Fundamental Theorem of Algebra

To justify this theorem, polynomials with low degrees will be examined.

Degree $1$

Consider a polynomial of degree $1.$ The solution to the equation $p_1(x)=0$ can be found by performing inverse operations. Recall that $p_1(x)$ is a polynomial of degree $1.$ Therefore, the value of $a$ is not $0.$ This means that $x=\text{-}\frac{b}{a}$ is a valid solution, and that polynomials of degree $1$ always have one solution.

Degree $2$

Now consider a polynomial of degree $2.$ This time, the Quadratic Formula can be used to find the solutions to $p_2(x)=0.$ There are three possible combinations for the resulting values of $x_1$ and $x_2.$

• The values of $x_1$ and $x_2$ can be two different real numbers.
• The values of $x_1$ and $x_2$ can be two different complex numbers.
• The values of $x_1$ and $x_2$ can be the same real number.

When the solution is a single real number, it is said that that solution has multiplicity $2.$ A solution of multiplicity $2$ counts as two solutions. Therefore, a polynomial of degree $2$ always has $2$ solutions and $p_2(x)$ can be written as a product of two factors.

Degree $3$

Next, consider a polynomial of degree $3.$ 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 $r_1$ be this solution. Using the Factor Theorem, it is possible to rewrite the equation $p_3(x)=0$ as the product of $(x-r_1)$ and a polynomial of degree $2.$ It has been already established that the equation $p_2(x)=0,$ where $p_2(x)$ is a polynomial of degree $2,$ always has two solutions. This means that the equation $p_3(x)$ has exactly $3$ solutions: $r_1,$ $r_2,$ and $r_3.$ Therefore, the left-hand side of the equation $p_3(x)=0$ can be written as a product of three factors.

Conclusion

Expanding upon what was determined so far, in general, if $n$ is a natural number, a polynomial equation $p_n(x)=0$ 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. It should be noted that this is an informal justification and should not be taken as a formal proof.