MathsCBSE Class 10 · Polynomials

Geometrical Meaning of Zeros of a Polynomial

Q: Does a point where the graph just touches the x-axis count as a zero?

Yes. The graph meets the x-axis there, so its x-coordinate is a zero. For a quadratic it means the two zeros are equal, so you count it once as a distinct zero.

Q: Can a quadratic polynomial have no zeros?

Yes. If the parabola lies entirely above or entirely below the x-axis, it never meets it, so there are no real zeros. An example is x squared plus 1.

Q: Do I need to draw graphs in the board exam for this topic?

You may be asked to draw a graph from a table of values, or to read zeros from a described graph. Practise making a neat table, plotting points and joining them with a smooth freehand curve.

Q: Is the division algorithm for polynomials in the 2026-27 syllabus?

No. The division algorithm for polynomials is not in the 2026-27 syllabus.

Q: Why can a polynomial of degree n have at most n zeros?

Its graph can meet the x-axis at most n times, so there can be at most n zeros. This lets you reject options such as a quadratic whose graph supposedly meets the x-axis three times.

Updated 13 October 2026 · Fact-checked

The zeros of a polynomial are the x-coordinates of the points where its graph meets the x-axis. A line meets it once, a parabola at most twice, and a cubic at most three times. Questions can ask you to count zeros from a described graph, read them from a plotted curve, or use them in case studies.

Understand Geometrical Meaning of Zeros of a Polynomial

A zero of $p (x)$ is a number $k$ with $p (k) = 0$ . Now think about the graph of $y = p (x)$ . Every point on it has the form $(x, p (x))$ . At a zero, the y-coordinate is $p (k) = 0$ , so the point $(k, 0)$ lies on the x-axis. That single observation is the whole idea: **the zeros of $p (x)$ are the x-coordinates of the points where the graph of $y = p (x)$ meets the x-axis.**

For a linear polynomial $a x + b$ with $a \neq = 0$ , the graph is a slanting straight line. It meets the x-axis exactly once, at $(- \frac{b}{a}, 0)$ . So a linear polynomial has exactly one zero.

For a quadratic $a x^{2} + b x + c$ with $a \neq = 0$ , the graph is a parabola. It opens upward if $a > 0$ and downward if $a < 0$ . Three cases are possible:

it cuts the x-axis at two points: two distinct zeros;
it touches the x-axis at one point: two equal zeros, one zero value;
it stays completely above or below the x-axis: no real zeros.

The parabola is symmetric about a vertical line called its axis. When it cuts the x-axis twice, this axis passes exactly midway between the two zeros, and the highest or lowest point (the vertex) lies on it.

For a cubic, the graph can meet the x-axis one, two or three times, never more. In general a polynomial of degree $n$ has **at most $n$ zeros**, because its graph can meet the x-axis at most $n$ times. Use this to reject impossible MCQ options.

Two cautions. First, only meetings with the x-axis count. The point where the graph crosses the y-axis is $(0, p (0))$ ; it gives the constant term, not a zero, unless it is the origin. Second, a reading from a hand-drawn graph is approximate, so it is good practice to confirm it algebraically by substituting or factorising. This links straight to the next topics: once you have the zeros, you can verify the sum and product rules, or build the polynomial back as $a (x - α) (x - β)$ .

Geometrical Meaning of Zeros of a Polynomial: formulas to remember

Geometrical meaning of a zero: $p (k) = 0 ⟺ (k, 0) lies on the graph of y = p (x)$; Zeros are x-coordinates of the points where the graph meets the x-axis, by cutting or touching.
Zero of a linear polynomial: $y = a x + b, a \neq = 0 meets the x -axis at (- \frac{b}{a}, 0)$; A line always gives exactly one zero.
Shape of a parabola: $y = a x^{2} + b x + c : a > 0 opens upward, a < 0 opens downward$; Match a polynomial to a described curve using the sign of a.
Maximum number of zeros: $de g p (x) = n \Rightarrow at most n zeros$; Linear: 1, quadratic: at most 2, cubic: at most 3.
y-intercept: $graph crosses the y -axis at (0, p (0)) = (0, c)$; Never count this point as a zero unless c = 0.
Axis of symmetry from the zeros: $x = \frac{α + β}{2}$; The vertex of the parabola lies on this line, midway between the two zeros.
Polynomial from graph points: $p (x) = a (x - α) (x - β)$; Read the zeros from where the graph meets the x-axis, then use one more point to find a.

