Zeros of a Polynomial: Exponents in algebraic expressions can be rational values. On the other hand, a polynomial is an algebraic statement with a whole number exponent on any variable. A polynomial’s zeros are the locations at which the polynomial turns zero. A polynomial with a value of zero \((0)\) is called a zero \((0)\) polynomial. In this article, let’s learn everything about the zeros of a polynomial in detail.

Polynomial

An expression of the form \(p(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n},\) , where \({a_n} = 0\), is known as  polynomial in \(x\) of degree \(n\).  Here  \({a_{0,}}{a_{1,}}{a_2}{…._{{a_n}}}\) are real numbers, and each power of \(x\) is a non-negative integer.

\(6{x^3} + 3x + 1\) is an example of a polynomial. It is an algebraic expression as well.

\(2y + 3\sqrt y \) is an algebraic expression but not a polynomial. – since the exponent on \(y\) is \(\frac{1}{2}\) which is not a whole number.

Degree of a Polynomial

If \(p(x)\) is a polynomial \(x\), the highest power of \(x\) in \(p(x)\) is called the degree of the polynomial \(p(x)\).

Example: The degree of the polynomial \({x^3} + 2x + 3\) is \(3\), as the highest power of \(x\) in the given expression is \(3.\)

Types of Polynomials

Polynomials are classified based on:

1. Number of Terms
2. Degree of Polynomial

Classification of Polynomials Based on the Number of Terms

The classification of polynomials based on the number of terms are explained below:

Classification	Description	Example
Monomial	A polynomial with just one term	\(2x,3pq,6{x^2}{y^2}\)
Binomial	A polynomial with two terms	\(3{x^2} + x,6x + 4\)
Trinomial	A polynomial with three terms	\(3{x^2} + x + 4\)

Types of Polynomials Based on Degree

The types of polynomials based on the degree are explained below:

Linear Polynomial

A polynomial of degree one is named a linear polynomial.

A linear polynomial is of the form \(p(x) = ax+b\), where \(a≠0\). 
Example: \(3x+1\)

Quadratic Polynomial

A quadratic may be a polynomial with the degree \(2\).

In general, any quadratic polynomial in \(x\) is of the form \(a{x^2} + bx + cx\) where \(a, b, c\) are real numbers and \(a≠0\).

Example: \(2{x^2} + 8x + 5\)

Cubic Polynomial

A polynomial of degree \(3\) is called a cubic polynomial.

The general form of a cubic polynomial is \(a{x^3} + b{x^2} + cx + d\)

where \(a, b, c, d\) are real numbers and \(a≠0\).

Example: \(2{x^3} – 12{x^2} + 22x – 1\)

Biquadratic Polynomial

A polynomial of degree \(4\) is called a biquadratic polynomial.

A biquadratic polynomial is of the form \(p(x) = a{x^4} + b{x^3} + c{x^2} + dx + e\) where \(a≠0\).

Examples: \(2{x^4} + 3{x^3} – 5{x^2} + x + 1\)

Zeros of a Polynomial

If \(p(x)\) is a polynomial \(x\), and if \(k\) is an actual number, then the value obtained by replacing \(x\) by \(k\) in \(p(x)\) is called the value of \(p(x)\) at \(x=k\) and is denoted by \(p(k)\).

Example: Find the value of  \({x^2} – 3x – 4\) at \(x=-1\) and \(x=4\)?

For \(x=-1\)
\(p( – 1) = {( – 1)^2} – 3( – 1) – 4\)
\( \Rightarrow 1 + 3 – 4\)
\( \Rightarrow 4 – 4\)
\( \Rightarrow 0\)
Then, for \(x=4\)
\(p(4) = {(4)^2} – 3(4) – 4\)
\( \Rightarrow 16 – 12 – 4\)
\( \Rightarrow 16 – 16\)
\( \Rightarrow 0\)

As \(p( – 1) = 0\) and \(p(4) = 0, – 1,\) and \(4\) are called the zeros of the quadratic polynomial \({x^2} – 3x – 4\). In generally, a real number k is said to be a zero of a polynomial \(p(x)\), if \(pk=0\).

