Polynomials

Polynomials

Polynomials usually carries 3 to 5 marks in the board paper: 5 in the 2026-27 sample paper, 4 in the 2025 paper and in the second 2026 paper, and 3 in the main 2026 paper.

The marks come as one or two 1-mark MCQs (often reading the number of zeroes from a graph), sometimes an Assertion-Reason item, and a 2- or 3-mark question on the relationship between zeroes and coefficients. The board has not set a 5-mark or case-study question from this chapter, so the case-based questions here are for practice in the same ideas.

Where marks are usually lost:

• sign slips in $-\frac{b}{a}$, especially when $b$ is itself negative;
• writing $x^2 + (\alpha + \beta)x + \alpha\beta$ instead of $x^2 - (\alpha + \beta)x + \alpha\beta$;
• using $(\alpha + \beta)^2 + 2\alpha\beta$ for $\alpha^2 + \beta^2$;
• counting the point on the $y$-axis as a zero when reading a graph.

Revise in 5 Minutes

Key facts

• A zero of $p(x)$ is a number $k$ with $p(k) = 0$.
• Geometrically, the zeroes of $y = p(x)$ are the $x$-coordinates of the points where the graph meets the $x$-axis. The point on the $y$-axis is $(0, p(0))$, not a zero.
• A polynomial of degree $n$ has at most $n$ zeroes. A quadratic has 0, 1 or 2 real zeroes.
• The graph of $ax^2 + bx + c$ is a parabola: it opens upwards if $a > 0$ and downwards if $a < 0$. It cuts the $y$-axis at $(0, c)$.

Zeroes and coefficients of $ax^2 + bx + c$ (zeroes $\alpha$, $\beta$)

Forming a polynomial: $k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]$, $k \neq 0$. So there are infinitely many quadratics with the same zeroes.

Useful forms

Quick checks

• Zeroes equal in size, opposite in sign: $b = 0$.
• Zeroes reciprocal to each other: $a = c$.
• Both zeroes negative (when the zeroes are real): sum negative and product positive.

Traps

• Losing the minus in $-\frac{b}{a}$ when $b$ is negative.
• Writing $x^2 + (\alpha + \beta)x + \alpha\beta$.
• Forgetting that $k$ in $kx^2 + \dots$ cannot be 0.
• Splitting the middle term wrongly with surds: check by expanding back.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

The zeroes of the polynomial $x^2 + (a + 1)x + b$ are 3 and $-5$. Find the values of $a$ and $b$.

Answer.

1. Sum of zeroes $= 3 + (-5) = -2 = -(a + 1)$, so $a = 1$ — 1 mark
2. Product of zeroes $= 3 \times (-5) = -15 = b$, so $b = -15$ — 1 mark

Question 2 (2 marks)

Find the zeroes of the polynomial $2x^2 - 3\sqrt{2}x + 2$ and check that their product is equal to $\frac{c}{a}$.

Answer.

1. $2x^2 - 2\sqrt{2}x - \sqrt{2}x + 2 = 2x(x - \sqrt{2}) - \sqrt{2}(x - \sqrt{2}) = (2x - \sqrt{2})(x - \sqrt{2})$ — 1 mark
2. Zeroes $\frac{\sqrt{2}}{2}$ (that is, $\frac{1}{\sqrt{2}}$) and $\sqrt{2}$; product $= \frac{\sqrt{2}}{2} \times \sqrt{2} = 1 = \frac{2}{2} = \frac{c}{a}$ — 1 mark

Question 3 (2 marks)

Aman was asked to write a quadratic polynomial whose zeroes are 5 and $-2$. He wrote $x^2 + 3x - 10$. Is he correct? If not, write a correct polynomial.

Answer.

Model answer:

No, Aman is not correct.

His polynomial factorises as $x^2 + 3x - 10 = (x + 5)(x - 2)$, so its zeroes are $-5$ and 2. He has added the sum of the zeroes in the middle term, where it should be subtracted.