How to solve Geometrical Meaning of Zeros of a Polynomial questions

Use this method for any question that describes or plots a graph and asks about zeros.

1Identify the type: a line (linear), a parabola (quadratic) or a curve with up to three x-axis meetings (cubic). Recall the maximum number of zeros.
2List only the points where the graph meets the x-axis, counting a touching point once.
3Ignore the y-axis crossing; note it separately as $(0, p (0))$ because it helps find unknowns.
4Write the zeros as the x-coordinates of those points, with signs.
5If you must draw the graph, make a table of at least 7 values of $x$ , plot, join with a smooth curve, and read where $y = 0$ .
6If the polynomial is asked, write $p (x) = a (x - α) (x - β)$ and use one more point (the y-intercept or the vertex) to find $a$ .
7Confirm by substituting each zero into $p (x)$ , or by factorising.

Geometrical Meaning of Zeros of a Polynomial in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Geometrical meaning of zeros

Suppose your school science club launches a water rocket from the playground. The teacher models its height above the ground, $h$ metres, after $t$ seconds as $h = - 5 t^{2} + 20 t$ . The graph of $h$ against $t$ is a downward parabola.

Using the zeros of $p (t) = - 5 t^{2} + 20 t$ , find when the rocket leaves and when it lands. Then find its greatest height and its height after 1 second.

Try it first, then show the solutionSolution

The rocket is on the ground when $h = 0$ , so we need the zeros of $p (t)$ .
$- 5 t^{2} + 20 t = - 5 t (t - 4) = 0$ , so $t = 0$ or $t = 4$ .
The graph meets the t-axis at $(0, 0)$ and $(4, 0)$ : launch at 0 s, landing at 4 s.
The axis of symmetry is midway: $t = \frac{0 + 4}{2} = 2$ .
Greatest height $= p (2) = - 5 (4) + 20 (2) = - 20 + 40 = 20$ m.
Height after 1 s $= p (1) = - 5 + 20 = 15$ m.

Answer: It leaves at 0 s, lands at 4 s, reaches a greatest height of 20 m at 2 s, and is 15 m high after 1 s.

Memory hook: Rocket on the ground means height zero: the zeros are take-off and landing.

Tricks to remember Geometrical Meaning of Zeros of a Polynomial

Short memory aids. Say each one aloud once, then use it on a question the same day.

X marks the zero
Zeros live on the X-axis only. The Y-axis gives you c, not a zero.
Use it when: Stops you counting the y-intercept in MCQs.
Degree = maximum hits
Line hits once, parabola at most twice, cubic at most thrice.
Use it when: Reject impossible options like a quadratic with 3 zeros.
Cup or cap
Positive a smiles (cup, opens up); negative a frowns (cap, opens down).
Use it when: Match the sign of a to a described parabola.
Cut, touch, miss
Cut gives 2, touch gives 1 value, miss gives 0 real zeros.
Use it when: Count zeros of a quadratic from its graph quickly.

Geometrical Meaning of Zeros of a Polynomial: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Counting zeros from a described graph · 1 mark · MCQBoard level

The graph of $y = p (x)$ cuts the x-axis at $(- 2, 0)$ , touches the x-axis at $(1, 0)$ without crossing it, and crosses the y-axis at $(0, 4)$ . The number of distinct zeros of $p (x)$ is
(a) 1
(b) 2
(c) 3
(d) 4

Try it first, then show the solutionSolution

Zeros are x-coordinates of points where the graph meets the x-axis, cutting or touching.
The graph meets the x-axis at $x = - 2$ and $x = 1$ .
The point $(0, 4)$ is on the y-axis, so it is not a zero.
Distinct zeros: $- 2$ and $1$ , that is 2.

Answer: (b) 2

Example 2: Assertion-reason: how many zeros a quadratic can have · 1 mark · Assertion-reasonBoard level

Assertion (A): A quadratic polynomial can have at most two zeros.
Reason (R): The graph of a quadratic polynomial is a parabola, which meets the x-axis at most twice.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

A parabola can cut the x-axis twice, touch it once, or miss it, so it meets the x-axis at most twice: R is true.
Each zero is the x-coordinate of a point where the graph meets the x-axis, so at most two meeting points means at most two zeros: A is true.
A follows directly from R, so R correctly explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Finding a linear polynomial from its graph · 2 marks · Very short answerBoard level

The graph of $y = a x + b$ is a straight line that meets the x-axis at $(3, 0)$ and the y-axis at $(0, - 6)$ . Find $a$ and $b$ , and state the zero of $a x + b$ .

