A root of a polynomial is a value that makes the polynomial equal to zero.
The Rational Root Theorem gives a short list of possible rational roots, and the Factor Theorem tells us
when a tested root gives a true factor.
1. What is a polynomial root?
If \(P(x)\) is a polynomial, then \(r\) is a root when:
\[
\begin{aligned}
P(r)=0.
\end{aligned}
\]
For example, if \(P(3)=0\), then \(x=3\) is a root.
2. The Factor Theorem
The Factor Theorem says:
\[
\begin{aligned}
P(r)=0
\quad\Longleftrightarrow\quad
x-r\text{ is a factor of }P(x).
\end{aligned}
\]
This means that checking roots and factoring are connected. If a candidate gives remainder zero, it creates a factor.
3. The Rational Root Theorem
Suppose:
\[
\begin{aligned}
P(x)=a_nx^n+\cdots+a_1x+a_0.
\end{aligned}
\]
If \(p/q\) is a rational root in lowest terms, then:
\[
\begin{aligned}
p&\text{ divides }a_0,\\
q&\text{ divides }a_n.
\end{aligned}
\]
So the possible rational roots are:
\[
\begin{aligned}
\pm\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}.
\end{aligned}
\]
4. Worked example
Find the rational roots of:
\[
\begin{aligned}
P(x)=2x^3-5x^2-4x+3.
\end{aligned}
\]
The leading coefficient is \(2\), and the constant term is \(3\).
Therefore, the possible rational roots are:
\[
\begin{aligned}
\pm 1,\quad \pm 3,\quad \pm \frac{1}{2},\quad \pm \frac{3}{2}.
\end{aligned}
\]
Test \(x=3\):
\[
\begin{aligned}
P(3)
&=
2(3)^3-5(3)^2-4(3)+3\\
&=
54-45-12+3\\
&=
0.
\end{aligned}
\]
Since \(P(3)=0\), the Factor Theorem says \(x-3\) is a factor.
5. Factoring after one root
Divide \(2x^3-5x^2-4x+3\) by \(x-3\):
\[
\begin{aligned}
2x^3-5x^2-4x+3
&=
(x-3)(2x^2+x-1).
\end{aligned}
\]
Now factor the quadratic:
\[
\begin{aligned}
2x^2+x-1
&=
(2x-1)(x+1).
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
\boxed{
2x^3-5x^2-4x+3
=
(x-3)(2x-1)(x+1)
}.
\end{aligned}
\]
The roots are:
\[
\begin{aligned}
x=3,\quad x=\frac{1}{2},\quad x=-1.
\end{aligned}
\]
6. Synthetic division and remainders
Synthetic division is a fast way to test candidates when the divisor is \(x-r\).
The last number in the synthetic division row is the remainder.
If the remainder is zero:
\[
\begin{aligned}
P(r)=0.
\end{aligned}
\]
Then \(r\) is a root and \(x-r\) is a factor.
7. Fractional roots
If a root is fractional, such as \(r=1/2\), then:
\[
\begin{aligned}
x-\frac{1}{2}
\end{aligned}
\]
is a factor over rational coefficients. To write a factor with integer coefficients, multiply by 2:
\[
\begin{aligned}
2x-1.
\end{aligned}
\]
That is why the root \(x=1/2\) corresponds to the factor \(2x-1\).
8. Repeated roots
A root may appear more than once. For example:
\[
\begin{aligned}
(x-2)^3
\end{aligned}
\]
has root \(x=2\) with multiplicity \(3\).
Multiplicity tells how many times the same factor appears.
9. When no rational roots are found
Not every polynomial has rational roots.
For example:
\[
\begin{aligned}
x^2-2
\end{aligned}
\]
has roots:
\[
\begin{aligned}
x=\pm\sqrt{2}.
\end{aligned}
\]
These are irrational, so the Rational Root Theorem will not find them as rational candidates.
10. Formula summary
The table below uses plain text formulas in table cells to avoid raw LaTeX rendering problems.
11. Common mistakes
- Testing only positive candidates and forgetting negative candidates.
- Using factors of the leading coefficient as numerators instead of denominators.
- Forgetting fractional candidates when the leading coefficient is not 1.
- Thinking every listed candidate is a root. Candidates must still be tested.
- Confusing root \(r\) with factor \(x-r\).
- Forgetting that root \(p/q\) gives integer factor \(qx-p\).
Key idea: the Rational Root Theorem gives candidates; the Factor Theorem confirms which candidates are real factors.
