Polynomials Practice Questions with Solutions, Class 10

Ques­tion 2#

The graph of a cubic poly­no­mial crosses the x-axis at $(-4, 0)$, touches it at $(0, 0)$, and does not meet it any­where else. Write down the num­ber of zeroes and the zeroes.

Cross­ing and touch­ing both count as meet­ing the x-axis. The graph meets it at two points, with x-coor­di­nates $-4$ and $0$. (A poly­no­mial with this graph is $x^2(x + 4)$; the zero $0$ is repeated, which is why the graph only touches there.)

Answer: two zeroes, $-4$ and $0$.

Ques­tion 3#

A parabola opens upwards and its ver­tex is $(1, 0)$. How many zeroes does the poly­no­mial have, and what are they?

The ver­tex is the low­est point, and it lies on the x-axis. Every other point is above the axis, so the graph touches the axis only at $(1, 0)$. An exam­ple is $(x - 1)^2$.

Answer: one zero, $x = 1$ (two equal zeroes).

Why this works: for a parabola, the posi­tion of the ver­tex com­pared with the x-axis, together with the direc­tion of open­ing, decides the num­ber of zeroes.

Ques­tion 4#

Where does the graph of $y = 3x + 6$ meet the x-axis? Hence state the zero of $3x + 6$.

On the x-axis $y = 0$, so $3x + 6 = 0$, giv­ing $x = -2$. Check: $3(-2) + 6 = 0$.

Answer: at $(-2, 0)$; the zero is $-2$.

Find­ing zeroes and ver­i­fy­ing the rela­tion­ships#

Ques­tion 5#

Find the zeroes of $x^2 - 3x - 10$ and ver­ify the rela­tion­ship between the zeroes and the coef­fi­cients.

Two num­bers with prod­uct $-10$ and sum $-3$ are $-5$ and $2$.

$$x^2 - 3x - 10 = x^2 - 5x + 2x - 10 = (x - 5)(x + 2)$$

The zeroes are $5$ and $-2$. With $a = 1$, $b = -3$, $c = -10$:

$$\displaystyle \begin{aligned} 5 + (-2) &= 3 = -\frac{-3}{1} \\ 5 \times (-2) &= -10 = \frac{-10}{1} \end{aligned}$$

Answer: zeroes $5$ and $-2$; sum $\displaystyle 3 = -\frac{b}{a}$ and prod­uct $\displaystyle -10 = \frac{c}{a}$.

Ques­tion 6#

Find the zeroes of $5v^2 + 15v$ and ver­ify the rela­tion­ship.

Take out the com­mon fac­tor: $5v^2 + 15v = 5v(v + 3)$. So the zeroes are $0$ and $-3$. Here $a = 5$, $b = 15$ and $c = 0$ (there is no con­stant term).

$$\displaystyle \begin{aligned} 0 + (-3) &= -3 = -\frac{15}{5} \\ 0 \times (-3) &= 0 = \frac{0}{5} \end{aligned}$$

Answer: zeroes $0$ and $-3$; both rela­tion­ships hold.

Ques­tion 7#

Find the zeroes of $t^2 - 11$ and ver­ify the rela­tion­ship.

$t^2 - 11 = (t - \sqrt{11})(t + \sqrt{11})$, so the zeroes are $\sqrt{11}$ and $-\sqrt{11}$. Here $a = 1$, $b = 0$, $c = -11$.

$$\displaystyle \begin{aligned} \sqrt{11} + (-\sqrt{11}) &= 0 = -\frac{0}{1} \\ \sqrt{11} \times (-\sqrt{11}) &= -11 = \frac{-11}{1} \end{aligned}$$

Answer: zeroes $\pm\sqrt{11}$; sum $0$, prod­uct $-11$, match­ing the coef­fi­cients.

Ques­tion 8#

Find the zeroes of $12x^2 - 7x + 1$ and ver­ify the rela­tion­ship.

We need two num­bers with prod­uct $12 \times 1 = 12$ and sum $-7$: these are $-3$ and $-4$.

$$\begin{aligned} 12x^2 - 7x + 1 &= 12x^2 - 3x - 4x + 1 \\ &= 3x(4x - 1) - 1(4x - 1) \\ &= (3x - 1)(4x - 1) \end{aligned}$$

The zeroes are $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{4}$.

$$\displaystyle \begin{aligned} \frac{1}{3} + \frac{1}{4} &= \frac{7}{12} = -\frac{-7}{12} \\ \frac{1}{3} \times \frac{1}{4} &= \frac{1}{12} = \frac{1}{12} \end{aligned}$$

Answer: zeroes $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{4}$; sum $\displaystyle \frac{7}{12}$ and prod­uct $\displaystyle \frac{1}{12}$, as the coef­fi­cients pre­dict.