For zeroes 5 and $-2$: sum $= 5 + (-2) = 3$ and product $= 5 \times (-2) = -10$.

So a correct polynomial is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - 3x - 10$. Check: $x^2 - 3x - 10 = (x - 5)(x + 2)$.

Marking scheme:

1. $x^2 + 3x - 10 = (x + 5)(x - 2)$ has zeroes $-5$ and 2, so Aman is not correct (his sum of zeroes is $-3$, not 3) — 1 mark
2. Sum $= 3$, product $= -10$; correct polynomial $x^2 - 3x - 10$ (or any non-zero multiple of it) — 1 mark

Question 4 (2 marks)

If $\alpha$ and $\beta$ are the zeroes of the polynomial $3x^2 + kx - 6$ and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{2}{3}$, find the value of $k$.

Answer.

1. $\alpha + \beta = -\frac{k}{3}$, $\alpha\beta = -2$; $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{k}{6}$ — 1 mark
2. $\frac{k}{6} = \frac{2}{3}$, so $k = 4$ — 1 mark

Question 5 (2 marks)

If $-4$ is one of the zeroes of the polynomial $3x^2 + px - 8$, find the value of $p$ and the other zero.

Answer.

1. $3(-4)^2 + p(-4) - 8 = 0$, so $48 - 4p - 8 = 0$ and $p = 10$ — 1 mark
2. Product of zeroes $= -\frac{8}{3}$, so other zero $= -\frac{8}{3} \div (-4) = \frac{2}{3}$ — 1 mark

Question 6 (3 marks)

Find the zeroes of the quadratic polynomial $6x^2 + x - 12$ and verify the relationship between the zeroes and the coefficients.

Answer.

1. $6x^2 + 9x - 8x - 12 = 3x(2x + 3) - 4(2x + 3) = (3x - 4)(2x + 3)$ — 1 mark
2. Zeroes $\frac{4}{3}$ and $-\frac{3}{2}$ — 1 mark
3. Sum $= \frac{4}{3} - \frac{3}{2} = -\frac{1}{6} = -\frac{b}{a}$; product $= \frac{4}{3} \times \left(-\frac{3}{2}\right) = -2 = \frac{-12}{6} = \frac{c}{a}$ — 1 mark

Question 7 (3 marks)

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 5x + 3$, find a quadratic polynomial whose zeroes are $\alpha + 2$ and $\beta + 2$.

Answer.

1. $\alpha + \beta = 5$ and $\alpha\beta = 3$ — 1 mark
2. New sum $= (\alpha + \beta) + 4 = 9$; new product $= \alpha\beta + 2(\alpha + \beta) + 4 = 3 + 10 + 4 = 17$ — 1 mark
3. Required polynomial $x^2 - 9x + 17$ — 1 mark

Question 8 (3 marks)

If $\alpha$ and $\beta$ are the real zeroes of the polynomial $kx^2 + 3x - 3$ such that $\alpha^2 + \beta^2 = 3$, find the value of $k$.

Answer.

Model answer:

Here $a = k$, $b = 3$, $c = -3$, so $\alpha + \beta = -\frac{3}{k}$ and $\alpha\beta = -\frac{3}{k}$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{9}{k^2} + \frac{6}{k}$. Setting this equal to 3 and multiplying by $k^2$: $9 + 6k = 3k^2$, that is, $k^2 - 2k - 3 = 0$.

So $(k - 3)(k + 1) = 0$, which gives $k = 3$ or $k = -1$.

Now check that the zeroes are real, as the question says. The zeroes of $kx^2 + 3x - 3$ are real only when $b^2 - 4ac = 9 + 12k \geq 0$.

For $k = 3$: $9 + 36 = 45 > 0$, so the zeroes are real (the polynomial is $3x^2 + 3x - 3$). For $k = -1$: $9 - 12 = -3 < 0$, so $-x^2 + 3x - 3$ has no real zeroes, and this value is rejected.

Hence $k = 3$.

