What are the relationship between the zeroes and coefficients of the quadratic polynomial ax2 bx c?

The number of zeroes of a polynomial equals the degree of the polynomial, and there is a well-defined mathematical relationship between the zeroes and the coefficients. In this lesson, let's explore the relationship between the zeroes and coefficients of a polynomial.

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Definition of Zeroes of a Polynomial
Relationship Between Zeroes and Coefficients of a Polynomial
Linear Polynomial
Quadratic Polynomial
Cubic Polynomial
Examples of Relationship Between Zeroes and Coefficients of Polynomials
FAQs on Relationship Between Zeroes and Coefficients of Polynomials
What Is the Relationship Between Zeroes and Coefficients of a Cubic Polynomial?
What Is the Relationship Between Zeroes and Coefficients of a Quadratic Polynomial?
What Is the Relationship Between Zeroes and Coefficients of a Linear Polynomial?
Relationship Between Zeros And Coefficients Of A Polynomial Example Problems With Solutions

Definition of Zeroes of a Polynomial

Zeroes of a polynomial are the solutions to the given polynomial equation when the polynomial is set as equal to zero. Polynomials are classified depending on the highest power of the variable in the given polynomial. Mathematically, if p(x) is a polynomial with variable x, and k is any real number and said to be the zero of the polynomial p(x) if p(x) at x = k is 0.

A coefficient is a number or quantity placed with a variable, usually an integer that is multiplied by the variable next to it. For the variables with no integer with them are assumed to have 1 as their coefficient. A coefficient can be positive or negative, real or imaginary, or in the form of decimals or fractions.

Relationship Between Zeroes and Coefficients of a Polynomial

The relationship between zeroes and the coefficients of polynomials can be defined based on the definite formulas as per the type of polynomial. The relation between the zeroes and the coefficients of a polynomial is given below:

Linear Polynomial

A linear polynomial is an expression of the form ax + b, having 1 as the degree of the polynomial. Here, “x” is a variable, “a” and “b” are constants. The zero of the polynomial = -b/a = – constant term/coefficient of x.

Quadratic Polynomial

The Quadratic polynomial is an expression of the form ax2 + bx + c having the highest degree 2. Here, “x” is a variable, “a”, "b", and “c” are constants and a ≠ 0. Let α and β be the two zeroes of the polynomial, then

The sum of zeroes, α + β is -b/a = – Coefficient of x/ Coefficient of x2
The product of zeroes, αβ is c/a = Constant term / Coefficient of x2

Cubic Polynomial

The cubic polynomial is an expression of the form ax3 + bx2 + cx + d having the highest degree 3. Here, “x” is a variable, “a”, "b", and “c” are constants, and a ≠ 0. Let α, β, and γ are the three zeroes of the polynomial, then

The sum of zeroes, α + β + γ is -b/a = – Coefficient of x2/ coefficient of x3
The sum of the product of zeroes, αβ+ βγ + αγ is c/a = Coefficient of x/Coefficient of x3
The product of zeroes, αβγ is -d/a = – Constant term/Coefficient of x3

Related Topics

Given below is a list of topics related to the relationship between zeroes and coefficients of polynomial.

Examples of Relationship Between Zeroes and Coefficients of Polynomials

Example 1: Determine the sum and the product of the zeroes of the quadratic polynomial 9x2 – 16 + 20.

Solution:

Given quadratic polynomial is 9x2 – 16 + 20.

Here, a = 9, b = -16, c = 20.

By the relationship between the zeroes and coefficients of the polynomial,

The sum of zeroes = -b/a = – Coefficient of x/ Coefficient of x2 = -(-16)/9 = 16/9

The product of zeroes = c/a = Constant term / Coefficient of x2 = 20/9

Therefore, the sum and the product of the zeroes of the given polynomial are 16/9 and 20/9.
Example 2: Find the zero of the polynomial 5x - 10.

Solution:

Given linear polynomial = 5x - 10

Here, a = 5, b = -10

By the relationship between the zeroes and coefficients of the polynomial,

The zero of a linear polynomial is -b/a = – constant term/coefficient of x = -(-10)/5 = 2.

Therefore, the zero of the given polynomial is 2.

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FAQs on Relationship Between Zeroes and Coefficients of Polynomials

The number of zeroes of a polynomial is determined by the degree of the polynomial, and thus there is a well-defined mathematical relationship between the zeroes and the coefficients.

What Is the Relationship Between Zeroes and Coefficients of a Cubic Polynomial?

