In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is a form of the algebraic equation. Some equations are formed with variables and exponents and coefficients, which is called as polynomial equations.
Polynomial equations are the expression which consists of variables and coefficients. It includes the operations of addition, subtraction and multiplication of the variables to form an equation. The exponents of variables are always positive integers in polynomial equations, that negative exponents and fraction exponents donâ€™t form a polynomial equation.
The polynomial equations are expressed in algebraic forms, which consist of variables and integer numbers and it also has multiple types which we have discussed here, in this article. This algebraic expression acts like a polynomial equation solver. Apart from these, we will also learn to solve the polynomial equations examples and problem questions.
Polynomial Equations Expression
Usually, the polynomial equation is expressed in the form of a_{n}(x^{n}). Here a is the coefficient, x is the variable and n is the exponent. As we have already discussed in the introduction part, the value of exponent should always be a positive integer.
If we expand the polynomial equation we get;
F(x) = a_{n}x^{n} + a_{n1}x^{n1} + a_{n2}x^{n2} + …….. + a_{1}x +a_{0}
This is the general expression and is also a polynomial equation solver. It can also be expressed as;
F(x) = \(\sum_{k=0}^{n}a_{k}n^k\)
Example of a polynomial equation is: 2x^{2 }+ 3x + 1 = 0, where 2x^{2 }+ 3x + 1 is basically a polynomial expression which has been set equal to zero, to form a polynomial equation.
Polynomial Equation Types
Polynomial Equation is basically of three types;
 Monomial Equations
 Binomial Equations
 Trinomial Equations
Monomial Equation: An equation which has only one variable term is called a Monomial equation. This is also called a Linear equation. It can be expressed in the algebraic form of;
ax + b = 0
For Example:
 4x+1=0
 5y=2
 8z3=0
Binomial Equations: An equation which has only two variable terms and is followed by one variable term is called a Monomial equation. This is also in a form of the quadratic equation. It can be expressed in the algebraic form of;
ax^{2 }+ bx + c = 0
For Example:
 2x^{2 }+ 5x + 20 = 0
 3x^{2 }– 4x + 12 = 0
Trinomial Equations: An equation which has only three variable terms and is followed by two variable and one variable term, is called a Monomial equation. This is also called as cubic equation. It can be expressed in the algebraic form of;
ax^{3 }+ bx^{2} + cx + d = 0
For Example:
 3x^{3 }+ 12x^{2} – 8x – 10 = 0
 9x^{3 }+ 5x^{2} – 4x – 2 = 0
Quadratic Polynomial Equation
A polynomial equations which has degree as two, is called as quadratic equation.
The expression for quadratic equation is:
ax^{2 }+ bx + c = 0 ; a ≠ 0
And a,b and c are real numbers.
The roots of quadratic equations gives two values for the variable x.
x = \(\frac{b\pm \sqrt{b^24ac}}{2a}\)
Cubic Polynomial Equation
A polynomial equation which has degree as three is called a cubic polynomial equation. Since the power of the variable is maximum up to 3, therefore, we get three values for variable say x.
It is expressed as;
a_{0} x^{3} + a_{1}x^{2} + a_{2}x + a_{3 }= 0, a ≠ 0
To get the value of x, we generally use, trial and error method, in which we start putting the value of x randomly, to get the given expression as 0. If for both sides of the polynomial equation, we get a 0 ,then the value of x is considered as one of the roots. After then we can find the other two values of x.
Let us take an example:
Problem: y^{3 }– y^{2 }+ y – 1 = 0 is a cubic polynomial equation. Find the roots of it.
Solution: y^{3 }– y^{2 }+ y – 1 = 0 is the given equation.
By trial and error method, start putting the value of x.
If y = 1, then,
(1)^{3 }– (1)^{2} 1 +1 = 0
1 + 1 1 + 1 = 0
4 ≠ 0
If y = 1, then,
1^{3 }– 1^{2 }+ 1 – 1 = 0
0 = 0
Therefore, one of the roots is 1.
y = 1
(y – 1) is one of the factors.
Now dividing the given equation by x+1 on both sides, we get,
y^{3 }– y^{2 }+ y – 1 = 0
Dividing both sides by y1, we get
(y1) (y^{2} + 1) = 0
Therefore, the roots are y = 1 which is a real number and y^{2 }+ 1 gives complex numbers or imaginary numbers.
Also: Get Polynomial Equation Solver here with us and solve all the polynomial equations.
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