Relationship between Zeroes and Coefficients of a Polynomial: Polynomials are algebraic expressions made up of variables and constants that have whole number exponents. We know that polynomials can be divided into different categories based on their degree, such as linear, quadratic, and cubic polynomials. Because the polynomial degree is equal to the number of zeros in the formula, the zeros of a polynomial can be simply computed using the polynomial degree.

Let us examine the relationship between the zeros of a polynomial and coefficients of the linear, quadratic and cubic polynomials with some additional solved examples in this article.

What is a Polynomial?

Polynomials are algebraic expressions consisting of variables and constants with a whole number exponents of the variables. This is how a polynomial looks:

A polynomial \(p(x)\) in one variable \(x\) is an algebraic expression in \(x\) of the form,

\(p(x) = {a_n}{x^n} + {a_{n – 1}}{x^{n – 1}} + …. + {a_2}{x^2} + {a_1}x + {a_0},\)

where \({a_0},{a_1},{a_2},…{a_n}\) are constants and \({a_n}\, \ne 0\).

\({a_0},{a_1},{a_2},…{a_n}\) are respectively the coefficients of \({x^0},{x^1},{x^2},…{x^n}\) and \(n\) is called the degree of the polynomial. Each of \({a_n}{x^n},{a_{n – 1}}{x^{n – 1}},….,{a_0}\), with \({a_n} \ne 0\) is called a term of the polynomial \(p(x)\).

It’s also worth noting that a polynomial can’t have negative or fractional exponents.

Examples of polynomials are \(3{y^2} + 2x + 5,{x^3} + 2{x^2} – 9x – 4,10{x^3} + 5x + y,4{x^2} – 5x + 7\) etc.

Zeros of a Polynomial

Let \(p(x)\) be a polynomial. If \(p(a)\) then we say that \(a\) is a zero of the polynomial \(p(x)\).

How to Find the Zeros of a Polynomial?

1. The zero of a polynomial is the value at which the polynomial becomes zero. Thus, in order to find zeros of the polynomial, we simply equate polynomial to zero and find the possible values of variables.
2. Let \(P(x)\) be a given polynomial. To find zeros, set this polynomial equal to zero. i.e. \(P(x) = 0\) Now, this becomes a polynomial equation. Solve this equation and find all the possible values of variables by factorizing the polynomial equation.
3. These are the values of \(x\) which make polynomial equal to zero; hence are called zeros of polynomial \(P(x)\). A number \(z\) is said to be a zero of a polynomial \(P(x)\) if and only if \(P(z) = 0\).

Factors of a Polynomial

Let \(p(x)\) be a polynomial of degree \(1\) or more and let \(a\) be any real number.

(i) If \(p(a) = 0\) then \((x – a)\) is a factor of \(p(x)\).

For example: suppose \(p(x) = ({x^2} + 2x + 1)\) is a polynomial
If \(p( – 1) = {( – 1)^2} + 2( – 1) + 1 = 1 – 2 + 1 = 2 – 2 = 0\)
Then \((x – ( – 1))\)\( = (x + 1)\)  is a factor of the polynomial \(p(x)\)

(ii) If \((x – a)\) is a factor of \(p(x)\) then \(p(a) = 0\)

For example: suppose \(p(x) = ({x^2} + 2x + 1)\) is a polynomial.
If \((x + 1)\)  is a factor of the polynomial \(p(x)\)
then, \(p( – 1) = {( – 1)^2} + 2( – 1) + 1 = 1 – 2 + 1 = 2 – 2 = 0\)

Now, let us see the zeros of linear, quadratic, and cubic polynomials and understand the relationship between the zeros and the coefficient of each polynomial.

Linear Polynomial

A polynomial whose highest power of the variable or the polynomial degree is \(1\) is a linear polynomial. The standard form of writing a linear polynomial is \(ax + b\) where \(a,b\) are constants and \(a \ne 0\).
\(a,b\) are respectively the coefficients of \({x^1},{x^0}\)

Example: \(x – 1,y + 1,a + 4\) etc.

Zeros of Linear Polynomials

There is only one zero in a linear polynomial. The zero of the linear polynomial is always real.