Zeros of a Linear Polynomial

Generally, if \(k\) is a zero of \(p(x)=ax+b\), then \(p(x)=ax+b=0\), i.e.,\(k = \frac{{ – b}}{a}\)

Therefore, the zero of the linear polynomial \(\frac{{ – b}}{a} = \frac{{ – \left({{\text{Constant}}\,{\text{term}}} \right)}}{{{\text{Coefficient}}\,{\text{of}}\,x}}\)

So, the zero of a linear polynomial is related to its coefficients.

Geometric Meaning of the Zeros of a Polynomial

Consider an example, the graph of \(y=2x+3\) is a straight line passing through the point \((-2, -1)\) and \((2, 7)\).

From the graph, \(y=2x+3\) intersects the \(x\)-axis at the point \(\left( {\frac{{ – 3}}{2},0} \right)\).

In general, for a linear polynomial \(ax + b, a≠0\), the graph of \(y = ax + b\) is \(a\). The straight line which intersects the \(x\)-axis at exactly one point, namely, \(\left( {\frac{{ – b}}{a},0} \right)\).

Therefore, the linear polynomial \(ax + b, a≠0\), has precisely one zero, namely, the \(x\)-coordinate of the point where the graph of \(y = ax + b\) intersects the \(x\)-axis.

Geometric Meaning of the Zeroes of a Quadratic Polynomial

Consider a quadratic polynomial \({x^2} – 3x – 4\).

List of few values for \(x\) as given in the table

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(Y = {x^2}\)	\(6\)	\(0\)	\(-4\)	\(-6\)	\(-6\)	\(-4\)	\(0\)	\(6\)

From the graph \(y = {x^2} – 3x – 4\) intersects the \(x\)-axis at the points \((4, 0)\) and \((-1, 0)\).

\(-1\) and \(4\) are the \(x\)-coordinates of the points where the graph of \(y = {x^2} – 3x – 4\) intersects the \(x\)-axis.

Therefore, the zeros of the quadratic polynomial \(a{x^2} + bx + cx,a \ne 0\)  are precisely the \(x\)-coordinates of the points where the parabola representing, \(y = a{x^2} + bx + cx\) , the \(x\)-intersects the \(x\)-axis.

Note: For any quadratic polynomial \(a{x^2} + bx + cx,a \ne 0,\) the graph of the corresponding equation \(y = a{x^2} + bx + cx\) has one of the two shapes, either  \( \cap \) or \( \cup \). It depends on whether \(a>0\) or \(a<0\), and these curves are called Parabolas.

Examples of Zeroes of Polynomial

Observe the graphs given here, Each is the graph of \(y=p(x)\), where \(p(x)\) is a polynomial. 
By observing the graph, we can find the number of zeroes of \(p(x)\).

(i) The number of zeroes is \(1\), the graph intersects the \(x\)-axis at one point only.
(ii) The number of zeroes is \(2\), the graph intersects the \(x\)-axis at two points.
(iii) The number of zeroes is \(3\), the graph intersects the \(x\)-axis at three points.
(iv) The number of zeroes is \(1\) as the graph intersects the \(x\)-axis at one point only.
(v) The number of zeroes is \(1\) as the graph intersects the \(x\)-axis at one point only.
(vi) The number of zeroes is \(4\) as the graph intersects the \(x\)-axis at four points.

Relationship b/w Zeroes and Coefficients of a Polynomial

Generally, \(\alpha \) and \(\beta \) are the zeroes of the quadratic polynomial \(p(x) = a{x^2} + bx + cx,a \ne 0\) then,

Sum of the zeroes \(\alpha  + \beta  = \frac{{ – b}}{a} = \frac{{ – \left({{\text{Coefficient}}\,{\text{of}}\,x} \right)}}{{{\text{Coefficient}}\, {\text{of}}\, {x^2}}}\)

Product of the zeroes \(\alpha  \times \beta  = \frac{c}{a} = \frac{{{\text{Constant}}\,{\text{term}}}}{{{\text{Coefficient}}\,{\text{of}}\,{x^2}}}\)

Solved Examples – Zeros of a Polynomial

Q.1. Find the zeroes of the polynomial \({x^2} + 7x + 10\)
Ans: Given polynomial is \({x^2} + 7x + 10\)
Here, we can find the zeroes of the polynomial by the method of splitting the middle terms.
\(\Rightarrow {x^2} + 5x + 2x + 10 = 0\)
\(\Rightarrow x(x + 5) + 2(x + 5) = 0\)
\( \Rightarrow (x + 2)(x + 5) = 0\)
\( \Rightarrow x = \, – 2\) and \(x =\, – 5\)
Therefore, the zeroes of the given polynomial \({x^2} + 7x + 10\) are \(-2\) and \(-5\).

