Summarize this article:
114 LearnersLast updated on October 30, 2025

The zeros of a polynomial are the values of x for which f(x) = 0. To find a zero, you can use trial values to divide the polynomial, factor the result, and solve the remaining cubic or quadratic equations.
The zeros of a cubic polynomial are the points at which the polynomial becomes zero. Since the highest degree of a cubic polynomial is three, it can have up to three zeros. A cubic polynomial always has at least one real solution, whereas a quadratic polynomial might not have any. Some or all zeros may be repeated.
In the cubic polynomial \(ax^3 + bx^2 + cx + d = 0 \), where a ≠ 0 and a, b, c are the coefficients of x3, x2, x, and d is the constant term, let α, β, and γ be the zeros.
Zeros of Cubic Polynomial Formulas
A cubic polynomial formula’s sum of zeros is \(α+β+γ= -b/a\)
A cubic polynomial formula’s sum of the products of zeros is \(αβ+βγ+γα=c/a\)
A cubic polynomial formula’s product of zeros is \(αβγ= -d/a\)
Properties of Zeros of a Cubic Polynomial
Since its degree is 3, a cubic polynomial can have at most three zeros.
Even if the other two are complex, every cubic polynomial with real coefficients will have at least one real zero.
For a cubic polynomial \(ax^3 + bx^2 + cx + d \) with zeros p, q, r:
\(p + q + r = -\frac{b}{a} \) it is the sum of zeros.
\(pq + qr + rp = \frac{c}{a} \) it is the sum of the products taken two at a time.
\(pqr = -\frac{d}{a} \) it is the product of the zeros.
Consider a cubic polynomial with the formula \(ax^3 + bx^2 + cx + d = 0 \), where a. A cubic polynomial can have three zeros because its highest power (or degree) is three. Since three is a cubic polynomial’s highest power (or degree), it is possible for it to contain three zeros.
By taking the actions listed below, we can quickly determine a cubic polynomial’s zeros:
Step 1: Assume that the three zeros of a given cubic polynomial are (p - q), p, and (p + q).
Step 2: Utilizing a cubic polynomial formula’s sum of zeros, find p.
Step 3: Factor the equation into a quadratic polynomial to determine the remaining two zeros.
For example, find the zeros of the cubic polynomial
\(x^3 - 12x^2 + 39x - 28 = 0 \)
Let the zeros of a given cubic polynomial be (p - q), p, and (p + q). Then, a cubic polynomial’s sum of zeros equals -b / a.
⇒\((p + q) + p + (p - q) = -12/1\)
⇒\(3p = 12\)
⇒\(p = 4\).
Now factorize the equation into a quadratic polynomial to determine the remaining two zeros.
\(x^3 - 12x^2 + 39x - 28 = (x - 4)(x^2 - 8x + 7) \)
\( x^3 - 12x^2 + 39x - 28 = (x - 4)(x - 1)(x - 7) \)
⇒x = 1, x = 4, and x = 7 are the three zeros of the given cubic polynomial.
Let α, β, and γ be the cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \), where a ≠ 0 and a, b, and c are the coefficients of \(x^3\), \(x^2\), and x, and d is the constant term. Next, a cubic polynomial’s product of zeros is provided as follows:
Product of zeros of a cubic polynomial = \( -\dfrac{\text{constant term}}{\text{coefficient of } x^3} \).
⇒ \( \alpha \beta \gamma = -\frac{d}{a} \).
Example: Determine the cubic polynomial \(\ kx^3 - 5x^2 - 12x + k = 0 \ \)
.
If the polynomial is \(\ kx^3 - 5x^2 - 12x + k = 0 \), then we have a = k, and d = k.
Product of zeros = -(constant term)/(coefficient of \(x^3\))
⇒\(\ \alpha \beta \gamma = -\frac{k}{k} = -1 \ \)
Therefore, the product of the zeros of a cubic polynomial is -1.
