Polynomial equations are one of the major concepts of Mathematics, where the relation between numbers and variables are explained in a pattern. In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is also a form of the algebraic equation.
The equations formed with variables and exponents and coefficients are called as polynomial equations. It can have a number of different exponents, where the higher one is called the degree of the exponent. We can solve polynomial by factoring them in terms of degree and variables present in the equation.
A polynomial function is an equation which consists of a single independent variable, where the variable can occur in the equation more than one time with different degree of the exponent. Students will also learn here to solve these polynomial functions. The graph of a polynomial function can also be drawn using turning points, intercepts, end behavior and the Intermediate Value Theorem.
Example of polynomial function: f(x) = 3x2+5x+19
Polynomial Equations Formula
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_{n-1}x^{n-1} + a_{n-2}x^{n-2} + …….. + 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.
Types of Polynomial Equation
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
- 8z-3=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 â‰ 0
And a,b and c are real numbers.
The roots of quadratic equations gives two values for the variable x.
x = \(\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
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 â‰ 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 â‰ 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 y-1, we get
(y-1) (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.
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