NCERT Exemplar Solutions Class 10 Maths Polynomials — Exercises 2.3 and 2.4

NCERT Exemplar Problems, Class 10 Mathematics — Polynomials, Exercises 2.3 and 2.4. All 16 questions solved.

Exercise 2.3 is ten polynomials to factorise; Exercise 2.4 is six longer questions that build on the same skill. They are taken together because Exercise 2.4 has only six questions, and because the second exercise is unreadable without the first.

One method does nearly all of it: splitting the middle term. To factorise $ax^2+bx+c$, find two numbers whose product is $ac$ and whose sum is $b$, split $bx$ into those two pieces, and group. Half of Exercise 2.3 has surds in the coefficients, which changes nothing about the method — the two numbers are simply surds too, and the product $ac$ comes out rational.

A note on the current syllabus. Exercise 2.4 Questions 3 to 6 all turn on dividing one polynomial by another. That topic is no longer in the Class 10 textbook: the chapter now ends at the relationship between zeroes and coefficients, and the words “quotient” and “remainder” do not appear in it. Questions 1 and 2 are squarely on syllabus. The division questions are still worth doing — the factor theorem behind them is assumed throughout Class 11 — but they will not come up in a board paper.

Key insight. Every question here is the same exchange in one direction or the other: zeroes ⟷ coefficients. Exercise 2.3 goes from coefficients to zeroes by factorising; Exercise 2.4 Question 1 goes the other way, building the polynomial from the sum and product. Knowing which direction you are travelling tells you which formula to reach for.

Exercise 2.3 — Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials.

Exercise 2.3, Question 1

$$4x^2 – 3x – 1$$

Solution. Here $ac = 4 \times (-1) = -4$ and $b = -3$. Two numbers with product $-4$ and sum $-3$: $-4$ and $+1$.

$$4x^2 – 4x + x – 1 = 4x(x – 1) + 1(x – 1) = (x-1)(4x+1)$$

The zeroes are $x = 1$ and $x = -\dfrac14$.

Verification. Sum $= 1 – \frac14 = \frac34$ and $-\dfrac{b}{a} = \dfrac{3}{4}$ ✓. Product $= 1 \times \left(-\frac14\right) = -\frac14$ and $\dfrac{c}{a} = -\dfrac14$ ✓.

Exercise 2.3, Question 2

$$3x^2 + 4x – 4$$

Solution. $ac = -12$, $b = 4$; the numbers are $6$ and $-2$.

$$3x^2 + 6x – 2x – 4 = 3x(x+2) – 2(x+2) = (x+2)(3x-2)$$

Zeroes: $x = -2$ and $x = \dfrac23$.

Verification. Sum $= -2 + \frac23 = -\frac43 = -\dfrac{b}{a}$ ✓. Product $= -\frac43 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 3

$$5t^2 + 12t + 7$$

Solution. $ac = 35$, $b = 12$; the numbers are $7$ and $5$.

$$5t^2 + 7t + 5t + 7 = t(5t+7) + 1(5t+7) = (5t+7)(t+1)$$

Zeroes: $t = -\dfrac75$ and $t = -1$.

Verification. Sum $= -\frac75 – 1 = -\frac{12}{5} = -\dfrac{b}{a}$ ✓. Product $= \frac75 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 4

$$t^3 – 2t^2 – 15t$$

Solution. This one is a cubic, and the first move is to take out the common factor of $t$ — which immediately hands you one zero.

$$t^3 – 2t^2 – 15t = t\left(t^2 – 2t – 15\right) = t(t-5)(t+3)$$

since $-15 = (-5)(3)$ and $-5 + 3 = -2$.

Zeroes: $t = 0$, $t = 5$, $t = -3$.

Verification against the cubic relations, with $a = 1$, $b = -2$, $c = -15$, $d = 0$:

$$\text{Sum} = 0 + 5 – 3 = 2 = -\frac{b}{a} \ ✓$$ $$\text{Sum of pairs} = (0)(5) + (5)(-3) + (-3)(0) = -15 = \frac{c}{a} \ ✓$$ $$\text{Product} = 0 = -\frac{d}{a} \ ✓$$

Exercise 2.3, Question 5

$$2x^2 + \frac{7}{2}x + \frac{3}{4}$$

Solution. Clear the fractions first — multiplying by $4$ changes the polynomial but not its zeroes, because a non-zero constant multiple vanishes exactly where the original does.

$$4\left(2x^2 + \frac72 x + \frac34\right) = 8x^2 + 14x + 3$$

Now $ac = 24$, $b = 14$; the numbers are $12$ and $2$.

$$8x^2 + 12x + 2x + 3 = 4x(2x+3) + 1(2x+3) = (2x+3)(4x+1)$$

Zeroes: $x = -\dfrac32$ and $x = -\dfrac14$.