For example: Consider a linear polynomial \(3x + 8\)
Then the zero of the linear polynomial is \(3x + 8 = 8\)
\( \Rightarrow 3x =\,- 8\)
\( \Rightarrow x =\,- \frac{8}{3}\)

Relationship between Zeroes and Coefficients of a Linear Polynomial

If \(\alpha \) is the zeros of a quadratic polynomial \(ax + b,\) then

The relationship between the zeros and coefficient of a linear polynomial

\( = \, – \frac{{{\rm{Constant}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,x}} = \, – \frac{b}{a}\)

For example: Consider a linear polynomial \(5x\, + 4\)
In order to find the zero of the linear polynomial
\(5x\, + 4 = 0 \Rightarrow x =\, – \frac{4}{5}\)

So the relationship between the zero and coefficient of the polynomial \( = \, – \frac{4}{5} = \, – \frac{{{\rm{Constant}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,x}} = \, – \frac{b}{a}\)

Quadratic Polynomial

A polynomial whose highest power of the variable or the polynomial degree is \(2\) is a quadratic polynomial. The standard form of writing a quadratic polynomial is \(a{x^2} + bx + c\) where \(a,b,c\) are constants and \(a \ne 0\)

\(a,b,c\) are respectively the coefficients of \({x^2},{x^1},{x^0}\)

Example: \({x^2} + x,{y^2} + 1,{a^2} + 8,\)

Zeros of Quadratic Polynomials

There are two zeros in a quadratic polynomial. The following cases are possible for the zeroes of a quadratic polynomial:

1. The two zeroes might be real and distinct.
2. The two zeroes might be real and equal.
3. The two zeros might be non-real (complex).

Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

If \(\alpha ,\,\beta \) are the zeros of a quadratic polynomial \(a{x^2} + bx + c,\)

The sum of zeros \( = \alpha + \beta = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}} =\, – \frac{{ – b}}{a}\)

The product of zeros \( = \alpha \beta = \, – \frac{{{\rm{Constant term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}} = \frac{c}{a}\)

For example: Consider a quadratic polynomial \(({x^2} – 8x + 12)\)
Now factorize the polynomial to get the zeros of the polynomial
\(({x^2} – 8x + 12) = {x^2} – 6x – 2x + 12\)
\( = x(x – 6) – 2(x – 6)\)
\(= (x – 2)(x – 6)\)

So the zeros of the polynomial are \(\alpha  = 2,\beta  = 6\)

The sum of zeros \( = \alpha + \beta = 2 + 6 = 8 = \frac{8}{1} = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}} =\, – \frac{b}{a}\)

The product of zeros \( = \alpha \beta = 2 \times 6 = 12 = \frac{{12}}{1} = \frac{{{\rm{Constant term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}} = \frac{c}{a}\)

Cubic Polynomial

A polynomial whose highest power of the variable or the polynomial degree is \(3\) is a cubic polynomial. The standard form of writing a cubic polynomial is \(a{x^3} + b{x^2} + cx + d\) where \(a,b,c,d\) are constants and \(a \ne 0\).

\(a,b,c,d\) are respectively the coefficients of \({x^3},{x^2},{x^1},{x^0}\)

Example: \({y^3} + 8,{x^3} – 27,5 + {a^3},{x^3} + {x^2} – x + 2,\), etc

Zeros of Cubic Polynomials

There are three zeros in a cubic polynomial. The following cases are possible for the zeroes of a cubic polynomial:

1. All three zeroes might be real and distinct.
2. All three zeroes might be real, and two of them might be equal.
3. All three zeroes might be real and equal.
4. One zero might be real, and the other two non-real (complex).

Relationship Between Zeroes and Coefficients of a Cubic Polynomial

If \(\alpha ,\beta \) and \(\gamma \) are the zeros of a cubic polynomial \(a{x^3} + b{x^2} + cx + d,\) then

The sum of zeros \( = \alpha + \beta + \gamma = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \, – \frac{b}{a}\)

The sum of the product of zeros \( = \alpha \beta + \beta \gamma + \gamma \alpha = \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \frac{c}{a}\)

The product of zeros \( = \alpha \beta \gamma = \frac{{{\rm{Constant}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = – \frac{d}{a}\)

For example: Consider a cubic polynomial \({x^3} – 7{x^2} + 4x + 12\)

By the long division method
\({x^3} – 7{x^2} + 4x + 12 = (x + 1)({x^2} – 8x + 12)\)
\( = (x + 1)(x – 2)(x – 6)\)
So, the roots are \( – 1,2,6\)

The sum of zeros \( =\, – 1 + 2 + 6 = 7 = \frac{7}{1} = \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} =\, – \frac{b}{a}\)

The sum of the product of zeros \({\rm{ = \alpha \beta  + \beta \gamma  + \gamma \alpha  =\, – 1 \times 2 + 2 \times 6 + 6 \times  – 1 =  – 2 + 12 – 6 = 4 = }}\frac{{\rm{4}}}{{\rm{1}}}{\rm{ = }}\frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} =\, {\rm{ – }}\frac{{\rm{c}}}{{\rm{a}}}\)