Try it first, then show the solutionSolution

$(0, - 6)$ lies on the line: $- 6 = a (0) + b$ , so $b = - 6$ . (½)
$(3, 0)$ lies on the line: $0 = 3 a + b$ . (½)
$3 a - 6 = 0$ , so $a = 2$ , and the polynomial is $2 x - 6$ . (½)
Check: zero $= - \frac{b}{a} = - \frac{- 6}{2} = 3$ , matching the x-axis point $(3, 0)$ . (½)

Answer: $a = 2$ , $b = - 6$ ; the zero of $2 x - 6$ is $3$ .

Example 4: Zeros, intercept and vertex of a parabola · 3 marks · Short answerHarder board level

For the graph of $y = x^{2} - 2 x - 8$ , find (i) the points where it meets the x-axis, (ii) the point where it meets the y-axis, and (iii) its lowest point. State the number of zeros.

Try it first, then show the solutionSolution

Set $y = 0$ : $x^{2} - 2 x - 8 = (x - 4) (x + 2) = 0$ , so $x = 4$ or $x = - 2$ . (½)
The graph meets the x-axis at $(4, 0)$ and $(- 2, 0)$ , so there are 2 zeros. (½)
Put $x = 0$ : $y = - 8$ , so it meets the y-axis at $(0, - 8)$ . (½)
Since $a = 1 > 0$ , the parabola opens upward and has a lowest point. (½)
Axis of symmetry: $x = \frac{4 + ( - 2 )}{2} = 1$ . (½)
$y = 1 - 2 - 8 = - 9$ , so the lowest point is $(1, - 9)$ . (½)

Answer: x-axis at $(- 2, 0)$ and $(4, 0)$ ; y-axis at $(0, - 8)$ ; lowest point $(1, - 9)$ ; 2 zeros.

Example 5: Drawing the graph and reading the zeros · 5 marks · Long answerHarder board level

Draw the graph of $y = x^{2} - x - 6$ for $- 3 \leq x \leq 4$ . From the graph, find the zeros of $x^{2} - x - 6$ . Verify them algebraically and check the sum and product of zeros against the coefficients.

Try it first, then show the solutionSolution

Table: $x = - 3, - 2, - 1, 0, 1, 2, 3, 4$ gives $y = 6, 0, - 4, - 6, - 6, - 4, 0, 6$ . (1)
Plot the points $(- 3, 6), (- 2, 0), (- 1, - 4), (0, - 6), (1, - 6), (2, - 4), (3, 0), (4, 6)$ . (1)
Join them with a smooth upward-opening curve, symmetric about $x = \frac{1}{2}$ . (1)
The curve meets the x-axis at $(- 2, 0)$ and $(3, 0)$ , so the zeros are $- 2$ and $3$ . (½)
Algebraic check: $x^{2} - x - 6 = (x - 3) (x + 2)$ , giving $3$ and $- 2$ . (½)
Sum $= 3 + (- 2) = 1 = - \frac{- 1}{1}$ and product $= - 6 = \frac{- 6}{1}$ . (1)

Answer: Zeros are $- 2$ and $3$ ; sum $1 = - \frac{b}{a}$ , product $- 6 = \frac{c}{a}$ , verified.

Example 6: Finding a polynomial from its graph and shifting it · 3 marks · Short answerHarder board level

The graph of a quadratic $y = p (x)$ meets the x-axis at $(- 1, 0)$ and $(5, 0)$ and the y-axis at $(0, - 5)$ . (i) Find $p (x)$ . (ii) How many zeros does $p (x) + 9$ have? Find them and describe the graph of $y = p (x) + 9$ .

Try it first, then show the solutionSolution

Zeros are $- 1$ and $5$ , so $p (x) = a (x + 1) (x - 5)$ . (½)
$p (0) = a (1) (- 5) = - 5 a = - 5$ , so $a = 1$ . (½)
$p (x) = x^{2} - 4 x - 5$ . (½)
$p (x) + 9 = x^{2} - 4 x + 4 = (x - 2)^{2}$ . (½)
It is zero only at $x = 2$ , so one zero value (two equal zeros). (½)
The graph of $p (x)$ shifted 9 units up: its lowest point $(2, - 9)$ moves to $(2, 0)$ , so it touches the x-axis there. (½)

Answer: $p (x) = x^{2} - 4 x - 5$ ; $p (x) + 9$ has one zero, $x = 2$ , where its graph touches the x-axis.