Verification against the original coefficients $a = 2$, $b = \frac72$, $c = \frac34$: sum $= -\frac32 – \frac14 = -\frac74 = -\dfrac{b}{a}$ ✓, product $= \frac38 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 6

$$4x^2 + 5\sqrt{2}\,x – 3$$

Solution. The surd does not change the method. We need two numbers with product $ac = 4(-3) = -12$ and sum $5\sqrt2$. Try multiples of $\sqrt2$: $6\sqrt2$ and $-\sqrt2$ give product $-6 \times 2 = -12$ ✓ and sum $5\sqrt2$ ✓.

$$4x^2 + 6\sqrt2\,x – \sqrt2\,x – 3 = 2x\left(2x + 3\sqrt2\right) – \frac{\sqrt2}{2}\left(2x + 3\sqrt2\right)$$

(the second grouping works because $\frac{\sqrt2}{2}\times 3\sqrt2 = 3$), so

$$= \left(2x + 3\sqrt2\right)\left(2x – \frac{\sqrt2}{2}\right)$$

Zeroes: $2x = -3\sqrt2 \implies x = -\dfrac{3\sqrt2}{2}$, and $2x = \dfrac{\sqrt2}{2} \implies x = \dfrac{\sqrt2}{4}$.

Verification. Sum $= \frac{\sqrt2}{4} – \frac{3\sqrt2}{2} = -\frac{5\sqrt2}{4} = -\dfrac{b}{a}$ ✓. Product $= -\frac{3 \times 2}{8} = -\frac34 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 7

$$2s^2 – \left(1 + 2\sqrt{2}\right)s + \sqrt{2}$$

Solution. The middle coefficient has already been split for you — the bracket $\left(1 + 2\sqrt2\right)$ is the sum of the two pieces. Check the product: $ac = 2\sqrt2$, and $(-1)\left(-2\sqrt2\right) = 2\sqrt2$ ✓.

$$2s^2 – 2\sqrt2\,s – s + \sqrt2 = 2s\left(s – \sqrt2\right) – 1\left(s – \sqrt2\right) = \left(s – \sqrt2\right)\left(2s – 1\right)$$

Zeroes: $s = \sqrt2$ and $s = \dfrac12$.

Verification. Sum $= \sqrt2 + \frac12 = \dfrac{1+2\sqrt2}{2} = -\dfrac{b}{a}$ ✓. Product $= \dfrac{\sqrt2}{2} = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 8

$$v^2 + 4\sqrt{3}\,v – 15$$

Solution. $ac = -15$ and $b = 4\sqrt3$. Two surds whose product is $-15$ and sum $4\sqrt3$: $5\sqrt3$ and $-\sqrt3$, since $5\sqrt3 \times \left(-\sqrt3\right) = -5 \times 3 = -15$ ✓.

$$v^2 + 5\sqrt3\,v – \sqrt3\,v – 15 = v\left(v + 5\sqrt3\right) – \sqrt3\left(v + 5\sqrt3\right) = \left(v + 5\sqrt3\right)\left(v – \sqrt3\right)$$

Zeroes: $v = -5\sqrt3$ and $v = \sqrt3$.

Verification. Sum $= -4\sqrt3 = -\dfrac{b}{a}$ ✓. Product $= -5 \times 3 = -15 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 9

$$y^2 + \frac{3}{2}\sqrt{5}\,y – 5$$

Solution. Clear the fraction by doubling: $2y^2 + 3\sqrt5\,y – 10$. Now $ac = -20$ and $b = 3\sqrt5$; the numbers are $4\sqrt5$ and $-\sqrt5$, whose product is $-4 \times 5 = -20$ ✓.

$$2y^2 + 4\sqrt5\,y – \sqrt5\,y – 10 = 2y\left(y + 2\sqrt5\right) – \sqrt5\left(y + 2\sqrt5\right) = \left(y + 2\sqrt5\right)\left(2y – \sqrt5\right)$$

Zeroes: $y = -2\sqrt5$ and $y = \dfrac{\sqrt5}{2}$.

