Zeroes of a Polynomial and Their Graphs: Class 10 Maths

Degree#

The degree of a poly­no­mial is the high­est power of $x$ that appears with a non-zero coef­fi­cient. For exam­ple, $2x^3 - 3x^2 + x - 5$ has degree $3$, and $7 - 2x$ has degree $1$.

Types by degree#

The con­di­tion $a \ne 0$ mat­ters: if the lead­ing coef­fi­cient were $0$, the degree would drop.

Value of a poly­no­mial#

If $p(x)$ is a poly­no­mial and $k$ is a real num­ber, then $p(k)$ is the num­ber you get by replac­ing $x$ with $k$. This is called the value of $p(x)$ at $x = k$.

Zero of a poly­no­mial#

A real num­ber $k$ is a zero of $p(x)$ if $p(k) = 0$. For a lin­ear poly­no­mial $ax + b$, solv­ing $ax + b = 0$ gives exactly one zero:

$$\displaystyle x = -\frac{b}{a}$$

The key result: zeroes are x-inter­cepts#

State­ment#

The zeroes of a poly­no­mial $p(x)$ are pre­cisely the x-coor­di­nates of the points where the graph of $y = p(x)$ meets the x-axis. So the num­ber of zeroes equals the num­ber of points where the graph meets the x-axis.

Why it is true#

Every point on the x-axis has y-coor­di­nate $0$. A point $(k, y)$ lies on the graph of $y = p(x)$ exactly when $y = p(k)$. So the graph meets the x-axis at $(k, 0)$ exactly when $p(k) = 0$, that is, exactly when $k$ is a zero of $p(x)$. Both state­ments describe the same set of num­bers, so they must match one for one.

Graph of a lin­ear poly­no­mial#

The graph of $y = ax + b$ with $a \ne 0$ is a straight line that is not par­al­lel to the x-axis, so it crosses the x-axis at exactly one point, $\displaystyle \left(-\frac{b}{a}, 0\right)$. A lin­ear poly­no­mial there­fore has exactly one zero.

Graph of a qua­dratic poly­no­mial#

The graph of $y = ax^2 + bx + c$ is a curve called a parabola.

• If $a \gt 0$, the parabola opens upwards (like a cup) and has a low­est point.
• If $a \lt 0$, it opens down­wards (like a cap) and has a high­est point.

The low­est or high­est point is called the ver­tex. A parabola can relate to the x-axis in only three ways:

1. It cuts the x-axis at two dis­tinct points: two zeroes.
2. It just touches the x-axis at one point (the ver­tex lies on the axis): one zero, which is counted twice as two equal zeroes.
3. It lies com­pletely above or com­pletely below the x-axis: no real zeroes.

So a qua­dratic poly­no­mial has at most two zeroes.

Graph of a cubic poly­no­mial#

The graph of a cubic $y = ax^3 + bx^2 + cx + d$ always goes off in oppo­site ver­ti­cal direc­tions at the two ends (one end up, the other down), so it must meet the x-axis at least once. It can meet the x-axis at one, two or three points. A cubic poly­no­mial there­fore has at most three zeroes.

The gen­eral rule#

A poly­no­mial of degree $n$ has at most $n$ zeroes, so its graph meets the x-axis at most $n$ times.

Method: read­ing zeroes from a graph#

1. Look only at the points where the curve meets the x-axis (cross­ing or touch­ing both count).
2. Count these points. That count is the num­ber of zeroes.
3. Read the x-coor­di­nate of each point. These are the zeroes.
4. Check your answer against the degree: a qua­dratic can­not have more than $2$ zeroes, a cubic not more than $3$.
5. If the equa­tion is given, con­firm each zero by sub­sti­tut­ing it and get­ting $0$.

Ignore where the graph meets the y-axis. That point gives $p(0)$, the con­stant term, not a zero.

