For a polynomial p(z) of degree n>0, the equation p(z)=0 has exactly n solutions, counting each repeating solution the number of times the solution is repeated. This means that a solution repeated two times counts as two solutions, a solution repeated three times counts as three solutions, and so on.
Proof
Informal Justification
To justify this theorem, polynomials with low degrees will be examined.
Degree 1
Consider a polynomial of degree 1.
p1(x)=ax+b
The solution to the equation p1(x)=0 can be found by performing inverse operations.
Recall that p1(x) is a polynomial of degree 1. Therefore, the value of a is not 0. This means that x=-ab is a valid solution, and that polynomials of degree 1 always have one solution.
Degree 2
Now consider a polynomial of degree 2.
p2(x)=ax2+bx+c
This time, the Quadratic Formula can be used to find the solutions to p2(x)=0.
ax2+bx+c=0⇓x1,x2=2a-b±b2−4ac
There are three possible combinations for the resulting values of x1 and x2.
The values of x1 and x2 can be two different real numbers.
The values of x1 and x2 can be two different complex numbers.
The values of x1 and x2 can be the same real number.
When the solution is a single real number, it is said that that solution has multiplicity2. A solution of multiplicity 2 counts as two solutions. Therefore, a polynomial of degree 2 always has 2 solutions and p2(x) can be written as a product of two factors.
p2(x)=(x−x1)(x−x2)
Degree 3
Next, consider a polynomial of degree 3.
p3(x)=ax3+bx2+cx+d
This time, the graph of an arbitrary cubic polynomial will be examined.
The end behavior of a cubic polynomial is always down-up or up-down. Therefore, it always has at least one real solution. Let r1 be this solution. Using the Factor Theorem, it is possible to rewrite the equation p3(x)=0 as the product of (x−r1) and a polynomial of degree 2.
ax3+bx2+cx+d=0⇕(x−r1)(a2x2+b2x+c2)=0
It has been already established that the equation p2(x)=0, where p2(x) is a polynomial of degree 2, always has two solutions. This means that the equation p3(x) has exactly 3 solutions: r1,r2, and r3. Therefore, the left-hand side of the equation p3(x)=0 can be written as a product of three factors.
p3(x)=0⇕(x−r1)(x−r2)(x−r3)=0
Conclusion
Expanding upon what was determined so far, in general, if n is a natural number, a polynomial equation pn(x)=0 has n solutions, even if some solutions have a multiplicity greater than 1. Also, the left-hand side can be written as a product of n factors.
ForAnyNaturalNumbernpn(x)=0⇕(x−r1)…(x−rn)=0
It should be noted that this is an informal justification and should not be taken as a formal proof.