Example 7: A cubic whose graph touches the x-axis · 3 marks · Short answerHarder board level

Find the points where the graph of $y = x^{3} - 2 x^{2} + x$ meets the x-axis. At which point does the graph cross the x-axis and at which does it only touch it? How many distinct zeros does the polynomial have?

Try it first, then show the solutionSolution

Take out the common factor: $x^{3} - 2 x^{2} + x = x (x^{2} - 2 x + 1)$ . (½)
$x^{2} - 2 x + 1 = (x - 1)^{2}$ , so $p (x) = x (x - 1)^{2}$ . (½)
$p (x) = 0$ gives $x = 0$ or $x = 1$ (repeated), so the graph meets the x-axis at $(0, 0)$ and $(1, 0)$ . (½)
$p (- 1) = - 1 - 2 - 1 = - 4 < 0$ and $p (\frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} > 0$ : the sign changes at $0$ , so the graph crosses the x-axis at $(0, 0)$ . (½)
$p (\frac{1}{2}) > 0$ and $p (2) = 2 (1)^{2} = 2 > 0$ : no sign change at $1$ , so the graph only touches the x-axis at $(1, 0)$ . (½)
Distinct zeros: $0$ and $1$ , that is 2, even though the degree is 3. (½)

Answer: The graph meets the x-axis at $(0, 0)$ , where it crosses, and at $(1, 0)$ , where it touches; 2 distinct zeros, $0$ and $1$ (with $1$ repeated).

Example 8: Case study: a garden fountain jet · 4 marks · Case-basedHOTS / competency

Suppose a fountain in a park shoots a jet of water from a nozzle at ground level. Taking the nozzle as the origin and the ground as the x-axis, the jet follows $y = - \frac{1}{2} x^{2} + b x$ , where $x$ and $y$ are in metres. The jet lands on the ground 8 m from the nozzle.
(i) Find the value of $b$ . (1 mark)
(ii) Find the greatest height of the jet. (1 mark)
(iii) A wall 2 m high stands on the ground at $x = 1$ . Does the jet clear it? At what other distance from the nozzle is the jet at the same height as above the wall? (2 marks)

Try it first, then show the solutionSolution

(i) The jet lands at $(8, 0)$ , so $8$ is a zero: $- \frac{1}{2} (64) + 8 b = 0$ . (½)
$- 32 + 8 b = 0$ , so $b = 4$ and the jet is $y = - \frac{1}{2} x^{2} + 4 x$ . (½)
(ii) Zeros are $0$ and $8$ , so the axis of symmetry is $x = \frac{0 + 8}{2} = 4$ . (½)
Greatest height $= - \frac{1}{2} (16) + 16 = 8$ m. (½)
(iii) At $x = 1$ : $y = - \frac{1}{2} + 4 = 3.5$ m, which is more than 2 m, so the jet clears the wall by 1.5 m. (1)
The jet is symmetric about $x = 4$ , and $x = 1$ is 3 m left of it, so the same height occurs at $x = 4 + 3 = 7$ . Check: $- \frac{49}{2} + 28 = 3.5$ m. (1)

Answer: (i) $b = 4$ (ii) 8 m (iii) yes, the jet is 3.5 m high at the wall, clearing it by 1.5 m; it is again 3.5 m high at 7 m from the nozzle.

Where students lose marks in Geometrical Meaning of Zeros of a Polynomial

The slips examiners see most often, and what to write instead.