Q.2. Find the zeroes of the polynomial \(p(x) = 6{x^4} – {x^3} – 15{x^2} + 2x – 7\). If  \(x=2\).
Ans: Given polynomial is \(p(x) = 6{x^4} – {x^3} – 15{x^2} + 2x – 7\)  and \(x=2\).
\(\Rightarrow 6{(2)^4} – {2^3} – 15{(2)^2} + 2(2) – 7\) 
\( \Rightarrow 6 \times 16 – 8 – 15(4) + 2(2) – 7\) 
\( \Rightarrow 6 \times 16 – 8 – 15(4) + 2(2) – 7\) 
\( \Rightarrow 96 – 8 – 60 + 4 – 7\) 
\( \Rightarrow 100 – 75 = 25\) 
Therefore, the zeroes of the given polynomial \(6{x^4} – {x^3} – 15{x^2} + 2x – 7\)  is \(25\)

Q.3. Find the zeroes of the quadratic polynomial  \(3{x^2} + 5x – 2\) and verify the relationship between the zeroes and the co-efficient.
Ans: The given polynomial is \(3{x^2} + 5x – 2\).
Here, we can find the zeroes of the polynomial by the method of splitting the middle terms.
\( \Rightarrow 3{x^2} + 5x – 2\)
\( \Rightarrow 3{x^2} + 6x – 1x – 2 = 0\)
\(\Rightarrow 3x(x + 2) – 1(x + 2) = 0\)
\( \Rightarrow (3x – 1)(x + 2) = 0\)
\( \Rightarrow 3x – 1 = 0,x + 2 = 0\)
\( \Rightarrow x = \frac{1}{3},x =\, – 2\)

Therefore, the zeroes of the given polynomial \(3{x^2} + 5x – 2\) are \(\frac{1}{3}\) and \(-2\)
Sum of the zeroes \( = \alpha  + \beta  = \frac{{ – b}}{a} = \frac{{ – \left({{\text{Coefficient}}\,{\text{of}}\,x} \right)}}{{{\text{Coefficient}}\,{\text{of}}\,{x^2}}}\)
\( \Rightarrow \frac{1}{3} + ( – 2) = \frac{{1 – 6}}{3} = \frac{{ – 5}}{3}\)
Product of the zeroes \(= \alpha  \times \beta  = \frac{c}{a} = \frac{{{\text{Constant}}\,{\text{term}}}}{{{\text{Coefficient}}\,{\text{of}}\,{x^2}}}\)
\( \Rightarrow \frac{1}{3} \times ( – 2) = \frac{{ – 2}}{3}\)
So, the sum and product of the zeroes are \(\frac{{ – 5}}{3}\) and \(\frac{{ – 2}}{3}\) respectively.
Hence, they verified the relationship between the zeroes and the coefficients.

Q.4. Find a quadratic polynomial, the sum, and product of whose zeroes are \(-3\) and \(2\), respectively. 
Ans: Consider the quadratic polynomial be \(a{x^2} + bx + c\) and its zeroes are \(\alpha \) and \(\beta \).
We have 
Sum of the zeroes \( = \alpha  + \beta  = \frac{{ – b}}{a} = \frac{{ – ({\text{ Coefficient of }}x)}}{{{\text{ Coefficient of }}{x^2}}} \Rightarrow  – 3\)
\( \Rightarrow \alpha + \beta = \frac{{ – b}}{a} = – 3 = \frac{{ – 3}}{1}\)
Product of the zeroes \(=\alpha \times \beta=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
\(\Rightarrow \alpha \times \beta = \frac{c}{a} = 2 = \frac{2}{1}\)
From the above,\(a = 1,b = 3,c = 2\)
Therefore, the required polynomial is
\({x^2} + 5x + 2\)

