NCERT Exemplar Solutions Class 10 Maths Polynomials — Exercises 2.1 and 2.2

NCERT Exemplar Problems, Class 10 Mathematics — Polynomials, Exercises 2.1 and 2.2. All 13 questions solved.

Exercise 2.1 is eleven multiple-choice questions and Exercise 2.2 is two questions carrying twelve sub-parts between them, so the two are taken together here. Almost everything on this page comes out of one pair of facts, and it is worth having them written down before starting.

For a quadratic $ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

For a cubic $ax^3 + bx^2 + cx + d$ with zeroes $\alpha$, $\beta$, $\gamma$:

$$\alpha+\beta+\gamma = -\frac{b}{a}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}$$

The alternating signs are the thing to get right, and most of the wrong options on this page are what you get by dropping one of them.

A note on the current syllabus. Exercise 2.2 Question 1, parts (i) to (iv), are about dividing one polynomial by another — degrees of quotients and remainders. That topic is gone from the rationalised textbook: the chapter now runs Introduction → Geometrical Meaning of the Zeroes → Relationship between Zeroes and Coefficients → Summary, and the words “quotient” and “remainder” do not appear in it at all. The four parts are still worth doing — the degree bookkeeping is a genuinely useful habit, and it is standard in Class 11 — but no board paper will set them.

Key insight. A zero tells you about the coefficients, not just about the graph. Read every one of these questions as “what do the signs of $-b/a$ and $c/a$ have to be?”, and the options separate themselves.

Choose the correct answer from the given four options in the following questions:

Exercise 2.1, Question 1

If one of the zeroes of the quadratic polynomial $(k-1)x^2 + kx + 1$ is $-3$, then the value of $k$ is

Solution. A zero is a value that makes the polynomial vanish, so substitute $x = -3$ and set the result to zero. There is no need for the sum-and-product relations here.

$$(k-1)(-3)^2 + k(-3) + 1 = 0$$ $$9k – 9 – 3k + 1 = 0$$ $$6k = 8 \implies k = \frac{4}{3}$$

Exercise 2.1, Question 2

A quadratic polynomial, whose zeroes are $-3$ and $4$, is

Solution. With zeroes $-3$ and $4$, the sum is $1$ and the product is $-12$, so the family of such polynomials is

$$k\left[x^2 – (\text{sum})x + (\text{product})\right] = k\left(x^2 – x – 12\right), \qquad k \ne 0$$

Option (C) is this with $k = \frac12$. Check: $\frac12\left(x^2 – x – 12\right) = \frac{x^2}{2} – \frac{x}{2} – 6$. ✓

The other three fail:

• (A) $x^2 – x + 12$ has the right sum but the product is $+12$, not $-12$.
• (B) $x^2 + x + 12$ has both signs wrong.
• (D) $2x^2 + 2x – 24 = 2\left(x^2 + x – 12\right)$ has zeroes $-4$ and $3$ — the right numbers with the signs exchanged. This is the option that catches a dropped minus in $-(\text{sum})$.

Exercise 2.1, Question 3

If the zeroes of the quadratic polynomial $x^2 + (a+1)x + b$ are $2$ and $-3$, then

Solution. Here $a$ in the relations is $1$ (the coefficient of $x^2$), and the letter $a$ in the question is buried inside the linear coefficient — worth separating before substituting.

$$\text{Sum} = 2 + (-3) = -1 = -(a+1) \implies a + 1 = 1 \implies a = 0$$ $$\text{Product} = 2 \times (-3) = -6 = b \implies b = -6$$

Exercise 2.1, Question 4

The number of polynomials having zeroes as $-2$ and $5$ is

Solution. Any polynomial with those two zeroes is divisible by $(x+2)(x-5)$, so it has the form

$$p(x) = k\,(x+2)(x-5)$$

and every non-zero $k$ gives a different polynomial — $k = 1, 2, 3, \tfrac12, -7, \dots$. There are infinitely many, so certainly more than three.

The question says “polynomials”, not “quadratic polynomials”, which widens it further: $(x+2)(x-5)(x^2+1)$ also has exactly those real zeroes. Either reading gives the same option.

