Cubic Polynomial
Cubic polynomial is a type of polynomial based on the degree i.e. the highest exponent of the variable. Hence, a cubic polynomial is a polynomial with the highest power of the variable or degree is 3. A polynomial is an algebraic expression with variables and constants with exponents as whole numbers. Let us learn more about cubic polynomials, the definition, the formulas, and solve a few examples.
1.  Definition of Cubic Polynomial 
2.  Solving Cubic Polynomial 
3.  Graph of Cubic Polynomial 
4.  Roots of Cubic Polynomial 
5.  FAQs on Cubic Polynomial 
Definition of Cubic Polynomial
A cubic polynomial is a polynomial with the highest exponent of a variable i.e. degree of a variable as 3. Based on the degree, a polynomial is divided into 4 types namely, zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial. The general form of a cubic polynomial is p(x): ax^{3} + bx^{2} + cx + d, a ≠ 0, where a, b, and c are coefficients and d is the constant with all of them being real numbers. An equation involving a cubic polynomial is called a cubic equation. Some of the examples of a cubic polynomial are p(x): x^{3} − 5x^{2} + 15x − 6, r(z): πz^{3} + (√2)^{10}.
Cubic Polynomial Formula
The cubic polynomial formula is in the general form of ax^{3} + bx^{2} + cx + d and the formula for the solution of the cubic equation is ax^{3} + bx^{2} + cx + d = 0.
Solving Cubic Polynomial
The general form of a cubic polynomial is ax^{3} + bx^{2} + cx + d, a ≠ 0. While solving a cubic polynomial, we always have to rearrange the equation to a cubic equation, breaking it down to a quadratic equation, and then solve it by using two different ways  factor theorem and synthetic division method. Let us see how to solve equations in both methods.
Cubic Polynomial With Synthetic Division
Synthetic division is a method used to perform the division operation on polynomials when the divisor is a linear factor. We can represent the division of two polynomials in the form: p(x)/q(x) = Q + R/(q(x))
where,
 p(x) is the dividend
 q(x) is the linear divisor
 Q is quotient
 R is remainder
While solving a cubic polynomial we use the synthetic division method and the steps are:
 Step 1: Check whether the cubic polynomial is in the standard form.
 Step 2: Write the coefficients in the dividend's place and write the zero of the linear factor in the divisor's place.
 Step 3: Bring the first coefficient down, multiply and write it below the next coefficient.
 Step 4: Add them and write the value below.
 Step 5: Repeat the previous 2 steps until you reach the last term.
 Step 6: Separate the last term thus obtained which is the remainder.
 Step 7: Now group the coefficients with the variables to get the quotient.
Cubic Polynomial With Factor Theorem
Factor theorem is a kind of polynomial remainder theorem that links the factors of a polynomial and its zeros. As per the factor theorem, (x – a) can be considered as a factor of the polynomial p(x) of degree n ≥ 1, if and only if p(a) = 0. Here, a is any real number. The formula of the factor theorem is p(x) = (x – a) q(x). It is important to note that all the following statements apply for any polynomial p(x):
 (x – a) is a factor of p(x).
 p(a) = 0.
 The remainder is zero when p(x) is divided by (x – a).
 The solution to p(x) = 0 is a and the zero of the function p(x) is a.
Factor theorem is mainly applicable for polynomials with the degree of zero, one, and two. For degrees like 3 and 4, such as a cubic polynomial, factor theorem is used along with synthetic division and the steps are as follows:
 Step 1: Use the synthetic division method to divide the given polynomial p(x) by the given binomial (x−a)
 Step 2: Once the division is completed the remainder should be 0. If the remainder is not zero, then it means that (xa) is not a factor of p(x).
 Step 3: Using the division algorithm, write the given polynomial as the product of (xa) and the quadratic quotient q(x)
 Step 4: If it is possible, factor the quadratic quotient further.
 Step 5: Express the given polynomial as the product of its factors.
For example: Consider this division: (x^{3}  2x^{3}  8x  35)/(x  5). The polynomial is of order 3. The divisor is a linear factor. Let's use synthetic division to find the quotient. Thus, the quotient is one order less than the given polynomial. It is x^{2} + 3x + 7 and the remainder is 0. (x^{3}  2x^{3}  8x  35)/(x  5) = x^{2} + 3x + 7.
Graph of Cubic Polynomial
A cubic polynomial function of the third degree has the form shown on the right and it can be represented as y = ax^{3} + bx^{2} + cx + d, where a, b, c, and d are real numbers and a ≠ 0. When a cubic polynomial cannot be solved with the abovementioned methods, we can solve it graphically. The points where the graph crosses the xaxis are considered the solution and are called the roots of a cubic polynomial. While plotting a graph for a cubic polynomial, we need to remember two important aspects:
 If the sign of a is positive, then the graph will be from down to up.
 If the sign of a is negative, then the graph will be up to down.
