Factor X 2 Xy Y 2

Introduction

Factoring polynomial expressions is a cornerstone of algebra that transforms seemingly complex equations into manageable pieces. One of the most frequently encountered forms in secondary‑school mathematics is the trinomial

[ x^{2}+xy+y^{2}. ]

Understanding how (and when) this expression can be factored not only sharpens algebraic intuition but also paves the way for solving quadratic equations, simplifying rational expressions, and tackling problems in number theory and geometry. This article walks you through the complete factorisation process for (x^{2}+xy+y^{2}), explores the underlying algebraic principles, presents alternative methods, and answers common questions that students often raise.

Why (x^{2}+xy+y^{2}) Is Special

At first glance, the trinomial looks similar to the classic perfect‑square form (a^{2}+2ab+b^{2}=(a+b)^{2}). That said, this subtle difference means the expression is not a perfect square in the usual sense, and the usual “split the middle term” technique does not work directly. Even so, the middle term here is only (xy), not (2xy). Yet the expression can still be factored—provided we allow complex numbers or special algebraic tricks.

The factorisation depends on the field (set of numbers) we are working in:

• Over the real numbers, (x^{2}+xy+y^{2}) is irreducible; it cannot be expressed as a product of two non‑constant real polynomials.
• Over the complex numbers, it splits into linear factors involving the primitive cube roots of unity.
• Over the integers modulo a prime (e.g., (\mathbb{Z}_p)), the factorisation may or may not exist depending on the value of (p).

The article will cover each of these scenarios, giving you a toolbox that works no matter which number system your problem lives in.

Factorising Over the Complex Numbers

The role of cube roots of unity

The complex numbers (\omega) and (\omega^{2}) are defined as the non‑real solutions to the equation

[ z^{3}=1 \quad\text{with}\quad z\neq1. ]

Explicitly,

[ \omega = -\frac12 + i\frac{\sqrt{3}}{2}, \qquad \omega^{2}= -\frac12 - i\frac{\sqrt{3}}{2}, ]

and they satisfy the relations

[ \omega^{3}=1,\qquad \omega^{2}+\omega+1=0,\qquad \omega^{2}= \overline{\omega}. ]

These properties are the key to breaking down (x^{2}+xy+y^{2}).

Derivation of the factorisation

Consider the polynomial in two variables as a quadratic in (x) (or (y)):

[ x^{2}+xy+y^{2}=0. ]

Treating (y) as a constant, solve for (x) using the quadratic formula:

[ x=\frac{-y\pm\sqrt{y^{2}-4y^{2}}}{2} =\frac{-y\pm y\sqrt{-3}}{2} =y\frac{-1\pm i\sqrt{3}}{2}. ]

The two roots are

[ x = y\omega \quad\text{and}\quad x = y\omega^{2}. ]

Thus the polynomial can be written as

[ x^{2}+xy+y^{2}= (x-\omega y)(x-\omega^{2} y). ]

Because (\omega) and (\omega^{2}) are complex conjugates, the product expands back to the original real expression, confirming the factorisation is correct.

Verifying the expansion

[ \begin{aligned} (x-\omega y)(x-\omega^{2} y) &= x^{2} - (\omega+\omega^{2})xy + \omega\omega^{2}y^{2}\ &= x^{2} - (-1)xy + ( \omega^{3})y^{2}\ &= x^{2}+xy+y^{2}, \end{aligned} ]

where we used (\omega+\omega^{2}=-1) and (\omega\omega^{2}= \omega^{3}=1).

Result:

[ \boxed{x^{2}+xy+y^{2}= (x-\omega y)(x-\omega^{2} y)}. ]

This factorisation is valid only in the complex field (\mathbb{C}).

Factorising Over the Real Numbers

Since the discriminant (b^{2}-4ac = 1^{2}-4\cdot1\cdot1 = -3) is negative, the quadratic has no real roots. Because of this, no factorisation into real linear factors exists. That said, we can still express the polynomial as a product of two irreducible quadratic factors when additional variables are introduced, or we can use trigonometric substitution to simplify related integrals.

For most high‑school curricula, the appropriate statement is:

(x^{2}+xy+y^{2}) is irreducible over the real numbers.

If a problem asks you to “factor” it in a real‑only context, the answer is simply “cannot be factored further.”

Factorising Modulo a Prime

When working in a finite field (\mathbb{Z}_p) (integers modulo a prime (p)), the factorisation depends on whether (-3) is a quadratic residue modulo (p).

Legendre symbol test

Define the Legendre symbol (\left(\frac{-3}{p}\right)).

• If (\left(\frac{-3}{p}\right)=1), then (-3) has a square root modulo (p); the discriminant becomes a square, and the polynomial splits into linear factors.
• If (\left(\frac{-3}{p}\right)=-1), the polynomial remains irreducible.