Form­ing qua­drat­ics and using the rela­tion­ships#

Ques­tion 9#

Find a qua­dratic poly­no­mial whose zeroes have sum $5$ and prod­uct $6$.

Using $x^2 - Sx + P$: $x^2 - 5x + 6$. Check: $x^2 - 5x + 6 = (x - 2)(x - 3)$, zeroes $2$ and $3$, sum $5$, prod­uct $6$.

Answer: $x^2 - 5x + 6$ (or any non-zero mul­ti­ple of it).

Ques­tion 10#

Find a qua­dratic poly­no­mial whose zeroes have sum $\displaystyle \frac{1}{3}$ and prod­uct $\displaystyle -\frac{2}{3}$. Find its zeroes.

$$\displaystyle x^2 - \frac{1}{3}x - \frac{2}{3}$$

Mul­ti­ply by $k = 3$: $3x^2 - x - 2$. To fac­torise, two num­bers with prod­uct $-6$ and sum $-1$ are $-3$ and $2$:

$$3x^2 - x - 2 = 3x^2 - 3x + 2x - 2 = (3x + 2)(x - 1)$$

The zeroes are $1$ and $\displaystyle -\frac{2}{3}$. Check: sum $\displaystyle = \frac{1}{3}$, prod­uct $\displaystyle = -\frac{2}{3}$.

Answer: $3x^2 - x - 2$; zeroes $1$ and $\displaystyle -\frac{2}{3}$.

Ques­tion 11#

Find a qua­dratic poly­no­mial with zeroes $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

$$\begin{aligned} S &= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \\ P &= (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \end{aligned}$$

So the poly­no­mial is $x^2 - 4x + 1$.

Answer: $x^2 - 4x + 1$.

Why this works: the irra­tional parts can­cel in both the sum and the prod­uct, so the coef­fi­cients are inte­gers.

Ques­tion 12#

One zero of $2x^2 - 9x + k$ is twice the other. Find $k$ and both zeroes.

Let the zeroes be $\alpha$ and $2\alpha$. From the sum,

$$\displaystyle \alpha + 2\alpha = -\frac{-9}{2} \ \Rightarrow\ 3\alpha = \frac{9}{2} \ \Rightarrow\ \alpha = \frac{3}{2}$$

From the prod­uct,

$$\displaystyle \alpha \times 2\alpha = \frac{k}{2} \ \Rightarrow\ 2\left(\frac{3}{2}\right)^2 = \frac{k}{2} \ \Rightarrow\ \frac{9}{2} = \frac{k}{2} \ \Rightarrow\ k = 9$$

Check: $2x^2 - 9x + 9 = (2x - 3)(x - 3)$, with zeroes $\displaystyle \frac{3}{2}$ and $3$, and $3$ is twice $\displaystyle \frac{3}{2}$.

Answer: $k = 9$; the zeroes are $\displaystyle \frac{3}{2}$ and $3$.

Ques­tion 13#

If $\alpha$ and $\beta$ are the zeroes of $2x^2 - 5x + 3$, find (a) $\alpha^2 + \beta^2$ and (b) $\displaystyle \frac{1}{\alpha} + \frac{1}{\beta}$ with­out find­ing the zeroes. Then check by find­ing them.

$\displaystyle \alpha + \beta = \frac{5}{2}$ and $\displaystyle \alpha\beta = \frac{3}{2}$.

$$\displaystyle \begin{aligned} \text{(a)}\ \alpha^2 + \beta^2 &= \left(\frac{5}{2}\right)^2 - 2 \times \frac{3}{2} = \frac{25}{4} - 3 = \frac{13}{4} \\ \text{(b)}\ \frac{1}{\alpha} + \frac{1}{\beta} &= \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{3/2} = \frac{5}{3} \end{aligned}$$

Check: $2x^2 - 5x + 3 = (x - 1)(2x - 3)$, so the zeroes are $1$ and $\displaystyle \frac{3}{2}$. Then $\displaystyle 1 + \frac{9}{4} = \frac{13}{4}$ and $\displaystyle 1 + \frac{2}{3} = \frac{5}{3}$.

Answer: (a) $\displaystyle \frac{13}{4}$, (b) $\displaystyle \frac{5}{3}$.

