Zeros of a cubic polynomial can be defined as the point at which the polynomial becomes zero. A cubic polynomial is a polynomial with the highest power of the variable or degree is 3. The general form of a cubic polynomial is

ax³ + bx² + cx + d = 0,

where a ≠ 0, and a, b, c are the coefficients of x³, x², x and d is the constant term.

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What is Zeros of a Cubic Polynomial?

Zeros of a cubic polynomial is the point at which the polynomial becomes zero. A cubic polynomial can have three zeros because its highest power (or degree) is three. A quadratic polynomial may have no real solution but a cubic polynomial always has at least one real solution. If a cubic polynomial does have three zeros, two or even all three of them may be repeated.

Let us say α, β, and γ be the zeros of the cubic polynomial ax³ + bx² + cx + d = 0, where a ≠ 0 and a, b, c, are the coefficients of x³, x², x and d is the constant term.

Sum of zeros of a cubic polynomial formula: α + β + γ = –b/a

Sum of product of zeros of a cubic polynomial formula: αβ + βγ + γα = c/a

Product of zeros of a cubic polynomial formula: αβγ = –d/a

Zeros of a cubic polynomial will have 4 possibilities:

1. All three zeroes are real and different
   → This means all the answers (zeroes) are real numbers and none of them are the same.
    
2. All three zeroes are real, but two are the same
   → This means two of the answers are the same, and the third one is different. All are real numbers.
    
3. All three zeroes are real and the same
   → This means all three answers are the same real number.
    
4. One zero is real, and the other two are complex (not real)
   → This means only one answer is a real number, and the other two are complex numbers (they include 'i', the imaginary number).

How to find Zeros of a Cubic Polynomial?

Consider a cubic polynomial of the form ax³ + bx² + cx + d = 0, where a. A cubic polynomial can have three zeros because its highest power (or degree) is three.

We can easily find zeros of a cubic polynomial by following the below steps:

1. Step 1: Let (p – q), p, (p + q) be the three zeros of a given cubic polynomial.
    
2. Step 2: Find p using the sum of zeros of a cubic polynomial formula.
    
3. Step 3: Find out the other two zeros by factoring the equation into a quadratic polynomial.

Example: Find the zeros of the cubic polynomial
x³ – 12x² + 39x – 28 = 0

Let (p – q), p, (p + q) be the zeros of a given cubic polynomial.

Then, Sum of zeros of a cubic polynomial = –b/a

⇒ (p + q) + p + (p – q) = –12/1
⇒ 3p = 12
⇒ p = 4.

Now, find out the other two zeros by factorizing the equation into a quadratic polynomial.

x³ – 12x² + 39x – 28 = (x – 4)(x² – 8x + 7)
x³ – 12x² + 39x – 28 = (x – 4)(x – 1)(x – 7)

⇒ x = 1, x = 4, and x = 7 are the three zeros of the given cubic polynomial.

Product of Zeros of a Cubic Polynomial

Let \(\alpha\), \(\beta\), and \(\gamma\) be the zeros of the cubic polynomial \(ax^{3}+bx^{2}+cx+d=0\), where \(a\neq 0\), and \(a\), \(b\), \(c\) are the coefficients of \(x^{3}\), \(x^{2}\), \(x\) and \(d\) is the constant term. Then, the product of zeros of a cubic polynomial is given as:

Product of zeros of a cubic polynomial = -(constant term)/(coefficient of \(x^{3}\))

\(\Rightarrow\) \(\alpha\beta\gamma=-\frac{d}{a}\).

Example: Find the product of zeros of the cubic polynomial \(kx^{3}-5x^{2}-12x+k=0\).

Given polynomial is \(kx^{3}-5x^{2}-12x+k=0\), then we have

\(a = k\), and \(d = k\).

Product of zeros = -\frac{d}{a}[/latex])

\(\Rightarrow\) \(\alpha\beta\gamma=-\frac{k}{k}=-1\)

Therefore, the product of zeros of a cubic polynomial is \(-1\).