Worked exam­ples#

Exam­ple 1: find­ing a value#

Find the value of $p(x) = 2x^3 - 3x^2 + x - 5$ at $x = 2$ and at $x = -1$. State its degree.

$$\begin{aligned} p(2) &= 2(8) - 3(4) + 2 - 5 = 16 - 12 + 2 - 5 = 1 \\ p(-1) &= 2(-1) - 3(1) + (-1) - 5 = -2 - 3 - 1 - 5 = -11 \end{aligned}$$

The high­est power is $x^3$, so the degree is $3$ (a cubic). Since nei­ther value is $0$, nei­ther $2$ nor $-1$ is a zero.

Exam­ple 2: zero of a lin­ear poly­no­mial#

Find the zero of $p(x) = 3x - 6$ and say where its graph meets the x-axis.

Set $3x - 6 = 0$, so $3x = 6$ and $x = 2$. Check: $p(2) = 6 - 6 = 0$. The graph is a straight line that meets the x-axis at $(2, 0)$. The line $y = 2x - 4$ in the first fig­ure has the same zero, since $2(2) - 4 = 0$.

Exam­ple 3: zeroes of a qua­dratic and its graph#

Find the zeroes of $p(x) = x^2 - 2x - 15$ and describe its graph.

Split the mid­dle term: two num­bers with prod­uct $-15$ and sum $-2$ are $-5$ and $3$.

$$\begin{aligned} x^2 - 2x - 15 &= x^2 - 5x + 3x - 15 \\ &= x(x - 5) + 3(x - 5) \\ &= (x - 5)(x + 3) \end{aligned}$$

So $p(x) = 0$ when $x = 5$ or $x = -3$. Here $a = 1 \gt 0$, so the parabola opens upwards and cuts the x-axis at $(-3, 0)$ and $(5, 0)$. Its ver­tex lies mid­way between the zeroes, at $x = 1$, where $p(1) = 1 - 2 - 15 = -16$, so the ver­tex is $(1, -16)$. The graph meets the y-axis at $(0, -15)$, which is not a zero.

Exam­ple 4: count­ing zeroes with­out full fac­toris­ing#

With­out draw­ing, decide how many zeroes each has: (a) $x^2 - 6x + 9$, (b) $x^2 + 2x + 5$.

(a) $x^2 - 6x + 9 = (x - 3)^2$. This is $0$ only at $x = 3$ and pos­i­tive every­where else, so the parabola touches the x-axis at $(3, 0)$. It has one zero ($3$, repeated).

(b) Com­plete the square:

$$x^2 + 2x + 5 = (x + 1)^2 + 4$$

A square is never neg­a­tive, so the value is always at least $4$. The low­est point is $(-1, 4)$, above the x-axis, so the graph never meets the axis. There are no real zeroes.

Exam­ple 5: a parabola that opens down­wards#

For $p(x) = -x^2 + 4x - 3$, find the zeroes, the ver­tex and the direc­tion in which the graph opens.

Take out $-1$ and fac­torise:

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

The zeroes are $1$ and $3$. Since $a = -1 \lt 0$, the parabola opens down­wards. The ver­tex is at $\displaystyle x = \frac{1 + 3}{2} = 2$, and $p(2) = -4 + 8 - 3 = 1$, so the ver­tex is $(2, 1)$, above the x-axis, which is why the curve crosses the axis twice. It meets the y-axis at $(0, -3)$. Com­pare with the sec­ond panel of the first fig­ure.

Exam­ple 6: zeroes of cubics from their graphs#

Find the zeroes of (a) $x^3 - 4x$, (b) $x^3 - x^2$ and match them with the cubic fig­ure.

(a) $x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$. The zeroes are $-2$, $0$ and $2$: three cross­ing points. As a check on the shape, $p(-1) = -1 + 4 = 3$ and $p(1) = 1 - 4 = -3$, so the curve is above the axis between $-2$ and $0$ and below it between $0$ and $2$.

(b) $x^3 - x^2 = x^2(x - 1)$. The zeroes are $0$ (repeated) and $1$. The graph touches the x-axis at the ori­gin and crosses at $(1, 0)$: two points, so two dis­tinct zeroes.