For example:

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• (p=7): (-3 \equiv 4 \pmod{7}) and (4) is a square (since (2^{2}=4)). Hence (x^{2}+xy+y^{2}) factors as ((x+2y)(x+5y)) modulo 7.
• (p=5): (-3 \equiv 2 \pmod{5}) and 2 is not a quadratic residue modulo 5, so the polynomial stays irreducible.

Explicit factorisation when possible

Assume we have found an element (r) such that (r^{2}\equiv -3 \pmod{p}). Then the roots of the quadratic in (x) are

[ x = \frac{-y \pm yr}{2} = y\frac{-1 \pm r}{2}. ]

Thus the factorisation in (\mathbb{Z}_p[x,y]) is

[ x^{2}+xy+y^{2}= \bigl(x - y\frac{-1+r}{2}\bigr)\bigl(x - y\frac{-1-r}{2}\bigr). ]

All arithmetic is performed modulo (p).

Applications of the Factorisation

1. Solving homogeneous quadratic equations

Consider the system

[ \begin{cases} x^{2}+xy+y^{2}=0,\ ax+by=0, \end{cases} ]

where (a,b) are real constants. Over (\mathbb{C}) the first equation gives the ratio (x/y = \omega) or (\omega^{2}). Substituting into the linear equation yields explicit solutions for ((x,y)) in terms of (a) and (b).

2. Simplifying rational expressions

If you encounter a fraction such as

[ \frac{x^{3}-y^{3}}{x^{2}+xy+y^{2}}, ]

recognise the numerator as a difference of cubes:

[ x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2}). ]

The denominator cancels, leaving the simple result (x-y). Understanding the factor (x^{2}+xy+y^{2}) is therefore essential for quick simplification.

3. Geometry: equilateral triangles

In the complex plane, the vertices of an equilateral triangle with one vertex at the origin can be written as (0,,z,,\omega z). The condition that the three points are equally spaced translates algebraically to

[ z^{2}+z(\omega z)+(\omega z)^{2}=0, ]

which reduces to the same trinomial after factoring out (z^{2}). Hence the factorisation reflects the symmetry of the equilateral triangle.

Frequently Asked Questions

Q1: Can I factor (x^{2}+xy+y^{2}) using the “ac method” (splitting the middle term)?

A: No. The “ac method” requires finding two numbers whose product is (ac = 1\cdot 1 = 1) and whose sum is (b = 1). The only integer pair ((1,1)) sums to 2, not 1, so the method fails. The correct approach is to recognise the connection to cube roots of unity.

Q2: Why does the discriminant being negative prevent factoring over the reals?

A: For a quadratic (ax^{2}+bx+c), real linear factors exist only when the discriminant (b^{2}-4ac) is non‑negative, because the roots are (\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). A negative discriminant yields imaginary square roots, meaning the polynomial cannot be expressed as a product of real linear terms.

Q3: Is there a way to “complete the square” for (x^{2}+xy+y^{2})?

A: Yes, but the result involves a mixed term:

[ x^{2}+xy+y^{2}= \left(x+\frac{y}{2}\right)^{2}+ \frac{3}{4}y^{2}. ]

This representation is useful for integration or for visualising the expression as the sum of a square and a positive term.

Q4: What happens if I replace (y) by (-y)?

A: The expression becomes (x^{2}-xy+y^{2}). Its discriminant is also (-3), so the factorisation over (\mathbb{C}) changes to

[ x^{2}-xy+y^{2}= (x-\omega^{2} y)(x-\omega y), ]

which is simply a permutation of the original factors.

Q5: Can the factorisation be used in polynomial division?

A: Absolutely. If you need to divide a higher‑degree polynomial by (x^{2}+xy+y^{2}), you can first replace the divisor with its complex linear factors, perform the division in (\mathbb{C}[x,y]), and then recombine the results if a real answer is required.

Conclusion

The trinomial (x^{2}+xy+y^{2}) serves as a compact illustration of how the nature of the coefficient field determines factorability. Over the real numbers it remains irreducible, reflecting a negative discriminant. Over the complex numbers, it splits neatly into linear factors that involve the primitive cube roots of unity (\omega) and (\omega^{2}):

[ \boxed{x^{2}+xy+y^{2}= (x-\omega y)(x-\omega^{2} y)}. ]

In modular arithmetic, the factorisation hinges on whether (-3) is a quadratic residue modulo the chosen prime, leading to either a linear factorisation or irreducibility. Recognising these patterns empowers you to simplify algebraic expressions, solve homogeneous equations, and understand geometric configurations such as equilateral triangles in the complex plane.

Mastering the factorisation of (x^{2}+xy+y^{2}) therefore not only enriches your algebraic toolkit but also deepens your appreciation of the interplay between number systems, symmetry, and polynomial structure. Keep practising with variations—changing coefficients, swapping signs, or working modulo different primes—to cement the concepts and become confident in handling any similar trinomial that appears in your mathematical journey.