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Last updated on September 14, 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 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, 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 ax3+bx2+cx +d with zeros p, q, r:
p+q+r=-ba it is the sum of zeros.
pq+qr+rp=ca it is the sum of the products taken two at a time.
pqr=-da it is the product of the zeros.
Consider a cubic polynomial with the formula ax3+bx2+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 x3-12x2+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.
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.
Let α, β, and γ be the cubic polynomial ax3+bx2+cx+d=0, where a ≠ 0 and a, b, and c are the coefficients of x3, x2, 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 = - ( constant term) / (coefficient of x3).
⇒ αβγ=-da.
Example: Determine the cubic polynomial kx3-5x2-12x+k=0.
If the polynomial is kx3-5x2-12x+k=0 , then we have a = k, and d = k.
Product of zeros = -(constant term)/(coefficient of x3)
⇒αβγ=-kk=-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 ax3+bx2+cx+d=0, where a ≠ 0, where a, b, and c are the coefficients of x3,x2, and x and d is the constant term. Next, a cubic polynomial’s sum of zeros is expressed as follows:
The sum of zeros of a cubic polynomial = -(coefficient of x2)/(coefficient of x3)
⇒α+β+γ=-ba.
Example: Determine the cubic polynomial 5x3-15x2-12x+27=0.
Sum of zeros of a cubic polynomial = -(coefficient of x2) / (coefficient of x3)
⇒α+β+γ=-ba.
Example: Determine the cubic polynomial 5x3-15x2-12x+27=0.
If the polynomial is 5x3-15x2-12x+27=0 , then we have
a= 5 and b = -15.
Sum of zero = -ba
⇒ α+β+γ=--155=3
Therefore, the given cubic polynomial’s sum of zeros is 3.
Consider a cubic polynomial with the formula ax3+bx2+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=b2c2-4ac3-4b3d-27a2d2+18abcd
Therefore,
D < 0.
For example,
Determine the type of zeros in the cubic polynomial x3+x=0. Applying the discriminant formula.
Given that x3+x=0 is a cubic polynomial, we have
a = 1, b = 0, c = 1, and d = 0.
We know, D=b2c2-4ac3-4b3d-27a2d2+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 ax3+bx2+cx+d=0, where a ≠ 0, and a, b, and c are the coefficients of x3, x2, x and d is the constant term.
Then, a cubic polynomial’s zeros and coefficients have the following relationship:
Sum of zeros = -ba
⇒α+β+γ=-ba.
Sum of the product of zeros = ca.
⇒αβ+βγ+γα=ca.
Product of zeros = -da
⇒αβγ=-da.
For Example:
Check the connection between zeros and coefficients of x3+5x2=6x=0 if the zeros are given as -6, -0, and -1.
Comparing the given cubic polynomial with ax3+bx2+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.
α+β+γ=-ba
⇒(-6)+0+1=-51
⇒ -5= -5.
Confirm the total product of the cubic polynomial’s zeros, i.e.
αβ+βγ+γα=ca
⇒(-6)(0)+(0)(1)+(1)(-6)=-6 1
⇒-6=-6
Confirm that the cubic polynomial’s zeros are multiplied, that is,
αβγ=-da
⇒(-6)(0)(1)=-01
⇒ 0=0.
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.
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.
Find the zeros of f(x)=x3-6x2+11x-6
f(x)=0 ⇒ x3-6x2+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, x2-5x+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 ⇒ x3+3x-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 x2+5x+6. After factoring this with a quadratic component, we obtain x2+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⇒x3-7x+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 ofx2+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 ⇒ 2x2-3x2-2x+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 2x2-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 ⇒ x3-4x2-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 x2-3x-4. The quadratic expression is then factored as follows: x2-3x-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.
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