The product of zeros \( = \alpha \beta \gamma = \, – 1 \times 2 \times 6 = – 12 = \, – \frac{{12}}{1} = \, – \frac{{{\rm{Constant term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} =\, – \frac{d}{a}\)

Solved Examples

Q.1. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as \(2, – 7, – 14\) respectively.
Ans: Let the polynomial be \(a{x^3} + b{x^2} + cx + d\) and the zeroes be \(\alpha ,\,\beta \,and\,\gamma \)
Then, the sum of zeros \(\alpha  + \beta  + \,\,\gamma  = \frac{{ – 2}}{1} = 2 =\, – \frac{b}{a}\)
The product of zeros \(\alpha \beta \,\gamma  =\, – 14 = \frac{{ – 2}}{1} =\, – \frac{d}{a}\)
\(\therefore a = 1,b =\, – 2,c =\, – 7\) and \(d = 14\)
So, one cubic polynomial which satisfies the given conditions will be \({x^3} – 2{x^2} – 7x + 14\)

Q.2. Find a quadratic polynomial with the sum and the product of its zeros as \(4,\,8\)
Ans: Let the polynomial be \(a{x^2} + bx + c\) and the zeroes be \(\alpha ,\beta \)
Then, the sum of zeros \(\alpha \beta  = \frac{8}{1} = \frac{c}{a}\)
The product of zeros \(\alpha  + \beta  = \frac{{ – 4}}{1} = 4 =  – \frac{b}{a}\)
\(\therefore a = 1,b =\, – 4,c = 8\)
So, one cubic polynomial which satisfies the given conditions will be \({x^2} – 4x + 8\)

Q.3. Find a linear polynomial whose zero is \(7\)
Ans: Let the polynomial be \(ax + b,\) and the zero is \(\alpha \)
Then, \(\alpha  =\, – \frac{{ – 7}}{1} = 7 =\, – \frac{b}{a}\)
\(\therefore \,a = 1,b =\, – 7\)
So, the linear polynomial which satisfies the given conditions will be \(x – 7\)

Q.4. Find the zeroes of the quadratic polynomial \({x^2} + 7x + 10\), and verify the relationship between the zeroes and the coefficients.
Ans: We have \({x^2} + 7x + 10 = {x^2} + 2x + 5x + 10\)
\( = x(x + 2) + 5(x + 2) = (x + 2)(x + 5)\)
So, the value of \({x^2} + 7x + 10\) is zero when \((x + 2) = 0\,or\,(x + 5) = 0,\) i.e., when \(x =\, – 2\) or \(x =\, – 5\)

Therefore, the zeroes \({x^2} + 7x + 10\) are \( – 2\) and \( – 5\)
The sum of zeros \( =\, – 2 + ( – 5) =\, – 7 =\, – \frac{{ – 7}}{1} =\, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}\)
The product of zeros \( = ( – 2)( – 5) = 10 = \frac{{10}}{1} =\, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}\)
Hence, the relationship between the zeros and coefficients of the given polynomial is verified.

Q.5. Find the zeroes of the cubic polynomial \(2{x^2} – 5{x^2} – 14x + 8\)and verify the relationship between the zeroes and the coefficients.
Ans: Given, a cubic polynomial \(2{x^2} – 5{x^2} – 14x + 8\) and By using the trial and error method, we will find one factor.
Like the possible values are the positive and negative factors of the constant term \(8\)
Therefore the possible factors are \( \pm 1, \pm 2, \pm 4, \pm 8\)

Now, the expression vanishes to become zero if we put \(x\, = \, – 2\)
So, \((x\, + 2)\)will be a factor in the expression.
Now, splitting the middle terms to take out the common factor, we can write
\(2{x^3} – 5{x^2} – 14x + 8 = 2{x^3} + 4{x^2} – 9{x^2} – 18x + 4x + 8\)
\( = 2{x^2}(x + 2) – 9x(x + 2) + 4(x + 2)\)
\( = (2{x^2} – 9x + 4)(x + 2)\)
\( = (x + 2)(2{x^2} – 8x – x + 4)\)
\( = (x + 2)[2x(x – 4) – 1(x – 4)]\) [Taking out \(2x\,\& \, – 1\) common]
\( = \,(x + 2)(x – 4)(2x – 1)\)

Therefore, the factorization of \( = \,(2{x^3} – 5{x^2} – 14x + 8)\,is\,(x + 2)(x – 4)(2x – 1)\)
The zeros of the cubic polynomial are \( – 2,4\,and\,\frac{1}{2}\)
Then, the sum of zeros\( =\, – 2 \times 4 + 4 \times \frac{1}{2} + \,\,\frac{1}{2} \times ( – 2) =\, – 8 + 2 – 1 =\, – \frac{{14}}{2} = \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}}\)
The product of zeros \( =\, – 2 \times 4 \times \frac{1}{2} =\, – \frac{8}{2} = \frac{{{\rm{Coefficient}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}}\)
Hence, the relationship between the zeros and coefficients of the given polynomial is verified.