Sum of Zeros of a Cubic Polynomial
Let α, β, and γ be the zeros of the cubic polynomial \(\ ax^3 + bx^2 + cx + d = 0 \ \) where a ≠ 0, where a, b, and c are the coefficients of \(x^3\),\(x^2\), and x and d is the constant term. Next, a cubic polynomial’s sum of zeros is expressed as follows:
\(\ \text{The sum of zeros of a cubic polynomial} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \ \)
⇒\(\ \alpha + \beta + \gamma = -\frac{b}{a} \ \).
Example: Determine the cubic polynomial \(\ 5x^3 - 15x^2 - 12x + 27 = 0 \ \).
\(\ \text{Sum of zeros of a cubic polynomial} = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \ \)
⇒\(\ \alpha + \beta + \gamma = -\frac{b}{a} \ \).
If the polynomial is \(\ 5x^3 - 15x^2 - 12x + 27 = 0 \ \), then we have
a= 5 and b = -15.
Sum of zero = -ba
⇒ \(\ \alpha + \beta + \gamma = -\left(-\frac{15}{5}\right) = 3 \ \)
Therefore, the given cubic polynomial’s sum of zeros is 3.
Consider a cubic polynomial with the formula \(\ ax^3 + bx^2 + cx + d = 0 \ \), where a ≠ 0. The nature of these zeros can be defined by the use of the discriminant of a cubic polynomial. The following relation proves it:
\(\ D = b^2c^2 - 4ac^3 - 4b^3d - 27a^2d^2 + 18abcd \ \)
Therefore,
D < 0.
For example,
Determine the type of zeros in the cubic polynomial \( x^3+x=0\). Applying the discriminant formula.
Given that \(x^3+x=0\) is a cubic polynomial, we have
a = 1, b = 0, c = 1, and d = 0.
We know, \(\ D = b^2c^2 - 4ac^3 - 4b^3d - 27a^2d^2 + 18abcd \ \)
So,
⇒\(\ D = (0)^2(1)^2 - 4(1)(1)^3 - 4(0)^3(0) - 27(1)^2(0)^2 + 18(1)(0)(1)(0) \ \)
⇒\(\ D = 0 - 4 - 0 - 0 + 0 \ \)
⇒\(D=-4<0\)
The given cubic polynomial has one real zero and two complex conjugates since
D < 0.
The values of the variable for which the polynomial’s value is zero are known as zeros of the polynomial. Let α, β, and γ be the zeros of the cubic polynomial \(\ ax^3 + bx^2 + cx + d = 0 \ \), where a ≠ 0, and a, b, and c are the coefficients of \(x^3\), \(x^2\), x and d is the constant term.
Then, a cubic polynomial’s zeros and coefficients have the following relationship:
Sum of zeros = -ba
⇒\(\ \alpha + \beta + \gamma = -\frac{b}{a} \ \).
Sum of the product of zeros = ca.
⇒\(\ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \ \).
Product of zeros = -da
⇒\(\ \alpha\beta\gamma = -\frac{d}{a} \ \).
For Example:
Check the connection between zeros and coefficients of \(\ x^3 + 5x^2 - 6x = 0 \ \) if the zeros are given as -6, -0, and -1.
Comparing the given cubic polynomial with \(\ ax^3 + bx^2 + cx + d = 0 \ \)
⇒a=1, b=5, c=-6, and d=0
Given the zeros: -6, -0, and -1
⇒α=-6,β=0, and γ=1
Check that the cubic polynomial’s zeros sum is correct, i.e.
\(\ \alpha + \beta + \gamma = -\frac{b}{a} \ \)
⇒\(\ (-6) + 0 + 1 = -\frac{5}{1} \ \)
⇒ -5= -5.
Confirm the total product of the cubic polynomial’s zeros, i.e.
\(\ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \ \)
⇒\( (-6)(0) + (0)(1) + (1)(-6) = -6 \)
⇒-6=-6
Confirm that the cubic polynomial’s zeros are multiplied, that is,
\( \alpha\beta\gamma = -\frac{d}{a} \)
⇒\((-6)(0)(1)=-01\)
⇒ 0 = 0.