Exercise 2.1, Question 5

Given that one of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ is zero, the product of the other two zeroes is

Solution. Let the zeroes be $0$, $\beta$, $\gamma$. Use the second relation, the sum of products in pairs:

$$0\cdot\beta + \beta\gamma + \gamma\cdot 0 = \frac{c}{a}$$

Every term containing the zero vanishes, leaving $\beta\gamma = \dfrac{c}{a}$ directly.

Option (C) is the trap: the product of all three zeroes is zero, but the question asks for the product of the other two, which is a different quantity. Reaching for the third relation $\alpha\beta\gamma = -d/a$ gives $0 = -d/a$, which tells you $d = 0$ and nothing about $\beta\gamma$.

Exercise 2.1, Question 6

If one of the zeroes of the cubic polynomial $x^3 + ax^2 + bx + c$ is $-1$, then the product of the other two zeroes is

Solution. Two facts are needed. First, the product of all three zeroes is $-c$ (since the leading coefficient is $1$), so with zeroes $-1$, $\beta$, $\gamma$:

$$(-1)\beta\gamma = -c \implies \beta\gamma = c$$

Second, $-1$ being a zero means $p(-1) = 0$:

$$(-1)^3 + a(-1)^2 + b(-1) + c = 0 \implies -1 + a – b + c = 0 \implies c = 1 – a + b$$

Therefore $\beta\gamma = c = b – a + 1$.

Exercise 2.1, Question 7

The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are

Solution. Do not compute the roots — read the signs off the coefficients.

$$\alpha + \beta = -99 < 0, \qquad \alpha\beta = 127 > 0$$

A positive product means the two zeroes have the same sign; a negative sum then forces that sign to be negative. So both are negative, which rules out (A) and (C) at once.

(D) needs the discriminant: $99^2 – 4(127) = 9801 – 508 = 9293 \ne 0$, so the zeroes are distinct. (For the record they are about $-97.70$ and $-1.30$.)

Exercise 2.1, Question 8

The zeroes of the quadratic polynomial $x^2 + kx + k$, $k \ne 0$,

Solution. Here $\alpha + \beta = -k$ and $\alpha\beta = k$. Suppose both zeroes were positive. Then the sum $-k$ would be positive, so $k < 0$; but the product $k$ would also have to be positive, so $k > 0$. The two requirements contradict each other, so both zeroes can never be positive whatever $k$ is.

The other options all fail on an example:

• (B) With $k = 5$: the zeroes are about $-3.62$ and $-1.38$, both negative. So they can both be negative.
• (C) and (D) The discriminant is $k^2 – 4k = k(k-4)$, which is zero at $k = 4$ (equal zeroes) and non-zero elsewhere — so they are neither always unequal nor always equal.

Exercise 2.1, Question 9

If the zeroes of the quadratic polynomial $ax^2 + bx + c$, $c \ne 0$, are equal, then

Solution. Equal zeroes means the discriminant vanishes:

$$b^2 – 4ac = 0 \implies ac = \frac{b^2}{4}$$

A square is never negative, and $c \ne 0$ (given) with $a \ne 0$ (it is a quadratic) makes $b^2/4$ strictly positive. So $ac > 0$, which says exactly that $a$ and $c$ have the same sign.

Nothing at all follows about $b$: the same polynomial with $b$ replaced by $-b$ still has equal zeroes, so options (B) and (D) cannot be determined either way.

Exercise 2.1, Question 10

If one of the zeroes of a quadratic polynomial of the form $x^2 + ax + b$ is the negative of the other, then it

Solution. Let the zeroes be $t$ and $-t$, with $t \ne 0$ (otherwise the two zeroes are the same number, not negatives of each other in any useful sense).

$$\text{Sum} = t + (-t) = 0 = -a \implies a = 0$$

so the linear term disappears. And

$$\text{Product} = t(-t) = -t^2 = b$$

Since $t \ne 0$, $t^2 > 0$ and so $b = -t^2 < 0$: the constant term is negative. Both halves of (A) hold.

The polynomial is therefore $x^2 – t^2$, which is the general shape of every example — $x^2 – 9$, $x^2 – 2$, and so on.

Exercise 2.1, Question 11

Which of the following is not the graph of a quadratic polynomial?

In words: (A) is an upward-opening parabola lying entirely above the $x$-axis; (B) and (C) are downward-opening parabolas cutting the axis at two points; (D) is an S-shaped curve that crosses the axis at three points.