Summary

We learned about polynomials in this article, how they look, and the different sorts of polynomials based on their degree like linear polynomial, quadratic polynomial, cubic polynomial. The zeros of linear, quadratic, and cubic polynomials and the relationship between zeros and coefficients of polynomials were then explored in length, with solved examples.

Frequently Asked Questions

We have provided some frequently asked questions here:

Q.1. What is a zero of a polynomial?
Ans: Let \(p(x)\) be a polynomial. if \(p(a) = 0\) then we say that \(a\) is a zero of the polynomial \(p(x)\)
For example: For equation \(p(x)=x^2-4\), zeroes are \(2\) and \(-2\) since \(p(2) = p( – 2) = 0\)

Q.2. What is the relationship between zeros and coefficients of a polynomial?
Ans: If \(\alpha \) is the zeros of a quadratic polynomial \(ax + b,\) then
The relationship between the zeros and coefficient of a linear polynomial
\( = \, – \frac{{{\rm{Constant}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,x}} = \, – \frac{b}{a}\)
If \(\alpha ,\beta \) are the zeros of a quadratic polynomial \(a{x^2} + bx + c\) then
The sum of zeros \( = \alpha + \beta = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}} = \frac{b}{a}\)

The product of zeros \( = \alpha \beta = \, – \frac{{{\rm{Coefficient}}\,{\rm{term}}}}{{{\rm{Constant}}\,{\rm{of}}\,{x^2}}} = \frac{c}{a}\)
If \({\rm{\alpha ,}}\,{\rm{\beta }}\) and \(\gamma \) are the zeros of a cubic polynomial \(a{x^3} + b{x^2} + cx + d\) then
The sum of zeros \( = \alpha + \beta + \gamma = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \, – \frac{b}{a}\)

The sum of the product of zeros \( = \alpha \beta + \beta \gamma + \gamma \alpha = \,- \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \frac{c}{a}\)
The product of zeros \( = \alpha \beta \gamma = \, – \frac{{{\rm{Coefficient}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \, – \frac{d}{a}\)

Q.3. What do you know about the zero of a zero polynomial?
Ans: Suppose we have given a zero polynomial. So consider the zero polynomial as \(P(x) = 0\)
Now in this, all coefficients of the polynomial are equal to zero. So since all coefficients are \(0,x\) can take any value, whether its integer, its rational number, or any other real number.
Any value of \(x\) can be a zero of a zero polynomial.

Q.4. What is the sum of two zeroes of the polynomial?
Ans: If \(\alpha ,\beta \) are the zeros of a quadratic polynomial \(a{x^2} + bx + c\)then
The sum of zeros \( = \alpha + \beta = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \, – \frac{b}{a}\)

If \(\alpha  + \beta ,and\,\gamma \)are the zeros of a cubic polynomial \(a{x^2} + b{x^2} + cx + d\) then
The sum of zeros \( = \alpha + \beta + \gamma = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^3}}} = \, – \frac{b}{a}.\)

Q.5. How do you find the relationship between zeros and coefficients?
Ans: Suppose we have a quadratic polynomial \({x^2} + 7x + 10\)
Then the factor of the quadratic polynomial is \({x^2} + 7x + 10 = (x + 2)(x + 5)\)
So, the value of \({x^2} + 7x + 10\) is zero when \((x + 2) = 0\,or\,(x + 5) = 0,i.e.,\) when \(x =  – 2\,or\,x =  – 5\)

Therefore, the zeroes of \({x^2} + 7x + 10\,are\, – 2\,and\, – 5\)
The sum of zeros \( = \, – 2 + \left( { – 5} \right) = \, – 7 = \, – \frac{{ – 7}}{1} = \, – \frac{{{\rm{Coefficient}}\,{\rm{of}}\,x}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}\)
The product of zeros \( = \,\left( { – 2} \right)\left( { – 5} \right) = \,10 = \frac{{10}}{1} = \frac{{{\rm{Constant}}\,{\rm{term}}}}{{{\rm{Coefficient}}\,{\rm{of}}\,{x^2}}}.\)

We hope this detailed article on relationship between zeroes and coefficients of a polynomial has helped you in your studies. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will assist you at the earliest.