Sum of Zeros of a Cubic Polynomial

Let \(\alpha\), \(\beta\), and \(\gamma\) be the zeros of the cubic polynomial \(ax^{3}+bx^{2}+cx+d=0\), where \(a\neq 0\), and \(a\), \(b\), \(c\) are the coefficients of \(x^{3}\), \(x^{2}\), \(x\) and \(d\) is the constant term. Then, the sum of zeros of a cubic polynomial is given as:

Sum of zeros of a cubic polynomial = -(coefficiant of \(x^{2}\))/(coefficiant of \(x^{3}\))

\(\Rightarrow\) \(\alpha+\beta+\gamma=-\frac{b}{a}\).

Example: Find the product of zeros of the cubic polynomial \(5x^{3}-15x^{2}-12x+27=0\).

Given polynomial is \(5x^{3}-15x^{2}-12x+27=0\), then we have

\(a = 5\), and \(b = -15\).

Sum of zeros = \(-\frac{b}{a}\)

\(\Rightarrow\) \(\alpha+\beta+\gamma=-\frac{-15}{5}=3\)

Therefore, the sum of zeros of the given cubic polynomial is \(3\).

Nature of Zeros of a Cubic Polynomial

Consider a cubic polynomial of the form \(ax^{3}+bx^{2}+cx+d=0\), where \(a\neq 0\). The nature of these zeros can be defined by the use of the discriminant of a cubic polynomial. It is given by the following relation:

\(D = b^{2}c^{2} – 4ac^{3} – 4b^{3}d – 27a^{2}d^{2} + 18abcd\)

Therefore,

(i). When \(D = 0\), the cubic polynomial has real zeros and at least one repeated zeros.

(ii). When \(D > 0\), the cubic polynomial has three real and distinct zeros.

(iii). When \(D < 0\), the cubic polynomial has a pair of complex conjugates and one real zeros.

Example: find the nature of zeros of the cubic polynomial \(x^{3}+x=0\) using the discriminant formula.

Given cubic polynomial is \(x^{3}+x=0\), we have

\(a = 1\), \(b = 0\), \(c = 1\), and \(d = 0\).

We know, \(D = b^{2}c^{2} – 4ac^{3} – 4b^{3}d – 27a^{2}d^{2} + 18abcd\)

\(\Rightarrow\) \(D = (0)^{2}(1)^{2} – 4(1)(1)^{3} – 4(0)^{3}(0) – 27(1)^{2}(0)^{2} + 18(1)(0)(1)(0)\)

\(\Rightarrow\) \(D = 0 – 4 – 0 – 0 + 0\)

\(\Rightarrow\) \(D = -4 < 0\)

Since \(D < 0\), then the given cubic polynomial has a pair of complex conjugates and one real zero.

Relation between Zeros and Coefficients of a Cubic Polynomial

Zeros of the polynomial are defined as the values of the variable for which the value of the polynomial is zero. Let \(\alpha\), \(\beta\), and \(\gamma\) be the zeros of the cubic polynomial \(ax^{3}+bx^{2}+cx+d=0\), where \(a\neq 0\), and \(a\), \(b\), \(c\), are the coefficients of \(x^{3}\), \(x^{2}\), \(x\) and \(d\) is the constant term.

Then the relation between zeros and coefficient of a cubic polynomial is:

1. Sum of zeros = \(-\frac{b}{a}\)

\(\Rightarrow\) \(\alpha+\beta+\gamma=-\frac{b}{a}\).

2. Sum of product of zeros = \(\frac{c}{a}\)

\(\Rightarrow\) \(\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}\).

3. Product of zeros = \(-\frac{d}{a}\)

\(\Rightarrow\) \(\alpha\beta\gamma=-\frac{d}{a}\).

Example: Verify the relationship between zeros and coefficients of \(x^{3}+5x^{2}-6x=0\) if the zeroes are given as \(-6\), \(-0\) and \(-1\).

Comparing the given cubic polynomial with \(ax^{3}+bx^{2}+cx+d=0\)

\(\Rightarrow\) \(a=1\), \(b=5\), \(c=-6\), and \(d=0\)

Given the zeros: \(-6\), \(-0\) and \(-1\)

\(\Rightarrow\) \(\alpha=-6\), \(\beta=0\), and \(\gamma=1\)

(i) Verify the sum of the zeros of the cubic polynomial, i.e.

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

\(\Rightarrow\) \((-6)+0+1=-\frac{5}{1}\)

\(\Rightarrow\) \(-5=-5\).

(ii) Verify the sum of the product of the zeros of the cubic polynomial, i.e.

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

\(\Rightarrow\) \((-6)(0)+(0)(1)+(1)(-6)=\frac{-6}{1}\)

\(\Rightarrow\) \(-6=-6\).

(iii) Verify the product of the zeros of the cubic polynomial, i.e.

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

\(\Rightarrow\) \((-6)(0)(1)=-\frac{0}{1}\)

\(\Rightarrow\) \(0=0\).

Solved Examples of Zeros of a Cubic Polynomial

Example 1: Find the cubic polynomial whose zeros are \(3\), \(5\) and \(-2\).

Solution: Given that the zeros of a cubic polynomial are \(3\), \(5\) and \(-1\) that means \((x+3)\), \((x+5)\), and \((x-2)\).

We know that the zeroes of a cubic polynomial are denoted by \(\alpha\), \(\beta\), and \(\gamma\).

(i). Sum of zeros = [coefficient of \(x^{2}\)/coefficient of \(x^{3}\)]

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

\(\Rightarrow\) \(3+5-2=-\frac{b}{a}\)

\(\Rightarrow\) \(6=-\frac{b}{a}\)

(ii). Product of zeros = [constant term/coefficient of \(x^{3}\)]

\(\alpha\times\beta\times\gamma=-\frac{d}{a}\)

\(\Rightarrow\) \(3\times5\times(-2)=-\frac{d}{a}\)

\(\Rightarrow\) \(-30=\frac{d}{a}\)

(iii). Sum of the product of zeros = [coefficient of \(x\)/coefficient of \(x^{3}\)]

\(\alpha\times\beta+\beta\times\gamma+\gamma\times\alpha=\frac{c}{a}\)

\(\Rightarrow\) \(3\times5+5\times(-2)+(-2)\times 3=-\frac{c}{a}\)

\(\Rightarrow\) \(15-10-6=\frac{c}{a}\)

\(\Rightarrow\) \(-1=\frac{c}{a}\)

On comparing the above solutions, we get

\(a = 1\), \(b = -6\), \(c = -1\), and \(d = -30\).

\(\therefore\) \(x^{3}-6x^{2}-x+30=0\)

Hence, the cubic polynomial is \(x^{3}-6x^{2}-x+30=0\).

Example 2: Find all the zeros of \(x^{3}+6x^{2}+11x+6=0\) if \((x+1)\) is a factor.

Solution: Since \((x+1)\) is a factor of a given cubic equation.

\(\therefore\) \((x+1)\) will completely divide the given equation.

Now, \(x^{3}+6x^{2}+11x+6=0\) = \((x^{2}+5x+6)(x+1)\)

\(\Rightarrow\) = \((x+2)(x+3)(x+1)\)

\(\therefore\) Zeros of \((x+1)(x+2)(x+3)=0\) are

\(x+1=0\) \(\Rightarrow\) \(x=-1\)

\(x+2=0\) \(\Rightarrow\) \(x=-2\)

\(x+3=0\) \(\Rightarrow\) \(x=-3\)

Hence, the zeros of \(x^{3}+6x^{2}+11x+6=0\) are \(1\), \(2\), and \(3\).

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