Marking scheme:

1. $\alpha + \beta = -\frac{3}{k}$, $\alpha\beta = -\frac{3}{k}$; $\alpha^2 + \beta^2 = \frac{9}{k^2} + \frac{6}{k} = 3$ — 1 mark
2. $9 + 6k = 3k^2$, so $k^2 - 2k - 3 = 0$ and $(k - 3)(k + 1) = 0$ — 1 mark
3. Real zeroes need $9 + 12k \geq 0$: $k = 3$ gives 45 (real zeroes), $k = -1$ gives $-3$ (no real zeroes); so $k = 3$ — 1 mark

Question 9 (3 marks)

Verify that 2 and $-\frac{1}{3}$ are the zeroes of the polynomial $3x^2 - 5x - 2$. Hence find a quadratic polynomial whose zeroes are the reciprocals of these zeroes.

Answer.

1. $p(2) = 12 - 10 - 2 = 0$ and $p\left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{5}{3} - 2 = 0$; so both are zeroes — 1 mark
2. Reciprocals $\frac{1}{2}$ and $-3$: sum $= -\frac{5}{2}$, product $= -\frac{3}{2}$ — 1 mark
3. $x^2 + \frac{5}{2}x - \frac{3}{2}$, or $2x^2 + 5x - 3$ — 1 mark

Question 10 (3 marks)

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 7x + 11$ and $\alpha > \beta$, find the value of $\alpha - \beta$ and of $\alpha^2 - \beta^2$.

Answer.

1. $\alpha + \beta = 7$, $\alpha\beta = 11$; $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 49 - 44 = 5$ — 1 mark
2. Since $\alpha > \beta$, $\alpha - \beta = \sqrt{5}$ — 1 mark
3. $\alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = 7\sqrt{5}$ — 1 mark

Long Answer and Case-Based Questions

Question 11 (4 marks)

The entrance gate of a school in Mysuru has an arch in the shape of a parabola, as shown in the figure. Taking the ground as the $x$-axis and a vertical pole as the $y$-axis, the arch is the graph of $p(x) = -x^2 + 2x + 8$, where $x$ and $y$ are in metres. The arch meets the ground at A and B.

(i) Find the zeroes of $p(x)$. (1 mark)

Answer.

1. $-x^2 + 2x + 8 = -(x^2 - 2x - 8) = -(x - 4)(x + 2)$; zeroes $-2$ and $4$ — 1 mark

(ii) How wide is the arch at ground level, that is, what is the distance AB? (1 mark)

Answer.

1. A and B are $(-2, 0)$ and $(4, 0)$, so AB $= 4 - (-2) = 6$ m — 1 mark

(iii) The highest point of the arch is exactly above the mid-point of AB. Find the height of the arch at this point. (2 marks)

Answer.

1. Mid-point of AB: $x = \frac{-2 + 4}{2} = 1$ — 1 mark
2. Height $= p(1) = -1 + 2 + 8 = 9$ m — 1 mark

OR

(iii) A smaller arch over a side gate meets the ground 2 m on either side of the $y$-axis, and its highest point is 4 m above the ground, on the $y$-axis. Find the quadratic polynomial whose graph is this arch. (2 marks)

Answer.

1. Zeroes $-2$ and $2$, so $q(x) = k(x + 2)(x - 2) = k(x^2 - 4)$ — 1 mark
2. $q(0) = -4k = 4$ gives $k = -1$; so $q(x) = -x^2 + 4$ — 1 mark

Question 12 (4 marks)

A small bakery in Indore sells cakes. The owner finds that if $x$ hundred cakes are made and sold in a week, the profit (in thousands of rupees) is $P(x) = -x^2 + 8x - 12$. A negative value of $P(x)$ means a loss, and the values of $x$ at which $P(x) = 0$ are called the break-even points.

(i) Find the zeroes of $P(x)$. How many cakes give the break-even points? (1 mark)

Answer.

