Polynomial equations are equations that come in the following form:
We say that number is the solution or a root of an algebraic equation nth degree if:
To find root of an equation is the same as finding the root of matching polynomial. You already know how to solve some simple polynomial equations.
Example 1. Find all roots of the following equation.
First root is zero with multiplicity 3, second and third -2 with multiplicity 1.
What happens if things get even more complicated and you can’t get quadratic equation and simply take roots? For more complicated tasks we’ll use Bézout’s theorem and few theorems for specific problems.
Bézout’s theorem says that number α is a root of a polynomial if and only if polynomial f is divisible by polynomial
Now we’ll show you shorten division of a polynomial with linear polynomial. If this procedure is harder for you to understand, feel free to divide it step by step. The results must be the same.
Example 2. Divide with
The first thing you absolutely must not forget is to line up the terms according to their exponents, going from the largest exponent to the smallest.
Now, draw a table that has two rows and columns as many as given expression has terms with one extra. In those columns and first row write the coefficients in order. In the left corner write the number you are dividing with.
The first number from the first row comes down.
You multiply that first number with the zero of the polynomial you are dividing with and add the next right number . This will be your next lower number. The procedure continues.
With this we came to an end with our algorithm. If the last number we got is equal to zero, those two polynomials are divisible. If there stands any other number, that number is their remainder.
Now what about the other factor? You’ll read it off of the table’s second row. Those numbers you got mark the coefficients with matching unknowns.
If you are dividing polynomial of nth degree with a linear polynomial, the first exponent will be . Here we divided with which means that our first exponent of second factor will be 3.
This leads us to the solution: