Higher Degree Equations

Solve x^5 - x = 0. You can factor out x to get x(x^4 - 1) = 0, factor again to get x(x^2-1)(x^2+1) = 0, and read off five roots: 0, 1, -1, i, -i. Not bad. Now try x^5 - x - 1 = 0. No obvious factors, no pattern, no formula to plug into. Yet this equation has five roots somewhere in the complex plane, and the coefficients know exactly where they are.

Here is the strange part: for quadratics, cubics, and quartics, there exist formulas that express the roots in terms of the coefficients using only arithmetic and radicals. At degree five, no such formula can exist — this was proved in the early 1800s, and it is one of the landmarks of abstract algebra. So for higher-degree equations, you need different strategies: spotting patterns in the coefficients, transforming the equation into a simpler form, and exploiting symmetry. This article is about those strategies.

Roots and coefficients: the general pattern

Consider the general polynomial equation of degree n:

with a_n \neq 0. Let the n roots be r_1, r_2, \ldots, r_n. Then:

The k-th formula relates the sum of all products of k roots taken at a time to the coefficient a_{n-k}. The signs alternate: minus for the sum of single roots, plus for pairwise products, minus for triples, and so on.

For a quartic (degree-4) equation ax^4 + bx^3 + cx^2 + dx + e = 0 with roots \alpha, \beta, \gamma, \delta, the four formulas are:

The second formula has \binom{4}{2} = 6 terms — every way to pick two roots from four. The third has \binom{4}{3} = 4 terms. These numbers grow quickly with the degree, but the structure stays the same.

Transformation of equations

A powerful technique for working with higher-degree equations is to transform the equation — change the variable to produce a new equation that is easier to work with. The roots of the new equation are related to the original roots by the same transformation.

Shifting: removing a term

If you substitute x = y + h into a polynomial equation, the roots shift: if r was a root of the original, then r - h is a root of the new equation. By choosing h carefully, you can eliminate one term.

To remove the x^{n-1} term from a degree-n equation, set h = -a_{n-1}/(n \cdot a_n). The sum of the new roots becomes zero, and the second-highest term vanishes. You saw this in the cubic article — replacing x by t - b/(3a) turns a general cubic into a depressed cubic. The same idea works at any degree.

Take x^4 + 4x^3 + x^2 - 6x + 2 = 0. To remove the x^3 term, substitute x = y - 1 (since h = -4/(4 \cdot 1) = -1):

Expanding: y^4 - 4y^3 + 6y^2 - 4y + 1 + 4y^3 - 12y^2 + 12y - 4 + y^2 - 2y + 1 - 6y + 6 + 2 = 0

Collecting: y^4 + (-4 + 4)y^3 + (6 - 12 + 1)y^2 + (-4 + 12 - 2 - 6)y + (1 - 4 + 1 + 6 + 2) = 0

The y^3 term is gone. Even better, this particular equation has no odd-power terms at all, so it is a quadratic in y^2: substitute u = y^2 to get u^2 - 5u + 6 = 0, which factors as (u - 2)(u - 3) = 0. So y^2 = 2 or y^2 = 3, giving y = \pm\sqrt{2} or y = \pm\sqrt{3}, and x = y - 1.

Scaling: clearing fractions or simplifying coefficients

Substituting x = ky multiplies each root by 1/k. This is useful for clearing fractions. If your equation has coefficients like 1, 3/2, 5/4, substituting x = y/2 and then multiplying through by the right power of 2 can produce integer coefficients.

Reciprocating: the substitution y = 1/x

Substituting y = 1/x replaces each root r by 1/r. This is useful when you need to find an equation whose roots are the reciprocals of the original roots. If the original equation is a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0, multiply through by x^{-n} after substituting to get a_n y^{-n} + \cdots — which, after multiplying by y^n, becomes a_0 y^n + a_1 y^{n-1} + \cdots + a_n = 0. The coefficients are the original coefficients in reverse order.

Reciprocal equations

A polynomial equation is called a reciprocal equation if its coefficient sequence reads the same forwards and backwards. That is, a_k = a_{n-k} for every k. For example:

The coefficients are 1, 3, 5, 3, 1 — a palindrome. This symmetry has a powerful consequence: if r is a root, then 1/r is also a root. Here is why. Divide the entire equation by x^2 (which is valid since x = 0 is clearly not a root):

Group the terms that pair up:

Now substitute t = x + 1/x. Notice that x^2 + 1/x^2 = t^2 - 2 (from squaring t and subtracting 2). The equation becomes:

A degree-4 equation has been reduced to a degree-2 equation. Solve the quadratic to find t, then for each value of t, solve x + 1/x = t (which is itself a quadratic: x^2 - tx + 1 = 0) to find x.

