Last updated on July 9th, 2025
Polynomial equations are important in math because they show how numbers and variables are connected through a specific pattern. These are a common type of algebraic equation that is formed using coefficients and variables arranged in a specific pattern. In this article, we will discuss polynomial equations, their formulas, types, and applications in detail.
A polynomial equation is an equation with a polynomial expression equals to zero. It is made up of variables, numbers, and basic operations like addition, subtraction, and multiplication. The highest power of the variable shows the degree of the polynomial. In a polynomial equation, the exponents must be nonnegative numbers; negative numbers or fractions are not allowed. If this condition is satisfied, the equation can be classified as a polynomial equation.
For example, 2x² + 3x + 1 is a polynomial. When we set it to zero, it becomes a polynomial equation:
2x² + 3x + 1 = 0
Look at more such equations:
A polynomial is an algebraic expression made up of variables and constants. We use addition, subtraction, and multiplication to work with polynomials. The variables in a polynomial have powers that are non-negative integers.
Example: 2x + 3
A polynomial equation is formed when a polynomial is made equal to a value using the equal sign (=).
Example: 2x + 3 = 0
Note:
A polynomial equation is in the form:
p(x) = 0
In algebra, it can be written in the form:
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ = 0
Where:
Depending on the degree of a polynomial equation, it is classified into different types. We will learn about each one step by step:
Linear Polynomial Equation
Linear polynomials are equations where the variable has the highest power of 1.
General Form: ax + b = 0
Examples:
4x + 5 = 0
7x – 9 = 0
Quadratic Polynomial Equation
Quadratic polynomial equations are equations where the variable has the highest power of 2.
General Form: ax² + bx + c = 0
Examples:
2x² – 4x + 6 = 0
5x² + 3x – 8 = 0
Cubic Polynomial Equation
If the highest power of the variable is 3, then it is said to be a cubic polynomial equation.
General Form: ax³ + bx² + cx + d = 0
Examples:
x³ + 2x² – 3 = 0
2x³ – x + 4 = 0
Biquadratic Polynomial Equation
Biquadratic polynomial equations are polynomial equations in which the highest power of the variable is 4.
General Form: ax⁴ + bx³ + cx² + dx + e = 0
Example: 4x⁴ + 2x² – 6x + 1 = 0.
To solve the equation p(x) = 0, we have to find the value or values of x that satisfy the equation. Such values are called roots or zeros because p(a) = 0, where ‘a’ is a root of the polynomial. Let’s learn how different polynomial equations are solved.
Forming Linear Polynomial Equations
To solve a linear polynomial equation, we set it equal to zero and find the value of the variable.
Example:
Solve 6x - 9 = 0
Step 1: Add 9 to both sides:
6x = 9
Step 2: Divide both sides by 6:
x = 9/6 = 3/2
Solving a Quadratic Polynomial from its Roots
To find a quadratic polynomial when its roots are known, we utilize the fact that the polynomial can be formed using the sum and product of the roots.
Find the quadratic polynomial whose roots are -3 and 5.
Step 1: Sum of roots = -3 + 5 = 2
Step 2: Product of roots = -3 × 5 = -15
Step 3: Use the formula
x² - (sum of roots)x + (product of roots) = 0
Substitute values:
x² - 2x - 15 = 0
Solving Cubic Polynomial Equations
To solve a cubic polynomial equation, we find its roots by factoring or by using techniques such as synthetic division and the factor theorem. Let’s solve it step by step.
General form:
ax³ + bx² + cx + d = 0, a ≠ 0
Steps:
1. Write the equation in standard form.
2. Use the factor theorem or synthetic division to factor into a linear factor and a quadratic factor.
3. Find the remaining roots by solving the quadratic equation
Example:
Solve x³ - 5x² + 8x - 4 = 0.
Step 1: The given equation is already in standard form.
Step 2: Substitute x = 1 as a root:
1 – 5 + 8 – 4 = 0, so x = 1 is a root.
Polynomial equations have several practical applications in different fields and help students develop problem-solving skills. Let’s now learn how they are applied in various real-life situations:
When working with polynomial equations, students sometimes misunderstand the concept of polynomial equations, which leads to several errors. Here are a few common mistakes and ways to avoid them:
Solve: 2x + 3 = 7
x = 2
Consider the given equation:
2x + 3 = 7
We subtract 3 from both sides, to isolate the term with x:
2x + 3 – 3 = 7 – 3
2x = 4Now, divide both sides by 2:
2x ÷ 2 = 4 ÷ 2
x = 2
Solve: x² – 9 = 0
x = ±3
Step-by-step solution:
Write the equation:
x² – 9 = 0
Add 9 to both sides:
x² = 9
Take the square root of both sides:
x = ±√9
x = ±3
So, x can have values (–3) and 3
Solve: x³ – 4x = 0
x = 0, x = 2, or x = –2
First, write the equation:
x³ – 4x = 0
Factor out the common term (x):
x(x² – 4) = 0
Factor the difference of squares:
x(x – 2)(x + 2) = 0
Set each factor to 0:
x = 0
x – 2 = 0 → x = 2
x + 2 = 0 → x = –2
So, x = 0, x = 2, or x = –2
Solve: x² + 5x + 6 = 0
x = –2 or x = –3
Write the equation:
x² + 5x + 6 = 0
We then factor the quadratic expression:
Look for two numbers that multiply to 6 and add to 5 → (2 and 3)
(x + 2)(x + 3) = 0
Set each factor to 0:
x + 2 = 0 → x = –2
x + 3 = 0 → x = –3
So, x = –2 or x = –3
Solve: 3x² – 7x – 6 = 0
x = 3 or x = -⅔
We first identify the coefficients:
a = 3, b = -7, c = -6
Use the quadratic formula
x = [-b ± √(b² – 4ac)] / 2a
Let’s substitute the values into the formula
x = [-(-7) ± √((-7)² – 4×3×-6)] / (2×3)
x = [7 ± √(49 + 72)] / 6
x = [7 ± √121] / 6
Simplify the square root
x = [7 ± 11] / 6
Find the two solutions
x = (7 + 11)/6 = 18/6 = 3
x = (7 –11)/6 = -4/6 = -2/3
So, x = 3 or x = -2/3.