Polynomial Equations

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Section 4.5 Polynomial Equations

A degree $n$ polynomial equation is one that is equivalent to

\begin{equation} a_n x^n + a_{n-1} x^{n-1} +\cdots a_1 x + a_0 = 0,\tag{4.1} \end{equation}

where $a_n, a_{n-1}, \ldots, a_0$ are real number coefficients and $a_n\neq 0\text{.}$ For example, a linear equation $ax+b=0$ is a degree $n=1$ polynomial equation, while a quadratic equation $ax^2+bx+c=0$ is a degree $n=2$ polynomial equation. Solutions to (4.1) are called zeros or roots of the polynomial. Finding the zeros can be quite challenging in practice, depending on the degree and complexity of the expression. For this section, we will rely on factoring and the Principle 4.19.

Example 4.26. Solving by Factoring.

Let’s solve

\begin{equation*} 2 x^5 = 18x. \end{equation*}

First, move terms to one side to make zero appear:

\begin{equation*} 2 x^5 - 18x = 0. \end{equation*}

Next factor,

\begin{equation*} 2x(x^4 - 9) = 0. \end{equation*}

The second factor is a difference of squares,

\begin{equation*} 2x(x^2 - 3)(x^2+3) = 0. \end{equation*}

\begin{equation*} 2x(x - \sqrt{3})(x+\sqrt{3})(x^2+3) = 0. \end{equation*}

So,

\begin{equation*} 2x =0 \quad \text{OR} \quad x - \sqrt{3} = 0 \quad \text{OR} \quad x+\sqrt{3} = 0 \quad \text{OR} \quad x^2+3 =0. \end{equation*}

The first three factors give solutions $x=0, \sqrt{3}, -\sqrt{3}\text{.}$ The last $x^2 + 3 = 0$ has no real number solutions as $x^2+3 \gt 0$ for all real numbers $x\text{.}$

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