This trick works for any even-degree reciprocal equation. For odd-degree reciprocal equations, x = -1 is always a root (substitute and verify: the palindrome structure guarantees cancellation). Factor out (x + 1), and the remaining equation is an even-degree reciprocal equation.

Equations with symmetric roots

Some polynomial equations have roots that are related by a known symmetry — for instance, roots in arithmetic progression or roots that come in pairs r, -r.

Roots in arithmetic progression

If the three roots of a cubic x^3 + bx^2 + cx + d = 0 are in arithmetic progression, you can write them as \alpha - d_0, \alpha, \alpha + d_0 for some common difference d_0. Their sum is 3\alpha = -b, so \alpha = -b/3. This tells you the middle root immediately, and the remaining two are found from the other Vieta formulas.

For four roots in arithmetic progression — say \alpha - 3d_0, \alpha - d_0, \alpha + d_0, \alpha + 3d_0 — the sum is 4\alpha, which equals -b/a. Again, the centre is pinned, and the common difference comes from the next formula.

Roots with equal-magnitude pairing

If an equation has the property that whenever r is a root, -r is also a root, then all the odd-power coefficients are zero (since the elementary symmetric functions of odd order vanish). This means the equation is actually a polynomial in x^2, which halves the effective degree — exactly the trick used for the quartic y^4 - 5y^2 + 6 = 0 in the transformation section above.

Two worked examples

Common confusions

• "Vieta's formulas work only for quadratics and cubics." They work for every degree. A degree-n polynomial has n Vieta relations, one for each elementary symmetric function of the roots. The formulas get longer as the degree rises, but the pattern is the same.

• "Transforming an equation changes the roots." It changes the values of the roots in a controlled way, but it does not change the number of roots or the relationships among them. If you shift by h, every root shifts by h. You can always reverse the transformation to recover the original roots.

• "Every reciprocal equation has palindrome coefficients." There is a subtlety. Some books define "reciprocal equations of the second kind" where the coefficient pattern is a_k = -a_{n-k} — an anti-palindrome. These also have the property that if r is a root, 1/r is too, but with additional sign structure. The standard definition uses the palindrome a_k = a_{n-k}.

• "If roots are in AP, you write them as a, a+d, a+2d, a+3d." That works, but it is less efficient than the symmetric form \alpha - 3d, \alpha - d, \alpha + d, \alpha + 3d (for four roots). The symmetric form makes the sum trivially 4\alpha, instantly giving you the centre. The asymmetric form forces you to expand messier sums.

• "The product of all roots of x^n + \cdots + a_0 = 0 is a_0." Watch the sign: the product is (-1)^n a_0. For even n, the product is +a_0; for odd n, it is -a_0. Forgetting the (-1)^n is a reliable source of sign errors.

Going deeper

The impossibility at degree five

For degree two, three, and four, formulas exist that express the roots using only +, -, \times, \div, and n-th roots applied to the coefficients. At degree five, no such formula exists — and this is not a conjecture or a limitation of current knowledge. It is a theorem, proved in the early 1800s.

The proof rests on a branch of mathematics called Galois theory, named after a young French mathematician who developed it before dying at age twenty. The core idea: the symmetries of the roots of a polynomial form a mathematical structure called a group. For degree two, three, and four, this group has a special property (it is "solvable") that allows the roots to be peeled off one at a time using radicals. For the general degree-five polynomial, the group of symmetries is not solvable, and no radical formula can work.

This does not mean fifth-degree equations cannot be solved. They can — using elliptic functions, numerical methods, or other techniques. It means only that radicals (square roots, cube roots, etc.) are not powerful enough, by themselves, to reach the answer.

Newton's identities

Vieta's formulas give the elementary symmetric functions of the roots. Newton's identities connect these to the power sums p_k = r_1^k + r_2^k + \cdots + r_n^k. The first few Newton identities for a monic polynomial x^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0 with elementary symmetric functions e_1, e_2, \ldots, e_n are:

Each power sum depends on all the previous power sums and elementary symmetric functions. These recurrences let you compute r_1^k + \cdots + r_n^k for any k without ever finding the individual roots — a technique that appears frequently in competition mathematics.

Where this leads next

Higher-degree equations sit at the crossroads of several algebraic ideas.

• Cubic Equations — the degree-three case in full detail, including Cardano's formula and the root-coefficient relationships.
• Repeated Roots — what happens when roots coincide, how to detect them, and the connection to derivatives.
• Polynomial Equations — the general theory: Fundamental Theorem of Algebra, existence and counting of roots.
• Sum and Product of Roots — Vieta's formulas for quadratics worked out in detail, with applications to symmetric expressions.
• Quadratic Equations — Advanced — deeper techniques for degree two that preview the transformation ideas used here.