Mastering the zeros of a cubic polynomial helps in understanding the relationship between factors and roots. It also strengthens algebraic problem-solving and graph interpretation skills.
Students may make some mistakes on the zeros of a cubic polynomial, like using the wrong trial values, and more. Let us look at the mistakes and learn how to quickly correct them.
Zeros of a cubic polynomial are used in areas like engineering, physics, and more. Let us see how the zeros of cubic polynomials help in real life.
Find the zeros of f(x)=x3-6x2+11x-6
\(\ f(x) = 0 \Rightarrow x^{3} - 6x^{2} + 11x - 6 = 0 \ \)
By trying with different values, we discover that x = 1 satisfies the equation. When x = 1 is substituted into the polynomial, gives 1-6+11-6=0 is obtained, confirming that x = 1 is a zero. The quotient is, \(\ x^{2} - 5x + 6 = (x - 2)(x - 3) \ \). As a result, the full factorization of the polynomial is \(\ f(x) = (x - 1)(x - 2)(x - 3) \ \), and its zeros are 1, 2, and 3.
Find the zeros f(x)=x3+3x2-4x-12=0
\(\ f(x) = 0 \Rightarrow x^{3} + 3x^{2} - 4x - 12 = 0 \ \)
We test x = 2 in the polynomial, and we get \(\ 8 + 12 - 8 - 12 = 0 \ \), which means that x = 2 is a zero.
Next, the cubic polynomial is then divided by (x - 2), resulting in the quotient \(\ x^{2} + 5x + 6 \ \). After factoring this with a quadratic component, we obtain \(\ x^{2} + 5x + 6 = (x + 2)(x + 3) \ \). that the function’s zeros are -3, -2, and 2, and its fully factorized form is \(\ f(x) = (x - 2)(x + 2)(x + 3) \ \).
Find the zeros of f(x)=x3-7x+6
\(\ f(x) = 0 \Rightarrow x^{3} - 7x + 6 = 0 \ \)
When we substitute x as 1, such as x = 1, here, we obtain 1 – 7 + 6 = 0, indicating that x = 1 is a zero.
Next, we divide the polynomial by (x - 1), resulting in the quotient of \(\ x^{2} + x - 6 \ \). This quadratic expression is factored as, \(\ f(x) = (x - 1)(x - 2)(x + 3) \ \) and the zeros are -3, 1, and 2.
Find the zeros of f(x)=2x2-3x2-2x+3=0
\(\ f(x) = 0 \Rightarrow 2x^{3} - 3x^{2} - 2x + 3 = 0 \ \)
When we try x = 1, we will get \(\ 2(1)^{3} - 3(1)^{2} - 2(1) + 3 = 2 - 3 - 2 + 3 = 0 \ \), which means x = 1 is a zero. When we divide the cubic polynomial by (x - 1), we will get the quotient that is \(\ 2x^{2} - x - 3 = (2x + 3)(x - 1) \ \). Thus, the fully factorized polynomial is, \(\ f(x) = (x - 1)^{2}(2x + 3) \ \) with the zero being 1 (repeated) and -32.
Find the zeros of f(x)=x3-4x2-x+4
\(\ f(x) = 0 \Rightarrow x^{3} - 4x^{2} - x + 4 = 0 \ \)
After experimenting with various values, we discovered that x = 1 satisfies the equation. When x = 1 is substituted into the polynomial, \(\ 1 - 4 - 1 + 4 = 0 \ \) is obtained, indicating that x = 1 is a zero. The polynomial is then divided by (x - 1), resulting in the quotient \(\ x^{2} - 3x - 4 \ \). The quadratic expression is then factored as follows: \(\ x^{2} - 3x - 4 = (x - 4)(x + 1) \ \). Therefore, the full factorization is\(\ f(x) = (x - 1)(x - 4)(x + 1) \ \), and the zeros of the polynomial are -1, 1, and 4.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






