NCERT Class 10 Mathematics — Polynomials, Exercise 2.2. Both questions solved, all twelve parts.

This exercise tests the link between the zeroes of a quadratic polynomial and its coefficients. If $\alpha$ and $\beta$ are the zeroes of $ax^2 + bx + c$ (with $a \ne 0$), then

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

Question 1 finds the zeroes by factorising — usually by splitting the middle term — and checks both relations. Question 2 runs the other way: given the sum $S$ and product $P$ of the zeroes, the polynomial is

$$k\left(x^2 – Sx + P\right), \qquad k \ne 0.$$

Key insight. The two relations work in both directions. Going forwards (question 1) they are a check on your factorisation; going backwards (question 2) they are the answer, with no zeroes to find at all. That is why question 2 can be answered even when the polynomial it produces has no real zeroes.

Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^2 – 2x – 8$   (ii) $4s^2 – 4s + 1$   (iii) $6x^2 – 3 – 7x$

(iv) $4u^2 + 8u$   (v) $t^2 – 15$   (vi) $3x^2 – x – 4$

Solution. To split the middle term of $ax^2 + bx + c$, look for two numbers whose product is $ac$ and whose sum is $b$. The zeroes are then the values that make each factor zero. Always identify $a$, $b$, $c$ from the polynomial in standard form before checking.

Question 1 (i)

$a = 1$, $b = -2$, $c = -8$. We need two numbers with product $-8$ and sum $-2$: $-4$ and $2$.

$$x^2 – 2x – 8 = x^2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4)(x + 2)$$

The zeroes are $4$ and $-2$. Check:

$$4 + (-2) = 2 = -\frac{-2}{1} = -\frac{b}{a}, \qquad 4 \times (-2) = -8 = \frac{-8}{1} = \frac{c}{a}$$

Question 1 (ii)

$a = 4$, $b = -4$, $c = 1$. This is a perfect square:

$$4s^2 – 4s + 1 = (2s – 1)^2$$

Both factors vanish at $s = \frac12$, so the zeroes are $\frac12$ and $\frac12$ — the same zero twice, and it must be counted twice. Check:

$$\frac12 + \frac12 = 1 = -\frac{-4}{4}, \qquad \frac12 \times \frac12 = \frac14 = \frac{1}{4}$$

Question 1 (iii)

The terms are printed out of order. In standard form the polynomial is $6x^2 – 7x – 3$, so $a = 6$, $b = -7$, $c = -3$. We need product $ac = -18$ and sum $-7$: that is $-9$ and $2$.

$$6x^2 – 7x – 3 = 6x^2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1)(2x – 3)$$

The zeroes are $-\frac13$ and $\frac32$. Check:

$$-\frac13 + \frac32 = \frac{-2 + 9}{6} = \frac76 = -\frac{-7}{6}, \qquad -\frac13 \times \frac32 = -\frac12 = \frac{-3}{6}$$

Question 1 (iv)

$a = 4$, $b = 8$, $c = 0$. There is no constant term, so $u$ is a common factor:

$$4u^2 + 8u = 4u(u + 2)$$

The zeroes are $0$ and $-2$. Check:

$$0 + (-2) = -2 = -\frac{8}{4}, \qquad 0 \times (-2) = 0 = \frac{0}{4}$$

Question 1 (v)

$a = 1$, $b = 0$ (there is no $t$ term), $c = -15$. A difference of two squares:

$$t^2 – 15 = \left(t – \sqrt{15}\right)\left(t + \sqrt{15}\right)$$

The zeroes are $\sqrt{15}$ and $-\sqrt{15}$. Check:

$$\sqrt{15} + \left(-\sqrt{15}\right) = 0 = -\frac{0}{1}, \qquad \sqrt{15} \times \left(-\sqrt{15}\right) = -15 = \frac{-15}{1}$$

Question 1 (vi)

$a = 3$, $b = -1$, $c = -4$. We need product $ac = -12$ and sum $-1$: $-4$ and $3$.

$$3x^2 – x – 4 = 3x^2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (x + 1)(3x – 4)$$

The zeroes are $-1$ and $\frac43$. Check:

$$-1 + \frac43 = \frac13 = -\frac{-1}{3}, \qquad -1 \times \frac43 = -\frac43 = \frac{-4}{3}$$

Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) $\dfrac14, -1$   (ii) $\sqrt2, \dfrac13$   (iii) $0, \sqrt5$   (iv) $1, 1$   (v) $-\dfrac14, \dfrac14$   (vi) $4, 1$

Solution. If the zeroes are $\alpha$ and $\beta$, the polynomial $(x – \alpha)(x – \beta)$ expands to

$$x^2 – (\alpha + \beta)x + \alpha\beta = x^2 – Sx + P.$$

Any non-zero multiple $k(x^2 – Sx + P)$ has the same zeroes, so we are free to choose $k$ to clear fractions. Note the minus sign in front of the sum.

(i) $S = \frac14$, $P = -1$:

$$x^2 – \frac14 x – 1, \quad\text{and taking } k = 4: \quad 4x^2 – x – 4$$

Check: $-\frac{b}{a} = \frac14$ and $\frac{c}{a} = \frac{-4}{4} = -1$.

(ii) $S = \sqrt2$, $P = \frac13$:

$$x^2 – \sqrt2\,x + \frac13, \quad\text{and taking } k = 3: \quad 3x^2 – 3\sqrt2\,x + 1$$

Check: $-\frac{b}{a} = \frac{3\sqrt2}{3} = \sqrt2$ and $\frac{c}{a} = \frac13$.

(iii) $S = 0$, $P = \sqrt5$: the $x$ term disappears.

$$x^2 – 0 \cdot x + \sqrt5 = x^2 + \sqrt5$$

(iv) $S = 1$, $P = 1$:

$$x^2 – x + 1$$

(v) $S = -\frac14$, $P = \frac14$. Subtracting a negative sum gives a plus sign:

$$x^2 + \frac14 x + \frac14, \quad\text{and taking } k = 4: \quad 4x^2 + x + 1$$

(vi) $S = 4$, $P = 1$:

$$x^2 – 4x + 1$$

A note on (iii), (iv) and (v). If you try to check these by finding the zeroes, you will not find any: $x^2 + \sqrt5$ is always positive, and $x^2 – x + 1$ and $4x^2 + x + 1$ cannot be factorised over the real numbers either. The answers are still correct. The question asks for a polynomial whose zeroes have the given sum and product, and the relations $\alpha + \beta = -\frac{b}{a}$, $\alpha\beta = \frac{c}{a}$ still hold — the zeroes are just not real numbers. They are complex numbers, which you meet in Class 11, and their sum and product come out exactly as required.

Common mistakes

• Question 1(iii), reading the coefficients in printed order. Taking $b = -3$ and $c = -7$ from $6x^2 – 3 – 7x$ makes both checks fail. Rewrite it as $6x^2 – 7x – 3$ before doing anything else.
• Question 1(ii), giving only one zero. $(2s – 1)^2$ has the zero $\frac12$ twice. Listing it once makes the sum $\frac12$ instead of $1$, and the check against $-\frac{b}{a}$ then fails.
• Question 1(iv), cancelling $u$. Dividing $4u^2 + 8u$ by $u$ to “simplify” throws away the zero $u = 0$. Take $u$ out as a factor; never divide it away.
• Question 1(v), forgetting that $b = 0$. A missing $t$ term means $b = 0$, so the zeroes must add to $0$ — which is the quickest way to see they are $\pm\sqrt{15}$ and not, say, $\pm 15$.
• Question 2, the sign of the sum. The polynomial is $x^2 – Sx + P$, not $x^2 + Sx + P$. With the wrong sign, 2(i) becomes $4x^2 + x – 4$, whose zeroes add to $-\frac14$.
• Question 2(iii)–(v), deciding the answer must be wrong. A polynomial that cannot be factorised has not been built incorrectly. Check it through $-\frac{b}{a}$ and $\frac{c}{a}$, which is what the question actually asks for.

Practise next

• Exercise 2.1 — reading the number of zeroes off a graph, the geometric side of the same idea.
• Chapter 4, Exercise 4.2 — solving quadratic equations by factorisation, which is question 1’s middle-term splitting used to solve equations.
• Class 11 Complex Numbers and Quadratic Equations, Exercise 4.1 — the complex numbers that turn out to be the zeroes of question 2(iii), (iv) and (v).