Mistake: Counting the y-axis crossing as a zero
Students count every point where the graph meets an axis.
Wrong working
Graph meets axes at $(- 2, 0)$ , $(3, 0)$ and $(0, - 6)$ , so 3 zeros.
Right working
Only $(- 2, 0)$ and $(3, 0)$ are on the x-axis, so 2 zeros; $(0, - 6)$ gives $p (0) = - 6$ .
Do this: Underline 'x-axis' in every graph question before counting.
Usually costs: The full 1 mark in an MCQ.
Mistake: Writing the zero as the point instead of the number
The graph gives coordinates, and students copy them as the answer.
Wrong working
Zeros are $(4, 0)$ and $(- 2, 0)$ .
Right working
The graph meets the x-axis at $(4, 0)$ and $(- 2, 0)$ , so the zeros are $4$ and $- 2$ .
Do this: A zero is a value of $x$ , never an ordered pair.
Usually costs: Half a mark for the final answer.
Mistake: Treating a touching point as two separate zeros or as no zero
Confusion between 'two equal zeros' and 'number of meeting points'.
Wrong working
$y = (x - 2)^{2}$ only touches the axis, so it has no zeros.
Right working
It meets the x-axis at $(2, 0)$ , so $2$ is a zero (repeated); number of distinct zeros is 1.
Do this: Touching counts as meeting. Count it once as a point.
Usually costs: 1 mark in MCQ or assertion-reason items.
Mistake: Joining plotted points with straight lines
Students use a ruler when drawing a parabola.
Wrong working
Points of $y = x^{2} - x - 6$ joined by a ruler, giving a V-shape.
Right working
Join the same points freehand with a smooth curve, symmetric about $x = \frac{1}{2}$ .
Do this: Only linear graphs use a ruler; curves are drawn freehand.
Usually costs: The 1 mark for the curve in a 5-mark graph question.

Exam tips

In graph MCQs, first note the degree, then count x-axis meetings; options above the degree are impossible.
When you draw a graph, take enough values on both sides of the zeros so the turning point shows clearly.
State the zeros as numbers in a final line, then confirm by factorising or substituting.
In case studies, the ground or water level is the x-axis, so the landing points are the zeros.
Use the midpoint of the two zeros to find the vertex quickly instead of making a long table.