Verification against the original $a = 1$, $b = \frac32\sqrt5$, $c = -5$: sum $= -2\sqrt5 + \frac{\sqrt5}{2} = -\frac{3\sqrt5}{2} = -\dfrac{b}{a}$ ✓, product $= -2\sqrt5 \times \frac{\sqrt5}{2} = -5 = \dfrac{c}{a}$ ✓.

Exercise 2.3, Question 10

$$7y^2 – \frac{11}{3}y – \frac{2}{3}$$

Solution. Multiply through by $3$: $21y^2 – 11y – 2$. Then $ac = -42$, $b = -11$; the numbers are $-14$ and $3$.

$$21y^2 – 14y + 3y – 2 = 7y(3y – 2) + 1(3y-2) = (3y-2)(7y+1)$$

Zeroes: $y = \dfrac23$ and $y = -\dfrac17$.

Verification against the original $a = 7$, $b = -\frac{11}{3}$, $c = -\frac23$: sum $= \frac23 – \frac17 = \frac{11}{21} = -\dfrac{b}{a}$ ✓, product $= -\frac{2}{21} = \dfrac{c}{a}$ ✓.

Exercise 2.4

Exercise 2.4, Question 1

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.

The construction is the same every time: a quadratic with zero-sum $S$ and zero-product $P$ is

$$x^2 – Sx + P$$

and any non-zero multiple of it will do — which is how the fractions get cleared.

(i) $-\dfrac{8}{3}$, $\dfrac{4}{3}$

$$x^2 + \frac83 x + \frac43 \quad\longrightarrow\quad \times 3 \quad\longrightarrow\quad 3x^2 + 8x + 4$$

Splitting: $ac = 12$, $b = 8$, so $6$ and $2$.

$$3x^2 + 6x + 2x + 4 = 3x(x+2) + 2(x+2) = (x+2)(3x+2)$$

Zeroes: $-2$ and $-\dfrac23$.

(ii) $\dfrac{21}{8}$, $\dfrac{5}{16}$

$$x^2 – \frac{21}{8}x + \frac{5}{16} \quad\longrightarrow\quad \times 16 \quad\longrightarrow\quad 16x^2 – 42x + 5$$

$ac = 80$, $b = -42$, so $-40$ and $-2$.

$$16x^2 – 40x – 2x + 5 = 8x(2x-5) – 1(2x-5) = (2x-5)(8x-1)$$

Zeroes: $\dfrac52$ and $\dfrac18$.

(iii) $-2\sqrt{3}$, $-9$

$$x^2 + 2\sqrt3\,x – 9$$

$ac = -9$, $b = 2\sqrt3$; the numbers are $3\sqrt3$ and $-\sqrt3$, product $-3 \times 3 = -9$ ✓.

$$x^2 + 3\sqrt3\,x – \sqrt3\,x – 9 = x\left(x + 3\sqrt3\right) – \sqrt3\left(x + 3\sqrt3\right) = \left(x+3\sqrt3\right)\left(x – \sqrt3\right)$$

Zeroes: $-3\sqrt3$ and $\sqrt3$.

(iv) $-\dfrac{3}{2\sqrt{5}}$, $-\dfrac{1}{2}$

$$x^2 + \frac{3}{2\sqrt5}x – \frac12 \quad\longrightarrow\quad \times 10 \quad\longrightarrow\quad 10x^2 + \frac{30}{2\sqrt5}x – 5 = 10x^2 + 3\sqrt5\,x – 5$$

using $\dfrac{15}{\sqrt5} = 3\sqrt5$. Now $ac = -50$, $b = 3\sqrt5$; the numbers are $5\sqrt5$ and $-2\sqrt5$, product $-10 \times 5 = -50$ ✓.

$$10x^2 + 5\sqrt5\,x – 2\sqrt5\,x – 5 = 5x\left(2x + \sqrt5\right) – \sqrt5\left(2x + \sqrt5\right) = \left(2x + \sqrt5\right)\left(5x – \sqrt5\right)$$

Zeroes: $-\dfrac{\sqrt5}{2}$ and $\dfrac{\sqrt5}{5}$.

Exercise 2.4, Question 2

Given that the zeroes of the cubic polynomial $x^3 – 6x^2 + 3x + 10$ are of the form $a$, $a+b$, $a+2b$ for some real numbers $a$ and $b$, find the values of $a$ and $b$ as well as the zeroes of the given polynomial.

