The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra is a powerful rule that tells us exactly how many roots (or zeros) a polynomial has.

Simply put, every polynomial of degree $n$ has exactly $n$ roots.

However, to get exactly $n$ roots, we must follow two rules:

1. Count multiplicities: If a root appears more than once, we count it each time it appears. For example, $(x-3)^2 = 0$ means $x=3$ is a root with a "multiplicity of 2" (it counts as two roots).
2. Include complex roots: Roots don't have to be real numbers. They can include imaginary numbers involving $i$ (where $i = \sqrt{-1}$).

Complex Conjugate Root Theorem

When a polynomial has real coefficients, its complex roots always come in pairs. If $a + bi$ is a root, its complex conjugate $a - bi$ must also be a root.

The Rational Root Theorem

Finding all $n$ roots can be tricky. The Rational Root Theorem gives us a starting point by listing all possible rational roots.

If a polynomial has a rational root $\frac{p}{q}$, then:

• $p$ must be a factor of the constant term (the number with no $x$).
• $q$ must be a factor of the leading coefficient (the number in front of the highest power of $x$).

Example 1: Finding All Zeros

Problem: Find all zeros of $f(x) = x^4 - 6x^3 + 14x^2 - 14x + 5$.

Step 1: Identify the number of roots. The highest power (degree) is $4$, so there are exactly $4$ roots.

Step 2: Use the Rational Root Theorem to find a real root. The constant term is $5$ and the leading coefficient is $1$. Possible rational roots: $\pm \frac{1, 5}{1} = 1, -1, 5, -5$. Let's test $x = 1$: $f(1) = (1)^4 - 6(1)^3 + 14(1)^2 - 14(1) + 5 = 1 - 6 + 14 - 14 + 5 = 0$ Since $f(1) = 0$, $x = 1$ is a root.

Step 3: Factor the polynomial. Using synthetic or long division to divide $f(x)$ by $(x-1)$, we get: $x^3 - 5x^2 + 9x - 5$ Test $x = 1$ again on this new polynomial: $(1)^3 - 5(1)^2 + 9(1) - 5 = 1 - 5 + 9 - 5 = 0$ So, $x = 1$ is a root again (it has a multiplicity of 2).

Step 4: Find the remaining roots. Divide $x^3 - 5x^2 + 9x - 5$ by $(x-1)$ to get a quadratic: $x^2 - 4x + 5$. Now, use the quadratic formula to solve $x^2 - 4x + 5 = 0$: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}$ $x = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2}$ $x = \frac{4 \pm 2i}{2} = 2 \pm i$

Final Answer: The $4$ roots are $1$, $1$, $2+i$, and $2-i$.

Example 2: Using Complex Conjugates

Problem: If $2$ and $1 + i$ are zeros of $x^3 - 4x^2 + 6x - 4$, find all zeros.

Step 1: Identify the degree. The degree is $3$, so there must be exactly $3$ roots in total.

Step 2: Apply the Complex Conjugate Theorem. We are given two roots: $2$ (a real root) and $1 + i$ (a complex root). Because the polynomial has real coefficients, complex roots must come in conjugate pairs. Since $1 + i$ is a root, its conjugate $1 - i$ must also be a root.

Final Answer: The three roots are $2$, $1 + i$, and $1 - i$. We have found all $3$ roots, perfectly matching the degree of the polynomial!