Cubic poly­no­mi­als#

Ques­tion 14#

Ver­ify that $1$, $-2$ and $3$ are zeroes of $p(x) = x^3 - 2x^2 - 5x + 6$, and ver­ify the rela­tion­ship between the zeroes and the coef­fi­cients.

$$\begin{aligned} p(1) &= 1 - 2 - 5 + 6 = 0 \\ p(-2) &= -8 - 8 + 10 + 6 = 0 \\ p(3) &= 27 - 18 - 15 + 6 = 0 \end{aligned}$$

So all three are zeroes. With $a = 1$, $b = -2$, $c = -5$, $d = 6$ and $\alpha = 1$, $\beta = -2$, $\gamma = 3$:

$$\displaystyle \begin{aligned} \alpha + \beta + \gamma &= 1 - 2 + 3 = 2 = -\frac{-2}{1} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= -2 - 6 + 3 = -5 = \frac{-5}{1} \\ \alpha\beta\gamma &= 1 \times (-2) \times 3 = -6 = -\frac{6}{1} \end{aligned}$$

Answer: all three val­ues give $p = 0$, and the sum $2$, pair­wise sum $-5$ and prod­uct $-6$ match $\displaystyle -\frac{b}{a}$, $\displaystyle \frac{c}{a}$ and $\displaystyle -\frac{d}{a}$.

Ques­tion 15#

Find a cubic poly­no­mial whose zeroes have sum $3$, sum of prod­ucts two at a time $-1$, and prod­uct $-3$. Show that $3$, $1$ and $-1$ are its zeroes.

Take $a = 1$. Then $-b = 3$, $c = -1$ and $-d = -3$, so $b = -3$, $c = -1$, $d = 3$:

$$p(x) = x^3 - 3x^2 - x + 3$$

Fac­torise by group­ing:

$$x^3 - 3x^2 - x + 3 = x^2(x - 3) - 1(x - 3) = (x - 3)(x - 1)(x + 1)$$

So the zeroes are $3$, $1$ and $-1$. Check: sum $3$; pairs $3 - 1 - 3 = -1$; prod­uct $-3$.

Answer: $x^3 - 3x^2 - x + 3$, with zeroes $3$, $1$, $-1$.

Ques­tion 16#

The zeroes of $x^3 - 6x^2 + 3x + 10$ are $a - b$, $a$ and $a + b$. Find $a$, $b$ and the zeroes.

Sum of zeroes:

$$\displaystyle (a - b) + a + (a + b) = 3a = -\frac{-6}{1} = 6 \ \Rightarrow\ a = 2$$

Prod­uct of zeroes:

$$\displaystyle \begin{aligned} (a - b)\,a\,(a + b) &= -\frac{10}{1} \\ 2(4 - b^2) &= -10 \\ 4 - b^2 &= -5 \\ b^2 &= 9 \ \Rightarrow\ b = \pm 3 \end{aligned}$$

Either sign gives the same three zeroes: $-1$, $2$, $5$. Check with the mid­dle rela­tion­ship: $\displaystyle (-1)(2) + (2)(5) + (5)(-1) = -2 + 10 - 5 = 3 = \frac{3}{1}$.

Answer: $a = 2$, $b = \pm 3$; the zeroes are $-1$, $2$ and $5$.

Why this works: writ­ing the zeroes as $a - b$, $a$, $a + b$ makes the $b$ terms can­cel in the sum, so $a$ comes out at once.

Key terms#

Zero of a poly­no­mial

A num­ber that makes the poly­no­mial equal to $0$.

x-inter­cept

A point where a graph meets the x-axis; its x-coor­di­nate is a zero.

Parabola

The graph of a qua­dratic poly­no­mial.

Ver­tex

The turn­ing point of a parabola.

Repeated zero

A zero that occurs twice, where the graph touches the x-axis instead of cross­ing it.

Lead­ing coef­fi­cient

The coef­fi­cient $a$ of the high­est power.

Sum and prod­uct of zeroes

$\displaystyle -\frac{b}{a}$ and $\displaystyle \frac{c}{a}$ for a qua­dratic.

Com­mon ques­tions#

How do I check my zeroes quickly in the exam?#

Add and mul­ti­ply them and com­pare with $\displaystyle -\frac{b}{a}$ and $\displaystyle \frac{c}{a}$. If either fails, recheck your fac­tori­sa­tion.

Does a touch­ing point count as one zero or two?#

When asked for the num­ber of zeroes from a graph, count the touch­ing point once. The poly­no­mial has a repeated zero there.

What if a coef­fi­cient is miss­ing?#

Take it as $0$. In $5v^2 + 15v$, $c = 0$, so the prod­uct of zeroes is $0$ and one zero must be $0$.

Is my qua­dratic wrong if it dif­fers from the answer key by a con­stant?#

No. $x^2 - 5x + 6$ and $2x^2 - 10x + 12$ have the same zeroes. Any non-zero mul­ti­ple is cor­rect unless the ques­tion fixes the lead­ing coef­fi­cient.

Where does the y-inter­cept come in?#

The graph meets the y-axis at $(0, p(0))$, and $p(0)$ is the con­stant term. It helps you sketch the graph but is not a zero unless it equals $0$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.