The relationship between zeroes and coefficients of a cubic polynomial is as follows:

The sum of zeroes = -b/a = – Coefficient of x2/ coefficient of x3
The sum of the product of zeroes = c/a = Coefficient of x/Coefficient of x3
The product of zeroes -d/a = – Constant term/Coefficient of x3

What Is the Relationship Between Zeroes and Coefficients of a Quadratic Polynomial?

The relationship between zeroes and coefficients of a quadratic polynomial is as follows:

The sum of zeroes = -b/a = – Coefficient of x/ Coefficient of x2
The product of zeroes = c/a = Constant term / Coefficient of x2

What Is the Relationship Between Zeroes and Coefficients of a Linear Polynomial?

The relationship between zeroes and coefficients of a linear polynomial is given as the zero of a polynomial = -b/a = – constant term/coefficient of x.

Consider quadratic polynomial
P(x) = 2x2 – 16x + 30.
Now, 2x2 – 16x + 30 = (2x – 6) (x – 3) = 2 (x – 3) (x – 5) The zeros of P(x) are 3 and 5.

Sum of the zeros = 3 + 5 = 8 = \(\frac { -\left( -16 \right) }{ 2 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Product of the zeros = 3 × 5 = 15 = \(\frac { 30 }{ 2 }\) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)
So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and α, β are two zeros of polynomial then \(\alpha +\beta =-\frac { b }{ a } \) \(\alpha \beta =\frac { c }{ a } \)

In general, it can be proved that if α, β, γ are the zeros of a cubic polynomial ax3 + bx2 + cx + d, then

\(\alpha +\beta +\gamma =\frac { -b }{ a } \) \( \alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) \( \alpha \beta \gamma =\frac { -d }{ a } \)

Note: \(\frac { b }{ a } \), \(\frac { c }{ a }\) and \(\frac { d }{ a } \) are meaningful because a ≠ 0.

Relationship Between Zeros And Coefficients Of A Polynomial Example Problems With Solutions

Example 1: Find the zeros of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeros and its coefficients.
Sol. We have, 6x2 – 13x + 6 = 6×2 – 4x – 9x + 6 = 2x (3x – 2) –3 (3x – 2) = (3x – 2) (2x – 3)

So, the value of 6x2 – 13x + 6 is 0, when

(3x – 2) = 0 or (2x – 3) = 0 i.e., When x = \(\frac { 2 }{ 3 } \) or \(\frac { 3 }{ 2 } \)

Therefore, the zeros of 6x2 – 13x + 6 are

\(\frac { 2 }{ 3 } \) and \(\frac { 3 }{ 2 } \) Sum of the zeros = \(\frac { 2 }{ 3 } \) + \(\frac { 3 }{ 2 } \) = \(\frac { 13 }{ 6 } \) = \(\frac { \left( -13 \right) }{ 6 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\) Product of the zeros

= \(\frac { 2 }{ 3 } \) × \(\frac { 3 }{ 2 } \) = \(\frac { 6 }{ 6 } \) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 2: Find the zeros of the quadratic polynomial 4x² – 9 and verify the relation between the zeros and its coefficients.
Sol. We have,
4x2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3)
So, the value of 4x2 – 9 is 0, when 2x – 3 = 0 or 2x + 3 = 0 i.e., when x = \(\frac { 3 }{ 2 } \) or x = \(\frac { -3 }{ 2 } \).

Therefore, the zeros of 4x2 – 9 are \(\frac { 3 }{ 2 } \) & \(\frac { -3 }{ 2 } \).

Sum of the zeros = \(\frac { 3 }{ 2 }\) \(-\frac { 3 }{ 2 } \) = 0 = \(-\frac { \left( 0 \right) }{ 4 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\) Product of the zeros

= \(\frac { 3 }{ 2 }\) × \(\frac { -3 }{ 2 }\) = \(\frac { -9 }{ 4 }\) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 3: Find the zeros of the quadratic polynomial 9x2 – 5 and verify the relation between the zeros and its coefficients.
Sol. We have,
9x2 – 5 = (3x)2 – (√5)2 = (3x – √5) (3x + √5)
So, the value of 9x2 – 5 is 0, when 3x – √5 = 0 or 3x + √5 = 0 i.e., when x = \(\frac { \sqrt { 5 } }{ 3 } \) or x = \(\frac { -\sqrt { 5 } }{ 3 } \). Sum of the zeros = \(\frac { \sqrt { 5 } }{ 3 } \) \(-\frac { \sqrt { 5 } }{ 3 } \) = 0 = \(-\frac { \left( 0 \right) }{ 9 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\) Product of the zeros

