Polynomial Equation Solver

A polynomial equation is an equation of the form:

$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$

where $n$ is a positive integer called the degree, $a_n \neq 0$, and $a_0, a_1, \ldots, a_n$ are constants (coefficients).

Polynomials are classified by degree:

• Degree 1: Linear ($ax + b = 0$)
• Degree 2: Quadratic ($ax^2 + bx + c = 0$)
• Degree 3: Cubic ($ax^3 + bx^2 + cx + d = 0$)
• Degree 4: Quartic ($ax^4 + \cdots = 0$)
• Degree 5+: Quintic and higher

The Fundamental Theorem of Algebra states that a polynomial of degree $n$ has exactly $n$ roots (counting multiplicity) in the complex numbers. For example, a cubic equation always has 3 roots, which may be real or complex.

Higher-degree polynomial equations arise in physics (projectile motion, oscillations), engineering (control systems), economics (optimization), and computer graphics (curve intersections).

Unlike quadratics, there is no single formula that works for all higher-degree polynomials. Here are the main strategies:

1. Rational Root Theorem

For $a_n x^n + \cdots + a_0 = 0$ with integer coefficients, any rational root $\frac{p}{q}$ must satisfy:

• $p$ divides $a_0$ (the constant term)
• $q$ divides $a_n$ (the leading coefficient)

Test candidates and use synthetic division to reduce the degree.

Example: $x^3 - 6x^2 + 11x - 6 = 0$

• Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6$
• Test $x = 1$: $1 - 6 + 11 - 6 = 0$ ✓
• Divide out $(x - 1)$ to get $x^2 - 5x + 6 = (x-2)(x-3)$

2. Factoring by Grouping

Rearrange terms into groups that share common factors.

Example: $x^3 + x^2 - 4x - 4 = x^2(x+1) - 4(x+1) = (x^2-4)(x+1) = (x+2)(x-2)(x+1)$

3. Substitution (Disguised Quadratics)

If only even powers appear, let $u = x^2$:

Example: $x^4 - 5x^2 + 4 = 0$ → let $u = x^2$: $u^2 - 5u + 4 = 0$ → $(u-1)(u-4) = 0$

So $x^2 = 1$ or $x^2 = 4$, giving $x = \pm 1, \pm 2$.

4. Synthetic Division

Once a root $r$ is found, divide by $(x - r)$ to reduce the polynomial's degree, then repeat.

5. Descartes' Rule of Signs

Count sign changes in $f(x)$ and $f(-x)$ to determine the maximum number of positive and negative real roots.

• Forgetting complex roots: A degree-$n$ polynomial always has $n$ roots over $\mathbb{C}$. If you only find real roots, complex roots come in conjugate pairs.
• Missing repeated roots: $x^3 - 3x + 2 = (x-1)^2(x+2)$ has $x = 1$ as a double root.
• Incomplete list of rational root candidates: Check all combinations of factors of $a_0$ over factors of $a_n$.
• Arithmetic errors in synthetic division: Double-check each step — one wrong number propagates through the entire calculation.
• Assuming all roots are rational: Many polynomials have irrational or complex roots that cannot be found by the Rational Root Theorem alone.