For com­par­i­son, $x^3 + 1$ is $0$ only at $x = -1$ among real num­bers, so its graph meets the x-axis only once.

Com­mon mis­takes#

• Count­ing the point where the graph meets the y-axis as a zero. Only points on the x-axis count.
• Say­ing a parabola that only touches the x-axis has no zero. Touch­ing counts: it has one zero.
• Claim­ing a qua­dratic graph meets the x-axis three times. A parabola can meet it at most twice; recheck your sketch.
• For­get­ting that a cubic graph must meet the x-axis at least once.
• Think­ing a down­ward parabola always has zeroes. If its high­est point is below the x-axis, it has none.
• Writ­ing the zero as the point $(2, 0)$ when the ques­tion asks for the zero. The zero is the num­ber $2$.
• Drop­ping the minus sign when sub­sti­tut­ing a neg­a­tive value, for exam­ple writ­ing $(-1)^3 = 1$.

Try these#

1. Find $p(-1)$ for $p(x) = x^3 - 2x^2 + 3$. What does your answer tell you? Answer: $p(-1) = 0$, so $-1$ is a zero.
2. Find the zeroes of $x^2 - x - 12$. Answer: $4$ and $-3$.
3. How many zeroes does $x^2 + x + 1$ have? Answer: none (its graph stays above the x-axis).
4. The graph of a cubic poly­no­mial cuts the x-axis at $x = -3$, $x = 1$ and $x = 4$. How many zeroes does it have? Answer: $3$.
5. Find the zero of $5x + 2$. Answer: $\displaystyle -\frac{2}{5}$.
6. Describe the graph of $y = -x^2 + 6x - 9$ and give its num­ber of zeroes. Answer: a down­ward parabola touch­ing the x-axis at $(3, 0)$; one zero, $3$.

Key terms#

Poly­no­mial

A sum of terms $ax^n$ where each $n$ is a whole num­ber and each $a$ is real.

Degree

The high­est power of the vari­able with a non-zero coef­fi­cient.

Lin­ear poly­no­mial

A poly­no­mial of degree 1; its graph is a straight line.

Qua­dratic poly­no­mial

A poly­no­mial of degree 2; its graph is a parabola.

Cubic poly­no­mial

A poly­no­mial of degree 3.

Value of a poly­no­mial

The num­ber $p(k)$ obtained by sub­sti­tut­ing $x = k$.

Zero

A num­ber $k$ with $p(k) = 0$; the x-coor­di­nate of a point where the graph meets the x-axis.

Parabola

The U-shaped (or inverted U-shaped) graph of a qua­dratic poly­no­mial.

Ver­tex

The low­est point of an upward parabola or the high­est point of a down­ward one.

Com­mon ques­tions#

Can a qua­dratic poly­no­mial have exactly one zero?#

Yes. When the parabola just touches the x-axis, as $x^2 - 4x + 4 = (x - 2)^2$ does at $x = 2$, there is one zero. It is often described as two equal zeroes.

Can a cubic poly­no­mial have no zeroes?#

No. One end of a cubic graph goes up and the other goes down, so the curve must cross the x-axis at least once.

Does the sign of $a$ change the num­ber of zeroes?#

Not by itself. The sign of $a$ only decides whether the parabola opens up or down. The num­ber of zeroes depends on where the ver­tex is com­pared with the x-axis.

What does the point where the graph meets the y-axis tell me?#

It is $(0, p(0))$, and $p(0)$ is the con­stant term. It is a zero only if the con­stant term is itself $0$.

How many zeroes can a poly­no­mial of degree $n$ have?#

At most $n$. A lin­ear poly­no­mial has exactly one, a qua­dratic at most two and a cubic at most three.

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. Stew­art, J., Redlin, L. and Wat­son, S. Pre­cal­cu­lus: Math­e­mat­ics for Cal­cu­lus. Cen­gage Learn­ing.