Q.5. Find the zeroes of the quadratic polynomial  \({x^2} – 2x – 8\), and verify the relationship between the zeroes and the co-efficient.
Ans: The given polynomial is \({x^2} – 2x – 8\).
Here, we can find the zeroes of the polynomial by the method of splitting the middle terms.
\( \Rightarrow {x^2} – 2x – 8 = 0\)
\( \Rightarrow {x^2} – 4x + 2x – 8 = 0\)
\( \Rightarrow x(x – 4) + 2(x – 4) = 0\)
\( \Rightarrow (x + 2)(x – 4) = 0\)
\( \Rightarrow x + 2 = 0,x – 4 = 0\)
\( \Rightarrow x =\, – 2,x = 4\)

Therefore, the zeroes of the given polynomial \({x^2} – 2x – 8\) are \(-2\) and \(4\).
Sum of the zeroes \(=\alpha+\beta=\frac{-b}{a}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of the zeroes \(=\alpha \times \beta=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
\(\Rightarrow ( – 2) \times 4 = \,- 8\)
So, the sum and product of the zeroes are \(2\) and \(-8\) respectively.
Hence, they verified the relationship between the zeroes and the coefficients.

Q.6. Find a quadratic polynomial, the sum, and product of whose zeroes are \(1\) and \(1\), respectively. 
Ans: Consider the quadratic polynomial be \(a{x^2} + bx + c\) and its zeroes are \(\alpha \) and \(\beta \) .
We have 
Sum of the zeroes \(=\alpha+\beta=\frac{-b}{a}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}} \Rightarrow 1\)
\( \Rightarrow \alpha + \beta = \frac{{ – b}}{a} = \,- 1 = \frac{{ – 1}}{1}\)
Product of the zeroes \(=\alpha \times \beta=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
\( \Rightarrow \alpha \times \beta = \frac{c}{a} = 1 = \frac{1}{1}\)
From the above, \(a = 1,b = \,- 1,c = 1\)
Therefore, the required polynomial is \({x^2} – x + 1\).

Summary

If \(p(x)\) is a polynomial \(x\), and if \(k\) is an actual number, then the value obtained by replacing \(x\) by \(k\) in \(p(x)\) is called the value of \(p(x)\) at \(x=k\) and is denoted by p(k). This article helps to understand in detail about zeros of a polynomial. It includes details about polynomials, the degree of the polynomial, types of polynomials, and zeroes of the polynomial. By studying this article helps in solving the problems based on the zeros of a polynomial.

FAQs on Zeros of a Polynomial

Q.1. What is zero of polynomials? Give an example?
Ans: If \(p(x)\) is a polynomial \(x\), and if \(k\) is an actual number, then the value obtained by replacing \(x\) by \(k\) in \(p(x)\) is called the value of \(p(x)\) at \(x=k\) and is denoted by \(p(k)\).
Example: Find the value of  \({x^2} – 3x + 2\) at \(x=1\) and \(x=2\)?
For, \(x=1\)
\(p(1) = {(1)^2} – 3(1) + 2\)
\( \Rightarrow 1 – 3 + 2\)
\( \Rightarrow 3 – 3\)
\(\Rightarrow 0\)
Then, for \(x = 2\)
\(p(2) = {(2)^2} – 3(2) + 2\)
\( \Rightarrow 4 – 6 + 2\)
\( \Rightarrow 6 – 6\)
\( \Rightarrow 0\)
As \(p(1) = 0\) and \(p(2) = 0,1\) and \(2\) are called the zeros of the quadratic polynomial \({x^2} – 3x + 2\)

Q.2. Why \(8\) is a polynomial?
Ans: \(8\) is a Polynomial. Because \(8\) is a constant polynomial. 

Q.3. Is Pi a polynomial?
Ans: Pi is not a constant. A polynomial should have at least one power of a variable.

Q.4. Can \(0\) be a polynomial?
Ans: \(0\) is not a polynomial. We know a polynomial is a mathematical equation with constants, variables, and exponents blended using simple basic mathematical operations like addition, subtraction, multiplication, and division.

Q.5. Write the relationship between zeroes and coefficients of a polynomial.
Ans: In general, if \(\alpha \) and \(\beta \) are the zeroes of the quadratic polynomial
\(p(x) = a{x^2} + bx + cx,a \ne 0\) then,
Sum of the zeroes \(=\alpha+\beta=\frac{-b}{a}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of the zeroes \(=\alpha \times \beta=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

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