Solution. The graph of a quadratic polynomial is a parabola, and a parabola meets the $x$-axis in at most two points — because the zeroes of $ax^2+bx+c$ are the roots of a quadratic equation, and a quadratic has at most two roots.

Graphs (A), (B) and (C) are all parabolas: (A) has no zeroes (discriminant negative), (B) and (C) each have two. Graph (D) crosses the axis three times, so its polynomial has at least three zeroes and cannot be quadratic — it is the shape of a cubic.

Notice that “not a parabola” and “meets the axis more than twice” are two different tests, and only the second is usable from a picture. Curvature is easy to mis-see; counting crossings is not.

Exercise 2.2, Question 1

Answer the following and justify.

(i) Can $x^2 – 1$ be the quotient on division of $x^6 + 2x^3 + x – 1$ by a polynomial in $x$ of degree $5$?

Solution. No. Ignoring the remainder for a moment, division gives

$$\text{dividend} = \text{divisor} \times \text{quotient} + \text{remainder}$$

The divisor has degree $5$ and the proposed quotient degree $2$, so the product has degree $5 + 2 = 7$. The remainder must have degree less than $5$, so it cannot cancel the degree-$7$ term. The right-hand side therefore has degree $7$, while the dividend has degree $6$. Impossible.

(ii) What will the quotient and remainder be on division of $ax^2 + bx + c$ by $px^3 + qx^2 + rx + s$, $p \ne 0$?

Solution. The divisor has degree $3$ and the dividend degree $2$. Since the divisor is already of higher degree, it “goes into” the dividend zero times:

$$\text{quotient} = 0, \qquad \text{remainder} = ax^2 + bx + c$$

This is the polynomial analogue of dividing $7$ by $10$ in whole numbers: quotient $0$, remainder $7$.

(iii) If on division of a polynomial $p(x)$ by a polynomial $g(x)$ the quotient is zero, what is the relation between the degrees of $p(x)$ and $g(x)$?

Solution. Quotient zero means $p(x) = 0 \cdot g(x) + r(x)$, so $p(x)$ is the remainder; and a remainder must have degree strictly less than the divisor. Hence

$$\deg p(x) < \deg g(x)$$

(iv) If on division of a non-zero polynomial $p(x)$ by a polynomial $g(x)$ the remainder is zero, what is the relation between the degrees of $p(x)$ and $g(x)$?

Solution. Remainder zero means $p(x) = g(x)\,q(x)$ with $q(x) \ne 0$ (because $p$ is non-zero). Degrees add on multiplication, so

$$\deg p(x) = \deg g(x) + \deg q(x) \ \ge\ \deg g(x)$$

that is, $\deg g(x) \le \deg p(x)$. Equality happens exactly when $q(x)$ is a non-zero constant.

(v) Can the quadratic polynomial $x^2 + kx + k$ have equal zeroes for some odd integer $k > 1$?

Solution. No. Equal zeroes require the discriminant to vanish:

$$k^2 – 4k = 0 \implies k(k-4) = 0 \implies k = 0 \text{ or } k = 4$$

Those are the only two values of $k$ that work, and neither is an odd integer greater than $1$.

Exercise 2.2, Question 2

Are the following statements “True” or “False”? Justify your answers.

(i) If the zeroes of a quadratic polynomial $ax^2+bx+c$ are both positive, then $a$, $b$ and $c$ all have the same sign.

False. If both zeroes are positive then their sum $-\dfrac{b}{a}$ is positive, so $\dfrac{b}{a} < 0$ — meaning $a$ and $b$ have opposite signs. (Their product $\dfrac{c}{a} > 0$ does put $a$ and $c$ on the same side, but $b$ is on the other.) Example: $x^2 – 5x + 6$ has zeroes $2$ and $3$, and its coefficients are $+1$, $-5$, $+6$.

(ii) If the graph of a polynomial intersects the $x$-axis at only one point, it cannot be a quadratic polynomial.

False. A quadratic with equal zeroes touches the axis at exactly one point — $x^2 – 4x + 4 = (x-2)^2$ meets it only at $x = 2$. “One intersection point” is perfectly possible for a quadratic; it is “three or more” that is impossible.

(iii) If the graph of a polynomial intersects the $x$-axis at exactly two points, it need not be a quadratic polynomial.

