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Section Solving Polynomial Equations
Solving polynomial equations is a foundational skill in mathematics, and it plays a critical role in differential equationsβparticularly when solving linear differential equations with constant coefficients. In that context, we transform the differential equation into a polynomial equation, known as the characteristic equation, and solve it to find exponential solutions.
Higher-degree polynomial equations have the form
\begin{align*}
\text{degree 3:} \quad \amp a\ x^3 + b\ x^2 + c\ x + d = 0, \\
\text{degree 4:} \quad \amp a\ x^4 + b\ x^3 + c\ x^2 + d\ x + e = 0, \\
\text{degree 5:} \quad \amp a\ x^5 + b\ x^4 + c\ x^3 + d\ x^2 + e\ x + f = 0,\\
\amp \vdots
\end{align*}
and it turns out that these equations can always be factored into simpler polynomials. In particular, a polynomial of degree \(n\) can always be factored into \(n\) linear factors:
\begin{align*}
\text{degree 3:} \quad \amp (x - x_1)(x - x_2)(x - x_3) = 0, \\
\text{degree 4:} \quad \amp (x - x_1)(x - x_2)(x - x_3)(x - x_4) = 0, \\
\text{degree 5:} \quad \amp (x - x_1)(x - x_2)(x - x_3)(x - x_4)(x - x_5) = 0, \\
\amp \vdots
\end{align*}
where \(x_1, x_2, x_3, ...\) are the solutions and can be real or complex. While knowing this is powerful, the process of factoring them can be quite challenging. However, there are some special forms and strategies that can help. A few are summarized below.
β³οΈ Techniques for Solving Higher-Degree Polynomials.
Recognizing Special Forms
Some polynomials can be factored using special patterns. Common forms include:
Common Factoring: \(ax^n + bx^{n-1} = x^{n-1}(ax + b)\)
Difference of Squares: \(a^2 - b^2 = (a - b)(a + b)\)
Sum/Difference of Cubes: \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)
Recognizing these forms can help simplify the factorization process.
Known Factors
If you know one factor, then you can βdivide-outβ the polynomial by this factor. For example, suppose we know that \(x = 1\) is a root of the polynomial
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = 0\text{.}
\end{equation*}
Then, we know that \((x - 1)\) is a factor, so
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = (x - 1)p(x)
\end{equation*}
where \(p(x)\) is some second-degree polynomial you multiply by \((x - 1)\) to get our original polynomial. We can find \(p(x)\) by dividing both sides by \(x - 1\text{,}\) like so
\begin{equation*}
p(x) = \frac{x^3 - 6x^2 + 11x - 6}{x - 1} = x^2 - 5x + 6
\end{equation*}
How?
Use polynomial divison to compute
\begin{equation*}
\frac{x^3 - 6x^2 + 11x - 6}{x-1} =\ ?
\end{equation*}
\(x^2\) \(-\) \(5x\) \(+\) \(6\)
\(x-1\) \(x^3\) \(-\) \(6x^2\) \(+\) \(11x\) \(-\) \(6\)
\(x^3\) \(-\) \(x^2\)
\(-5x^2\) \(+\) \(11x\) \(-\) \(6\)
\(-5x^2\) \(+\) \(5x\)
\(6x\) \(-\) \(6\)
\(6x\) \(-\) \(6\)
\(0\)
Therefore,
\begin{equation*}
\frac{x^3 - 6x^2 + 11x - 6}{x-1} = x^2 - 5x + 6\text{.}
\end{equation*}
Therefore,
\begin{align*}
x^3 - 6x^2 + 11x - 6 \amp = (x - 1)(x^2 - 5x + 6)\\
\amp = (x - 1)(x - 2)(x - 3)
\end{align*}
Rational Root Theorem
If a polynomial has rational roots, they must be of the form:
\begin{equation*}
\pm \frac{\text{factor of constant term,}\ a_0}{\text{factor of leading coefficient,}\ a_n}.
\end{equation*}
This gives us a list of possible solutions to test. For example, if we have the polynomial
\begin{equation*}
2x^3 - 3x^2 - 11x + 6 = 0
\end{equation*}
then the possible factors of \(a_0 = 6\) are \(1, 2, 3, 6\) and the possible factors of \(a_n = 2\) are \(1, 2\text{.}\) Therefore, the possible fractional solutions are
\begin{equation*}
\pm \frac{\text{factor of 6}}{\text{factor of 2}} = \pm \frac11, \pm \frac12, \pm \frac21, \pm \frac22, \pm \frac31, \pm \frac32, \pm \frac61, \pm \frac62
\end{equation*}
We can test each of these values to find up to 3 solutions. Once we find one, we can use the previous technique to help find more.