The graph of a cubic polynomial looks like this:
Cubic Polynomial Function
Cubic polynomials can be solved in the similar manner as quadratic equations. But to make it to a much simpler form, we can use some of these special products:
 Perfect cube (2 forms): a^{3} ± 3a^{2}b + 3ab^{2} ± b^{3} = (a ± b)^{3}
 Difference of the cubes: a^{3} − b^{3} = (a − b)(a^{2} + ab +b^{2})
 Sum of the cubes: a^{3} + b^{3} = (a + b)(a^{2} − ab + b^{2})
Let us solve this cubic polynomial function y^{3} – 2y^{2} – y + 2. We will start by factorizing the equation:
y^{3} – 2y^{2} – y + 2 = y^{2}(y – 2) – (y – 2)
= (y^{2} – 1) (y – 2)
= (y + 1) (y – 1) (y– 2)
y = 1, 1 and 2.
Roots of Cubic Polynomials
The solution to a cubic equation is called the roots of the cubic equation. In most cases, there are 3 roots of a cubic polynomial but sometimes we do get two or only one. When a cubic polynomial is solved graphically, we get to the accurate roots or when we solve the equation with the formula, we derive at the roots. Let us say p,q, and r are 3 roots for the equation ax^{3} + bx^{2} + cx + d. The formulas are:
 p + q + r =  b/a
 pq + qr + rp = c/a
 pqr =  d/a
For example: Solve the following cubic equation, x^{3}  12x^{2} + 39x  28 = 0.
Solution: Let us consider 3 roots a  b, a, and a + b.
p = a  b, q = a, r = a + b
From the equation, x^{3}  12x^{2} + 39x  28 = 0 we know,
a = 1, b =  12, c = 39, d =  28
Sum of the roots =  b/a
a  b + a + a + b =  (12)/1
3a = 12
a = 4.
We can find out the two roots by factorizing the equation into a quadratic equation. Look at the image below.
x^{2}  8x + 7 = 0
x^{2}  7x  x + 7 = 0
x(x  7)  1 (x  7) = 0
(x  1) (x  7) = 0
x 1 = 0 and x  7 = 0
x = 1 and x = 7
Therefore, the roots are 1, 4 and 7.
Related Topics
Listed below are a few topics that are related to the cubic polynomial, take a look.
Cubic Polynomials Examples

Example 1: Check whether 2y + 1 is a factor of the polynomial 4y^{3} + 4y^{2} – y – 1 or not using the factor theorem.
Solution:
Let's equate the given binomial 2y + 1 = 0.
∴ y = 1/2
Substitute y = 1/2 in the given polynomial equation 4y^{3} + 4y^{2} – y – 1.
⟹ 4( 1/2)^{3} + 4(1/2)^{2} – (1/2) – 1
= 1/2 + 1 + 1/2 – 1
= 0
The remainder = 0, thus, 2y + 1 is a factor of the polynomial equation 4y^{3} + 4y^{2} – y – 1.

Example 2: Classify the given polynomials as cubic polynomials:
p(x): 5x^{2 }+ 6x + 1
q(z): z^{2 }− 1
p(y): y^{3} − 6y^{2} + 11y − 6
q(y): 81y^{3} − 1Solution: From the four polynomials, only two are cubic polynomials. They are: p(y): y^{3} − 6y^{2} + 11y − 6 and q(y): 81y^{3} − 1.
FAQs on Cubic Polynomials
What is Cubic Polynomial With Example?
A cubic polynomial is a type of polynomial with the degree of 3 i.e. the highest exponent of the variable is 3. The general form of a cubic polynomial is written as p(x): ax^{3} + bx^{2} + cx + d, a ≠ 0, where a, b, and c are coefficients and d is the constant with all of them being real numbers.
How Do You Find Cubic Polynomials?
Cubic polynomials can be solved by converting a cubic equation into a quadratic equation. Solving a cubic polynomial is done in two ways  factor theorem and synthetic division.
How to Use Factor Theorem for Cubic Polynomials?
The follows steps are used to solve cubic polynomials by factor theorem:
 Use the synthetic division method to divide the given polynomial p(x) by the given binomial (x−a)
 Once the division is completed the remainder should be 0. If the remainder is not zero, then it means that (xa) is not a factor of p(x).
 Using the division algorithm, write the given polynomial as the product of (xa) and the quadratic quotient q(x)
 If it is possible, factor the quadratic quotient further.
 Express the given polynomial as the product of its factors.
How to Use Synthetic Division Method for Cubic Polynomials?
Cubic polynomials are solved by the synthetic method using general steps like taking the coefficients alone, bringing the first down, multiplying with the zero of the linear factor, and adding with the next coefficient, and repeat until the end.
What is the Formula to Form a Cubic Polynomial?
The formula to form a cubic polynomial is ax^{3} + bx^{2} + cx + d = 0.
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