1. $-x^2 + 8x - 12 = -(x - 2)(x - 6)$; zeroes 2 and 6, that is, 200 and 600 cakes — 1 mark

(ii) Find the profit when 400 cakes are sold in a week. (1 mark)

Answer.

1. $P(4) = -16 + 32 - 12 = 4$, so the profit is ₹ 4000 — 1 mark

(iii) Verify the relationship between the zeroes and the coefficients of $P(x)$. (2 marks)

Answer.

1. Sum of zeroes $= 2 + 6 = 8$ and $-\frac{b}{a} = -\frac{8}{-1} = 8$ — 1 mark
2. Product of zeroes $= 2 \times 6 = 12$ and $\frac{c}{a} = \frac{-12}{-1} = 12$; verified — 1 mark

OR

(iii) For the owner's second shop, the profit polynomial $Q(x)$ is quadratic, its break-even points are at 300 and 500 cakes, and $Q(0) = -30$. Find $Q(x)$. (2 marks)

Answer.

1. Zeroes 3 and 5, so $Q(x) = k(x - 3)(x - 5) = k(x^2 - 8x + 15)$ — 1 mark
2. $Q(0) = 15k = -30$ gives $k = -2$; $Q(x) = -2x^2 + 16x - 30$ — 1 mark

Question 13 (4 marks)

In a maths lab activity, Tanya plotted the graph of a quadratic polynomial $y = p(x)$ on graph paper, as shown in the figure.

(i) How many zeroes does $p(x)$ have? (1 mark)

Answer.

1. The graph meets the $x$-axis at two points, so $p(x)$ has 2 zeroes — 1 mark

(ii) Write the zeroes of $p(x)$. (1 mark)

Answer.

1. The graph meets the $x$-axis at $(-1, 0)$ and $(3, 0)$; zeroes $-1$ and 3 — 1 mark

(iii) The graph cuts the $y$-axis at $(0, -3)$. Find $p(x)$. (2 marks)

Answer.

1. $p(x) = k(x + 1)(x - 3) = k(x^2 - 2x - 3)$ — 1 mark
2. $p(0) = -3k = -3$ gives $k = 1$; $p(x) = x^2 - 2x - 3$ — 1 mark

OR

(iii) Find a quadratic polynomial whose zeroes are twice the zeroes of $p(x)$. (2 marks)

Answer.

1. New zeroes $-2$ and 6: sum $= 4$, product $= -12$ — 1 mark
2. Required polynomial $x^2 - 4x - 12$ — 1 mark

Question 14 (4 marks)

In a maths club game in Patna, Asha thinks of the quadratic polynomial $p(x)$ whose zeroes are $-1$ and 4, with the coefficient of $x^2$ equal to 1. Bala thinks of the polynomial $q(x) = 2x^2 - 4x - 6$.

(i) Write Asha's polynomial $p(x)$. (1 mark)

Answer.

1. Sum $= 3$, product $= -4$; $p(x) = x^2 - 3x - 4$ — 1 mark

(ii) Find the zeroes of Bala's polynomial $q(x)$. (1 mark)

Answer.

1. $q(x) = 2(x^2 - 2x - 3) = 2(x - 3)(x + 1)$; zeroes 3 and $-1$ — 1 mark

(iii) Which zero do $p(x)$ and $q(x)$ share? Find a quadratic polynomial whose zeroes are the two zeroes that are not shared. (2 marks)

Answer.

1. Shared zero: $-1$; the other zeroes are 4 and 3 — 1 mark
2. Sum $= 7$, product $= 12$; required polynomial $x^2 - 7x + 12$ — 1 mark

OR

(iii) If $\alpha$ and $\beta$ are the zeroes of Asha's polynomial $p(x)$, find the value of $\alpha^2\beta + \alpha\beta^2$ without finding the zeroes separately. (2 marks)

Answer.

1. $\alpha + \beta = 3$, $\alpha\beta = -4$; $\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$ — 1 mark
2. $= -4 \times 3 = -12$ — 1 mark