Solution. The three zeroes are in arithmetic progression, and that is the whole gift: the middle one can be found from the sum alone.

$$a + (a+b) + (a+2b) = 3(a+b) = -\frac{b_{\text{coeff}}}{a_{\text{coeff}}} = 6 \implies a + b = 2$$

So the middle zero is $2$. Check it: $p(2) = 8 – 24 + 6 + 10 = 0$ ✓. Now divide out $(x-2)$:

$$x^3 – 6x^2 + 3x + 10 = (x-2)\left(x^2 – 4x – 5\right) = (x-2)(x-5)(x+1)$$

The zeroes are $-1$, $2$, $5$ — which is indeed an arithmetic progression with common difference $3$.

Matching to the form $a$, $a+b$, $a+2b$ gives two valid readings, depending on which end you start from:

$$a = -1,\ b = 3 \quad (\text{giving } -1, 2, 5) \qquad\text{or}\qquad a = 5,\ b = -3 \quad (\text{giving } 5, 2, -1)$$

Exercise 2.4, Question 3

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3 + \sqrt{2}\,x^2 – 10x – 4\sqrt{2}$, find its other two zeroes.

Solution. A zero at $x = \sqrt2$ means $\left(x – \sqrt2\right)$ is a factor, so divide it out. Long division gives

$$6x^3 + \sqrt2\,x^2 – 10x – 4\sqrt2 = \left(x – \sqrt2\right)\left(6x^2 + 7\sqrt2\,x + 4\right)$$

(which you can confirm by expanding, and the remainder is exactly $0$). Now factorise the quadratic: $ac = 24$ and $b = 7\sqrt2$, so the numbers are $4\sqrt2$ and $3\sqrt2$, product $12 \times 2 = 24$ ✓.

$$6x^2 + 4\sqrt2\,x + 3\sqrt2\,x + 4 = 2x\left(3x + 2\sqrt2\right) + \sqrt2\left(3x + 2\sqrt2\right) = \left(3x + 2\sqrt2\right)\left(2x + \sqrt2\right)$$

The other two zeroes are

$$x = -\frac{2\sqrt2}{3} \qquad\text{and}\qquad x = -\frac{\sqrt2}{2}$$

Exercise 2.4, Question 4

Find $k$ so that $x^2 + 2x + k$ is a factor of $2x^4 + x^3 – 14x^2 + 5x + 6$. Also find all the zeroes of the two polynomials.

Solution. “Is a factor” means the division leaves no remainder, so carry out the division keeping $k$ as a letter and then force the remainder to vanish. Dividing gives

$$\text{quotient } = 2x^2 – 3x – (8 + 2k), \qquad \text{remainder } = (7k + 21)x + \left(2k^2 + 8k + 6\right)$$

A remainder of degree one is zero only when both its coefficients are zero:

$$7k + 21 = 0 \implies k = -3$$ $$2k^2 + 8k + 6 = 2(k+1)(k+3) = 0 \implies k = -1 \text{ or } k = -3$$

Only $k = -3$ satisfies both, so $k = -3$.

With $k = -3$ the quotient becomes $2x^2 – 3x – 2 = (x-2)(2x+1)$, and the divisor becomes $x^2 + 2x – 3 = (x+3)(x-1)$. Hence

$$2x^4 + x^3 – 14x^2 + 5x + 6 = (x+3)(x-1)(x-2)(2x+1)$$

• Zeroes of $2x^4+x^3-14x^2+5x+6$: $1$, $-3$, $2$, $-\dfrac12$.
• Zeroes of $x^2+2x-3$: $1$ and $-3$.

The second pair being a subset of the first is exactly what “is a factor” means, and is a free check on the whole calculation.

Exercise 2.4, Question 5

Given that $x – \sqrt{5}$ is a factor of the cubic polynomial $x^3 – 3\sqrt{5}\,x^2 + 13x – 3\sqrt{5}$, find all the zeroes of the polynomial.