True. Two crossings tell you the polynomial has two distinct real zeroes; they say nothing about its degree. The quartic $x^4 – 1 = (x^2-1)(x^2+1)$ meets the axis only at $x = \pm 1$, because $x^2+1$ contributes no real zero at all. The cubic $x^3 – x^2 = x^2(x-1)$ meets it at exactly two points as well, $x = 0$ and $x = 1$. Neither is quadratic.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

True. Let the zeroes of $ax^3+bx^2+cx+d$ be $0$, $0$, $\gamma$. Then

$$\frac{c}{a} = 0\cdot 0 + 0\cdot\gamma + \gamma\cdot 0 = 0 \implies c = 0$$ $$-\frac{d}{a} = 0\cdot 0\cdot \gamma = 0 \implies d = 0$$

so the polynomial is $ax^3 + bx^2 = x^2(ax+b)$ — no linear term and no constant term.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

True. Write the zeroes as $-\alpha$, $-\beta$, $-\gamma$ with $\alpha, \beta, \gamma > 0$. Then

$$p(x) = a(x+\alpha)(x+\beta)(x+\gamma) = a\left[x^3 + (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x + \alpha\beta\gamma\right]$$

Every bracketed coefficient is a sum or product of positive numbers, hence positive, so all four coefficients are $a$ times something positive — all the same sign as $a$.

(vi) If all three zeroes of a cubic polynomial $x^3 + ax^2 – bx + c$ are positive, then at least one of $a$, $b$ and $c$ is non-negative.

False. Let the zeroes be $p, q, r > 0$ and read off all three relations, being careful with the printed signs of this particular polynomial:

$$p+q+r = -a > 0 \implies a < 0$$ $$pq+qr+rp = -b > 0 \implies b < 0$$ $$pqr = -c > 0 \implies c < 0$$

All three are strictly negative, so none of them is non-negative and the statement fails. This part is the mirror image of part (v), and the $-b$ printed in the polynomial is what makes the conclusion come out the other way.

(vii) The only value of $k$ for which the quadratic polynomial $kx^2 + x + k$ has equal zeroes is $\dfrac{1}{2}$.

False. Equal zeroes need the discriminant to vanish:

$$1^2 – 4(k)(k) = 0 \implies 4k^2 = 1 \implies k = \pm\frac{1}{2}$$

Both $\frac12$ and $-\frac12$ work, so $\frac12$ is not the only value. The word “only” is what makes the statement false; drop it and it would be true.

Common mistakes

• Dropping the minus in $-(\text{sum})$ in Question 2. Option (D), $2x^2+2x-24$, has zeroes $-4$ and $3$ — the given numbers with their signs flipped. It is placed precisely to catch this.
• Reaching for the wrong relation in Question 5. The product of the other two zeroes comes from $\alpha\beta+\beta\gamma+\gamma\alpha = c/a$, not from the triple product. The triple product gives $0 = -d/a$, which is true and useless.
• Computing the roots in Question 7. The signs of the sum and product settle it in two lines; the actual roots are irrational and take much longer to find.
• Reading Question 8 as asking what is possible. It asks what is impossible. Both zeroes negative happens often ($k = 5$); both positive never happens, because the sum and the product would need $k$ to be negative and positive at once.
• Missing that “only one intersection” includes a touch in 2.2(ii). A repeated zero meets the axis without crossing it, and that is still one point of intersection.
• Forgetting the printed minus sign in 2.2(vi). The polynomial is $x^3+ax^2-bx+c$, so the middle relation reads $pq+qr+rp = -b$, and $b$ comes out negative. Using $+b$ by habit reverses the answer.
• Treating “the only value” as decoration in 2.2(vii). $k = \frac12$ genuinely does give equal zeroes; the statement is false only because $k = -\frac12$ does too.

Practise next

• Exemplar Exercises 2.3 and 2.4 — the same relations used to find zeroes and to build polynomials from them, including the irrational-coefficient cases.
• NCERT textbook Exercise 2.1 — counting zeroes from a graph, which is Question 11 above as the textbook sets it.
• NCERT textbook Exercise 2.2 — verifying the sum-and-product relations, and constructing a quadratic from a given sum and product.
• Exemplar Exercise 1.1 if the sign reasoning in Questions 7 to 10 felt unfamiliar; it drills the same “read the answer off the structure” habit on Real Numbers.