Ma · Mathematics

How Do You Solve Polynomial Equations of Degree Greater Than 4?

Published By the 9to5Science editorial team

At a glance

  1. 1. Check the structure

    Look for common factors, substitutions, symmetry, or a factorization that reduces the degree.

  2. 2. Choose an exact method

    Use radicals when the polynomial has a suitable special form or its degree is at most four.

  3. 3. Find numerical roots

    For general higher-degree polynomials, use a numerical method or a computer algebra system.

  4. 4. Verify the result

    Substitute roots back into the polynomial and check whether all roots, including complex ones, are needed.

Polynomial equations of degree five or higher are not automatically impossible to solve. What changes is that there is no general formula using arithmetic operations and radicals that gives the roots of every such equation. For a particular polynomial, factoring, recognizing a special form, or using numerical methods may still work.

What changes after degree four?

The quadratic, cubic, and quartic formulas solve every polynomial of degrees two, three, and four using coefficients, arithmetic, and radicals (square roots, cube roots, and so on). The Abel–Ruffini theorem says that no corresponding general radical formula exists for degree five or higher.

This theorem is about a formula that works for all polynomials of a given degree. It does not say that every individual quintic or higher-degree equation lacks an exact solution. Some can be solved in radicals because they factor into lower-degree polynomials or have other helpful structure. Some individual higher-degree polynomials cannot be solved by radicals, even though their roots can be defined exactly in other ways.

The fundamental theorem of algebra still applies: a degree-nn polynomial has exactly nn complex roots when multiplicities are counted. The issue is finding a useful expression for those roots, not whether they exist.

First try to reduce the degree

Before turning to numerical methods, look for common factors, familiar identities, and substitutions. For example, consider

x5−5x3+4x=0.x^5-5x^3+4x=0.

Factor out xx, then treat the remaining expression as a quadratic in x2x^2:

x5−5x3+4x=x(x4−5x2+4)=x(x2−1)(x2−4)=x(x−1)(x+1)(x−2)(x+2).\begin{aligned} x^5-5x^3+4x &= x(x^4-5x^2+4) \\ &= x(x^2-1)(x^2-4) \\ &= x(x-1)(x+1)(x-2)(x+2). \end{aligned}

So the five roots are −2,−1,0,1,2-2,-1,0,1,2. The equation has degree five, but its structure makes exact solution straightforward. The same idea often works when a polynomial is a quadratic in x2x^2, has a common factor, or can be factored into simpler polynomials.

Other useful clues include symmetry between positive and negative powers, a substitution that lowers the degree, and known roots suggested by integer coefficients. If you find a root rr, polynomial division by x−rx-r reduces the remaining problem by one degree.

Use a numerical method when needed

For a polynomial with no convenient factorization, numerical root-finding can give decimal approximations. One common method is Newton’s method. To solve f(x)=0f(x)=0, start with a guess x0x_0 and repeat

xk+1=xk−f(xk)f′(xk).x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}.

For example, take the quintic

f(x)=x5−x−1.f(x)=x^5-x-1.

Starting at x0=1.2x_0=1.2, Newton’s method uses f′(x)=5x4−1f'(x)=5x^4-1 and gives

x1=1.2−1.25−1.2−15(1.2)4−1≈1.16922,x2≈1.16731.\begin{aligned} x_1 &= 1.2-\frac{1.2^5-1.2-1}{5(1.2)^4-1} \approx 1.16922, \\ x_2 &\approx 1.16731. \end{aligned}

Continuing the iteration gives the real root x≈1.167303978x\approx1.167303978. This is a numerical approximation, not a radical formula. A calculator or computer algebra system can also approximate the other complex roots if they are needed.

Newton’s method depends on the starting value and can fail or converge to a different root. For dependable computation, use software designed for polynomial roots, check its numerical tolerances, and substitute the reported approximations into the original polynomial. For a real root, a sign change across an interval can help confirm that at least one root lies there.

Exact answers and numerical answers

“Solving exactly” can mean different things. A root may be specified exactly as “the unique real root of x5−x−1=0x^5-x-1=0,” even if it has no expression using radicals. It can also be represented with a convergent series or special functions. Those representations are different from the familiar quadratic formula.

For practical work, choose a method based on what the problem requires:

  • If it asks for an exact factorization, inspect the polynomial’s structure and try algebraic techniques.
  • If it asks for roots to a specified number of decimal places, use a numerical solver and report the precision.
  • If it asks whether a general formula exists, the answer for degree five and above is no—not one that solves every polynomial using radicals.

Common mistakes

  • Assuming every quintic is unsolvable. Many higher-degree equations factor or have special forms that make exact solutions possible.
  • Assuming roots do not exist because there is no formula. Every polynomial has its full set of complex roots, counting multiplicities.
  • Calling a decimal approximation exact. State an approximation sign and the number of digits or error tolerance.
  • Using Newton’s method without checking the result. A starting guess can lead to the wrong root or fail to converge; verify the approximation in the original equation.

References

  1. Algebra

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