Practice questions on Geometrical Meaning of Zeros of a Polynomial

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The graph of a quadratic polynomial opens downward and does not meet the x-axis. Which of these could be the polynomial?
(a) $x^{2} + 2 x + 3$
(b) $- x^{2} + 2 x - 3$
(c) $- x^{2} + 4 x - 4$
(d) $- x^{2} + 2 x + 3$
Hint
Downward means $a < 0$ . Then check which one is never zero by writing it as minus a square minus a number.
Answer
(b). $- x^{2} + 2 x - 3 = - [(x - 1)^{2} + 2]$ , always negative, so no zeros. (a) opens up, (c) $= - (x - 2)^{2}$ touches at 2, (d) $= - (x - 3) (x + 1)$ cuts twice.
Q2 · 2 marks · Very short answerHarder board level
The graph of $y = x^{2} + k x + 9$ touches the x-axis at exactly one point, which lies on the negative side of the x-axis. Find $k$ and the point of contact.
Hint
Touching once means the polynomial is a perfect square $(x - m)^{2}$ .
Answer
$x^{2} + k x + 9 = (x \pm 3)^{2}$ , so $k = \pm 6$ . The point is on the negative side, so the zero is $- 3$ : $(x + 3)^{2}$ , $k = 6$ , point $(- 3, 0)$ .
Q3 · 3 marks · Short answerHarder board level
The graph of $y = x^{3} + k x^{2}$ passes through $(1, 3)$ . Find $k$ and the points where the graph meets the x-axis. Explain why this cubic meets the x-axis at only two distinct points.
Hint
Substitute $(1, 3)$ first, then take out $x^{2}$ as a common factor.
Answer
$1 + k = 3$ , so $k = 2$ and $y = x^{3} + 2 x^{2} = x^{2} (x + 2)$ . Zeros: $0$ (repeated) and $- 2$ , so the graph meets the x-axis at $(0, 0)$ and $(- 2, 0)$ . Near $x = 0$ , $x^{2} (x + 2) > 0$ on both sides, so the graph only touches at the origin and cuts at $(- 2, 0)$ . The repeated zero $0$ gives one meeting point, so there are only 2 distinct zeros.
Q4 · 3 marks · Short answerHarder board level
The graph of $y = x^{2} - (k + 1) x + 2 k - 2$ touches the x-axis at exactly one point. Find $k$ and the point of contact.
Hint
Touching means both zeros equal some $α$ . Use sum $2 α$ and product $α^{2}$ .
Answer
Let both zeros be $α$ . Sum: $2 α = k + 1$ , so $α = \frac{k + 1}{2}$ . Product: $α^{2} = 2 k - 2$ , so $\frac{( k + 1 ) ^{2}}{4} = 2 k - 2$ , giving $k^{2} + 2 k + 1 = 8 k - 8$ , that is $k^{2} - 6 k + 9 = (k - 3)^{2} = 0$ , so $k = 3$ . Then $α = 2$ and $y = x^{2} - 4 x + 4 = (x - 2)^{2}$ . Point of contact: $(2, 0)$ .
Q5 · 3 marks · Short answerHarder board level
The arch of a gateway is the graph of a quadratic that meets the ground (x-axis) at $(2, 0)$ and $(6, 0)$ , with highest point 8 m above the ground. Find the polynomial and the height of the arch at $x = 3$ .
Hint
The highest point lies on $x = 4$ , midway between the zeros. Use $p (x) = a (x - 2) (x - 6)$ .
Answer
$p (4) = a (2) (- 2) = - 4 a = 8$ , so $a = - 2$ and $p (x) = - 2 x^{2} + 16 x - 24$ . $p (3) = - 18 + 48 - 24 = 6$ m.
Q6 · 4 marks · Case-basedHOTS / competency
Suppose a school garden has a curved water channel. Its cross-section is modelled by $y = p (x) = x^{2} - 6 x + c$ , where the ground level is the x-axis and $x$ , $y$ are in decimetres. The channel meets the ground level at two points that are 2 dm apart.
(i) Find the sum of the zeros of $p (x)$ . (1 mark)
(ii) Find the zeros of $p (x)$ and the value of $c$ . (1 mark)
(iii) Find the zeros of $p (x) - c$ and the depth of the channel below ground level. (2 marks)
Hint
Sum of zeros is $- \frac{b}{a}$ . If the zeros are $α$ and $β$ with $α - β = 2$ , solve with the sum. The deepest point is midway between the zeros.
Answer
(i) Sum $= - \frac{- 6}{1} = 6$ . (ii) $α + β = 6$ and $α - β = 2$ give $α = 4$ , $β = 2$ ; $c = α β = 8$ . Check: $x^{2} - 6 x + 8 = (x - 2) (x - 4)$ . (iii) $p (x) - c = x^{2} - 6 x = x (x - 6)$ , zeros $0$ and $6$ . Deepest point on $x = 3$ : $p (3) = 9 - 18 + 8 = - 1$ , so the channel is 1 dm deep.
Q7 · 3 marks · Short answerHOTS / competency
A graph cuts the x-axis at $(- 4, 0)$ and $(1, 0)$ and crosses the y-axis at $(0, - 4)$ . Aman says the polynomial has 3 zeros: $- 4$ , $1$ and $- 4$ . Find his error, state the correct number of zeros, and find the quadratic polynomial.
Hint
Which of the three points is not on the x-axis?
Answer
He counted the y-intercept $(0, - 4)$ as a zero. There are 2 zeros, $- 4$ and $1$ . $p (x) = a (x + 4) (x - 1)$ , $p (0) = - 4 a = - 4$ , so $a = 1$ and $p (x) = x^{2} + 3 x - 4$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Where are the zeros of p(x) on the graph of y = p(x)?Show answer
At the x-coordinates of points where the graph meets the x-axis.
2.How many zeros does a linear polynomial have?Show answer
Exactly one, $- \frac{b}{a}$ .
3.What is the graph of a quadratic polynomial called?Show answer
A parabola.
4.Which way does $y = - 3 x^{2} + x + 1$ open?Show answer
Downward, since $a = - 3 < 0$ .
5.Maximum number of zeros of a cubic?Show answer
3.
6.Where does $y = x^{2} - 5 x + 7$ cross the y-axis?Show answer
At $(0, 7)$ .

Geometrical Meaning of Zeros of a Polynomial: frequently asked questions

Does a point where the graph just touches the x-axis count as a zero?

Yes. The graph meets the x-axis there, so its x-coordinate is a zero. For a quadratic it means the two zeros are equal, so you count it once as a distinct zero.

Can a quadratic polynomial have no zeros?

Yes. If the parabola lies entirely above or entirely below the x-axis, it never meets it, so there are no real zeros. An example is x squared plus 1.

Do I need to draw graphs in the board exam for this topic?

You may be asked to draw a graph from a table of values, or to read zeros from a described graph. Practise making a neat table, plotting points and joining them with a smooth freehand curve.

Is the division algorithm for polynomials in the 2026-27 syllabus?

No. The division algorithm for polynomials is not in the 2026-27 syllabus.

Why can a polynomial of degree n have at most n zeros?

Its graph can meet the x-axis at most n times, so there can be at most n zeros. This lets you reject options such as a quadratic whose graph supposedly meets the x-axis three times.