(t - 5) = 0$$
$$(2t + 1)(t - 5) = 0$$
Setting each factor to zero:
$$2t + 1 = 0 \implies t = -\frac{1}{2}$$
$$t - 5 = 0 \implies t = 5$$
Therefore, the zeroes of the polynomial are $5$ and $-\frac{1}{2}$.
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