Chapter 2.5, Problem 1E

To determine

The ________ of ________ states that if $f(x)$ is a polynomial of degree n (n > 0), then f has at least one zero in the complex number system.

Answer to Problem 1E

Fundamental theorem, algebra

Explanation of Solution

The fundamental theorem of algebra states that if $f(x)$ is a polynomial of degree n (n > 0), then f has at least one zero in the complex number system.

For example, consider a polynomial function $f(x)=x^3+10x^2+33x+34$ .

Finding the roots of above polynomial function,

The leading coefficient of given polynomial is 1 and the constant term is 34.

So, possible rational zeros are: $\frac{\text{Factors of}34}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 17,\pm 34}{\pm 1}=\pm 1,\pm 2,\pm 17,\pm 34$

Using synthetic division to check whether $x=-2$ , is a zero or not.

  $\begin{array}{ccccc}-2& 1& 10& 33& 34\\ & & -2& -16& -34\\ & 1& 8& 17& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\hspace{1em}\hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=-2$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x+2\right)(x^2+8x+17)$

Now, consider $x^2+8x+17.$

The above polynomial function is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-(8)\pm \sqrt{{(8)}^2-4(1)(17)}}{2(1)}\\ \Rightarrow x=\frac{-8\pm \sqrt{64-68}}{2}\\ \Rightarrow x=\frac{-8\pm \sqrt{-4}}{2}\\ \Rightarrow x=\frac{-8\pm 2\sqrt{-1}}{2}\\ \Rightarrow x=-4\pm i\\ \Rightarrow x=-4+i,-4-i\end{array}$

Therefore, the zeros of given polynomial function are $x=-2,-4+i\text{and}-4-i.$

Here, the polynomial has 2 complex zeros in complex number system.