Q. \(x^2+xy+y^2\)

Q: Is x^2+xy+y^2 always nonnegative?

Yes. Write x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge 0, with equality only when x=y=0.

Q: What is the minimum value of x^2+xy+y^2 given x^2+y^2=1?

The minimum is \frac{1}{4}. Using x=\cos\theta, y=\sin\theta, the expression becomes 1+\frac{1}{2}\sin 2\theta, minimized when \sin 2\theta=-1.

Q: How to rewrite x^2+xy+y^2 as a sum of squares?

x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2. This makes nonnegativity immediate.

Q: Does x^2+xy+y^2 relate to complex numbers?

Yes. Note x^2+xy+y^2=\left|x+y\omega\right|^2 where \omega=\frac{-1+i\sqrt{3}}{2} is a primitive cube root of unity, since \omega^2+\omega+1=0.

Q: What is the factorization over complex numbers?

Treat as quadratic in x: x^2+xy+y^2= \left(x-\frac{-1+i\sqrt{3}}{2}y\right)\left(x-\frac{-1-i\sqrt{3}}{2}y\right).

Answer

Factor the expression by using the identity

\[
x^2+xy+y^2=\left(x-\omega y\right)\left(x-\omega^2 y\right)
\]

where \( \omega \) is a complex cube root of unity satisfying \( \omega^2+\omega+1=0 \).

Over real numbers, it cannot be factored into linear factors, but it can be rewritten as a positive quadratic form:

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

Final result: \(x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\).

Detailed Explanation

We are asked to work with the expression \(x^2+xy+y^2\). A common goal with a polynomial like this is to factor it (rewrite it in a product form) if possible.

Step 1: Look for a standard factorization pattern.

The expression \(x^2+xy+y^2\) resembles the result of multiplying terms of the form \((x+ay)(x+by)\), where \(a\) and \(b\) are numbers chosen so that the product matches the middle term \(xy\).

Step 2: Recall a known identity.

A well-known factorization is:

\[
x^2+xy+y^2=(x+y)^2-xy
\]

Let’s verify it by expanding \((x+y)^2-xy\).

Step 3: Expand \((x+y)^2-xy\).

\[
(x+y)^2-xy = (x^2+2xy+y^2)-xy
\]
\[
= x^2+(2xy-xy)+y^2
\]
\[
= x^2+xy+y^2
\]

This matches exactly, so the identity is correct.

Step 4: Check if it can factor further over the real numbers.

To factor \(x^2+xy+y^2\) into two linear factors over the real numbers, it usually would need to split in a way that produces real coefficients. Here, the polynomial is a quadratic in \(x\) with discriminant:

\[
x^2+xy+y^2 = x^2 + (y)x + y^2
\]

The discriminant is:

\[
\Delta = (y)^2 – 4(1)(y^2)
\]
\[
\Delta = y^2 – 4y^2 = -3y^2
\]

This is negative for \(y \neq 0\), meaning it does not factor into real linear factors (it would require complex numbers).

Final result (factor over integers is not possible, but the difference-of-squares form is):

The expression can be rewritten as

\[
x^2+xy+y^2=(x+y)^2-xy
\]

This is a correct and commonly used simplification.

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Algebra FAQ

Can \(x^2+xy+y^2\) be factored over the reals?

Generally no. Over reals it does not factor nicely since the discriminant of \(t^2+t+1=0\) is negative. It is irreducible over \(\mathbb{R}\).

Is \(x^2+xy+y^2\) always nonnegative?

Yes. Write \(x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge 0\), with equality only when \(x=y=0\).

What is the minimum value of \(x^2+xy+y^2\) given \(x^2+y^2=1\)?

The minimum is \( \frac{1}{4}\). Using \(x=\cos\theta, y=\sin\theta\), the expression becomes \(1+\frac{1}{2}\sin 2\theta\), minimized when \(\sin 2\theta=-1\).

How to rewrite \(x^2+xy+y^2\) as a sum of squares?

\(x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\). This makes nonnegativity immediate.

Does \(x^2+xy+y^2\) relate to complex numbers?

Yes. Note \(x^2+xy+y^2=\left|x+y\omega\right|^2\) where \(\omega=\frac{-1+i\sqrt{3}}{2}\) is a primitive cube root of unity, since \(\omega^2+\omega+1=0\).

What is the factorization over complex numbers?

Treat as quadratic in \(x\): \(x^2+xy+y^2= \left(x-\frac{-1+i\sqrt{3}}{2}y\right)\left(x-\frac{-1-i\sqrt{3}}{2}y\right)\).

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