= \(\left( \frac { \sqrt { 5 } }{ 3 } \right) \) × \(\left( \frac { -\sqrt { 5 } }{ 3 } \right) \) = \(\frac { -5 }{ 9 } \) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 4: If α and β are the zeros of ax2 + bx + c, a ≠ 0 then verify the relation between the zeros and its coefficients.
Sol. Since a and b are the zeros of polynomial ax2 + bx + c.
Therefore, (x – α), (x – β) are the factors of the polynomial ax2 + bx + c.
⇒ ax2 + bx + c = k (x – α) (x – β)
⇒ ax2 + bx + c = k {x2 – (α + β) x + αβ}
⇒ ax2 + bx + c = kx2 – k (α + β) x + kαβ …(1)
Comparing the coefficients of x2, x and constant terms of (1) on both sides, we get a = k, b = – k (α + β) and c = kαβ ⇒ α + β = \(\frac { -b }{ k } \) and αβ = \(\frac { c }{ k } \)

α + β = \(\frac { -b }{ a } \) and αβ = \(\frac { c }{ a } \) [∵ k = a]

Sum of the zeros = \(\frac { -b }{ a } \) = \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)
Product of the zeros = \(\frac { c }{ a } \) = \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)

Example 5: Prove relation between the zeros and the coefficient of the quadratic polynomial
ax2 + bx + c.
Sol. Let a and b be the zeros of the polynomial ax2 + bx + c α = \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\) ….(1) β = \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\) ….(2) By adding (1) and (2), we get α + β = \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\) + \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\) = \(\frac{ -2b }{ 2a } \) = \(\frac{ -b }{ a } \) = \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\) Hence, sum of the zeros of the polynomial

ax2 + bx + c is \(\frac{ -b }{ a } \)

By multiplying (1) and (2), we get αβ = \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\) × \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\) = \(\frac{{{b}^{2}}-{{b}^{2}}+4ac}{4{{a}^{2}}}\) = \(\frac{4ac}{4{{a}^{2}}}\) = \(\frac{ c }{ a } \) = \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)

Hence, product of zeros = \(\frac{ c }{ a } \)

Example 6: find the zeroes of the quadratic polynomial x2 – 2x – 8 and verify a relationship between zeroes and its coefficients.
Sol. x2 – 2x – 8 = x2 – 4x + 2x – 8 = x (x – 4) + 2 (x – 4) = (x – 4) (x + 2)

So, the value of x2 – 2x – 8 is zero when

x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2.

So, the zeroes of x2 – 2x – 8 are 4, – 2.

Sum of the zeroes = 4 – 2 = 2 = \(-\frac { \left( -2 \right) }{ 1 } \) = \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\) Product of the zeroes

= 4 (–2) = –8 = \(\frac { -8 }{ 1 } \) = \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)

Example 7: Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients. 2x3 + x2 – 5x + 2 ; , 1, – 2
Sol. Here, the polynomial p(x) is 2x3 + x2 – 5x + 2
Value of the polynomial 2x3 + x2 – 5x + 2 when x = 1/2 = \(2{{\left( \frac{1}{2} \right)}^{3}}+{{\left( \frac{1}{2} \right)}^{2}}-5\left( \frac{1}{2} \right)+2\) = \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\) = 0 So, 1/2 is a zero of p(x). On putting x = 1 in the cubic polynomial

2x3 + x2 – 5x + 2

= 2(1)3 + (1)2 –¬ 5(1) + 2 = 2 + 1 – 5 + 2 = 0 On putting x = – 2 in the cubic polynomial

2x3 + x2 – 5x + 2

= 2(–2)3 + (–2)2 – 5 (–2) + 2 = – 16 + 4 + 10 + 2 = 0 Hence, \(\frac{ 1 }{ 2 } \), 1, – 2 are the zeroes of the given polynomial. Sum of the zeroes of p(x) = \(\frac{ 1 }{ 2 } \) + 1 – 2 = \(\frac{ -1 }{ 2 } \) = \(\frac{-\text{ coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}\) Sum of the products of two zeroes taken at a time = \(\frac{ 1 }{ 2 } \) × 1 + \(\frac{ 1 }{ 2 } \) × (–2) + 1 × (–2) = \(\frac{ 1 }{ 2 } \) – 1 – 2 = \(\frac{ -5 }{ 2 } \) = \(\frac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}}\) Product of all the three zeroes = \(\frac{ 1 }{ 2 } \) × (1) × (–2) = –1

= \(\frac{ -2 }{ 2 } \) = \(\frac{-\text{ constant term }}{\text{coefficient of }{{x}^{3}}}\)