Solution. Divide by the known factor:

$$x^3 – 3\sqrt5\,x^2 + 13x – 3\sqrt5 = \left(x – \sqrt5\right)\left(x^2 – 2\sqrt5\,x + 3\right)$$

The quadratic does not factorise over the rationals, so use the formula — or complete the square, which is quicker here:

$$x^2 – 2\sqrt5\,x + 3 = \left(x – \sqrt5\right)^2 – 5 + 3 = \left(x-\sqrt5\right)^2 – 2$$

Setting this to zero gives $\left(x – \sqrt5\right)^2 = 2$, so $x – \sqrt5 = \pm\sqrt2$.

$$x = \sqrt5 + \sqrt2 \qquad\text{and}\qquad x = \sqrt5 – \sqrt2$$

Exercise 2.4, Question 6

For which values of $a$ and $b$ are the zeroes of $q(x) = x^3 + 2x^2 + a$ also the zeroes of the polynomial $p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x + b$? Which zeroes of $p(x)$ are not the zeroes of $q(x)$?

Solution. “Every zero of $q$ is a zero of $p$” is, for these degrees, the same as saying $q(x)$ divides $p(x)$ exactly. So divide and kill the remainder.

Dividing $p(x)$ by $q(x)$ leaves

$$\text{remainder} = -(a+1)x^2 + 3(a+1)x + (b – 2a)$$

All three coefficients must vanish. The first two both give $a = -1$; substituting into the third,

$$b – 2(-1) = 0 \implies b = -2$$

So $a = -1$ and $b = -2$, making

$$q(x) = x^3 + 2x^2 – 1, \qquad p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x – 2$$

Factorising both:

$$q(x) = (x+1)\left(x^2 + x – 1\right)$$ $$p(x) = (x+1)\left(x^2+x-1\right)(x-1)(x-2) = q(x)\,(x-1)(x-2)$$

Every zero of $q$ is therefore a zero of $p$, as required, and the extra factor $(x-1)(x-2)$ supplies two more. The zeroes of $p$ that are not zeroes of $q$ are

$$x = 1 \qquad\text{and}\qquad x = 2$$

A misprint in the answer key. The printed key gives $a = -1$, $b = -2$ correctly, then says “1 and 2 are the zeroes of $q(x)$ which are not the zeroes of $p(x)$” — the statement the wrong way round. Neither $1$ nor $2$ is a zero of $q$: $q(1) = 2$ and $q(2) = 15$. And no zero of $q$ can fail to be a zero of $p$, because making every zero of $q$ a zero of $p$ is precisely what the first half of the question forced. The correct sentence is the one given above: $1$ and $2$ are zeroes of $p(x)$ that are not zeroes of $q(x)$.

Common mistakes

• Looking for rational splittings in Questions 6, 8 and 9 of Exercise 2.3. The two numbers are surds, but their product $ac$ is rational — that is the fact that makes the search finite. In Question 8, $ac = -15$ and the pair is $5\sqrt3$, $-\sqrt3$.
• Verifying the relations against the cleared polynomial instead of the original. In Questions 5, 9 and 10 the fractions are cleared to make the factorising easy, but the check must use the coefficients as printed. The zeroes are the same either way; the values of $-b/a$ and $c/a$ are not.
• Forgetting the third zero in Question 4 of Exercise 2.3. Taking out the common $t$ is factorising, and $t = 0$ is a zero. The verification against the cubic relations catches it immediately if it is dropped.
• Solving only one of the two remainder conditions in Exercise 2.4 Question 4. A linear remainder has two coefficients and both must vanish. The constant one alone admits $k = -1$, which is not a solution; checking both is what pins $k = -3$.
• Assuming a quadratic factor must factorise in Exercise 2.4 Question 5. $x^2 – 2\sqrt5x + 3$ has no rational splitting; completing the square gives $\left(x-\sqrt5\right)^2 = 2$ in one line.
• Reading the key’s sentence for Exercise 2.4 Question 6. It has $q$ and $p$ interchanged. Work out which polynomial has the extra factors — it is always the one of higher degree — and the direction is obvious.
• Treating “find a quadratic polynomial” in Exercise 2.4 Question 1 as having one answer. $x^2 – Sx + P$ and every non-zero multiple of it work. Clearing the fractions is choosing a convenient multiple, not correcting an error.

Practise next

• Exemplar Exercises 2.1 and 2.2 — the same sum-and-product relations used to reason about signs rather than to compute, which is how they are tested in the objective band.
• NCERT textbook Exercise 2.2 — the same “find the zeroes and verify the relations” drill with rational coefficients throughout, and the construction of Question 1 above.
• NCERT textbook Exercise 2.1 for the graphical reading of zeroes, worth doing alongside Question 4 of Exercise 2.3 where a cubic has three of them.
• Exemplar Exercises 1.3 and 1.4 — the irrational numbers appearing as zeroes here are the ones proved irrational there.