Use Technology
Factoring higher-order polynomials can be very difficult to do by and this is one skill that may be better suited for a computer. There are many software packages that can factor polynomials for you. For example, the Wolfram Alpha website will do it with ease.
π Example 316 . Example: Solving a Polynomial Equation.
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = 0
\end{equation*}
Solution .
Try rational roots. Testing \(x = 1\) gives:
\begin{equation*}
1^3 - 6(1)^2 + 11(1) - 6 = 0.
\end{equation*}
So \((x - 1)\) is a factor. Divide the polynomial to find:
\begin{equation*}
x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6).
\end{equation*}
Factoring the quadratic:
\begin{equation*}
(x - 1)(x - 2)(x - 3) = 0.
\end{equation*}
The roots are \(x = 1, 2, 3\text{.}\)
Here are a few quick examples. These illustrate techniques like factoring out common terms, recognizing perfect squares, or applying the Rational Root Theorem. All of these help us find roots that lead to solutions of linear homogeneous equations.
π Example 317 .
Factor and solve the following polynomial equations:
\(\displaystyle 4\ x^2 - 9 = 0 \)
\(\displaystyle x^2 - 3 = 0 \)
\(\displaystyle x^3 + 3\ x^2 - 4\ x = 0 \)
\(\displaystyle x^5 + 10\ x^4 = 0 \)
\(\displaystyle x^4 - 100\ x^2 = 0 \)
\(\displaystyle x^5 - 4\ x^3 = 0 \)
\(\displaystyle x^4 - 32 = 0 \)
Solution .
The degree of each equation tells you how many factors you should have.
\(4\ x^2 - 9 = 0 \)
\((2\ x + 3)(2\ x - 3) = 0 \)
\(\leftarrow\) difference of squares
\(2\ x + 3 = 0, \quad 2\ x - 3 = 0 \)
\(\leftarrow\) set each factor to 0
\(\ds x_1 = -\frac32, \quad x_2 = \frac32 \)
\(\leftarrow\) solutions
\(x^2 - 3 = 0 \)
\((x + \sqrt{3})(x - \sqrt{3}) = 0 \)
\(\leftarrow\) difference of squares
\(\ds x_1 = -\sqrt{3}, \quad x_2 = \sqrt{3} \)
\(\leftarrow\) solutions
\(x^3 + 3\ x^2 - 4\ x = 0 \)
\(x\ (x^2 + 3\ x - 4) = 0 \)
\(\leftarrow\) common factor
\(x\ (x + 4)(x - 1) = 0 \)
\(\leftarrow\) standard quadratic factoring
\(\ds x_1 = 0, \quad x_2 = -4, \quad x_3 = 1 \)
\(\leftarrow\) solutions
\(x^5 + 10\ x^4 = 0 \)
\(x^4\ (x + 10) = 0 \)
\(\leftarrow\) common factor
\(\ds x_1 = 0\ (\text{4 repeats}), \quad x_2 = -10 \)
\(\leftarrow\) solutions
Technically, \(x^4 = (x-0)^4\) and represents 4 repeated factors.
\(x^4 - 100\ x^2 = 0 \)
\(x^2\ (x^2 - 100) = 0 \)
\(\leftarrow\) common factor
\(x^2\ (x + 10)(x - 10) = 0 \)
\(\leftarrow\) difference of squares
\(\ds x_1 = 0\ (\text{2 repeats}), \quad x_2 = -10, \quad x_3 = 10 \)
\(\leftarrow\) solutions
\(x^5 - 4\ x^3 = 0 \)
\(x^3\ (x^2 - 4) = 0 \)
\(\leftarrow\) common factor
\(x^3\ (x + 2)(x - 2) = 0 \)
\(\leftarrow\) difference of squares
\(\ds x_1 = 0\ (\text{3 repeats}), \quad x_2 = -2, \quad x_3 = 2 \)
\(\leftarrow\) solutions
\(x^4 - 25 = 0 \)
\((x^2 + 5)(x^2 - 5) = 0 \)
\(\leftarrow\) difference of squares
\((x^2 + 5)(x - \sqrt{5})(x + \sqrt{5}) = 0 \)
\(\leftarrow\) difference of squares
\(x^2 + 5 = 0, \quad x - \sqrt{5} = 0, \quad x + \sqrt{5} = 0 \)
\(\leftarrow\) set each factor to 0
\(\ds x_1 = -i\sqrt{5}, \quad x_2 = i\sqrt{5}, \quad x_3 = -\sqrt{5}, \quad x_4 = \sqrt{5} \)
\(\leftarrow\) solutions
An important concept to remember is that any polynomial can be factored into the product of linear factors, allowing for complex solutions. This is known as the Fundamental Theorem of Algebra. However, factoring higher-degree polynomials can sometimes be challenging and may require the use of technology, such as computer algebra systems or graphing calculators, to find complex or irrational roots.
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