Q. \[ x^2 – y^2 \]

Q: How do I factor x^2-y^2?

Use difference of squares: x^2-y^2=(x-y)(x+y).

Q: How do I rewrite x^2-y^2 using a common factor?

x^2-y^2=(x-y)(x+y) or x^2-y^2=(x-y)(x+y) (no single monomial factor unless you factor further).

Q: What is x^2-y^2 expanded when (x-y)(x+y) is multiplied?

(x-y)(x+y)=x(x+y)-y(x+y)=x^2+xy-xy-y^2=x^2-y^2.

Q: What are zeros of x^2-y^2 if x and y are real?

x^2-y^2=0\Rightarrow x^2=y^2\Rightarrow x=y or x=-y.

Q: How do I compute \frac{x^2-y^2}{x-y} (assuming x\neq y)?

\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y.

Q: How do I compute \frac{x^2-y^2}{x+y} (assuming x\neq -y)?

\frac{x^2-y^2}{x+y}=\frac{(x-y)(x+y)}{x+y}=x-y.

Q: Can x^2-y^2 be written as a difference of two squares?

Yes: it is already x^2-y^2=(x)^2-(y)^2.

Answer

Factor the difference of squares.

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

Detailed Explanation

We want to simplify the expression \(x^2 – y^2\).

Step 1: Recognize the formula

The expression \(x^2 – y^2\) matches the difference of squares pattern:

\[
a^2 – b^2 = (a-b)(a+b)
\]

Step 2: Match variables

Compare \(x^2 – y^2\) with \(a^2 – b^2\):

\[
a = x,\quad b = y
\]

Step 3: Substitute into the formula

Replace \(a\) with \(x\) and \(b\) with \(y\):

\[
x^2 – y^2 = (x-y)(x+y)
\]

Final Answer

\[
\boxed{(x-y)(x+y)}
\]

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Algebra FAQ

How do I factor \(x^2-y^2\)?

Use difference of squares: \(x^2-y^2=(x-y)(x+y)\).

How do I rewrite \(x^2-y^2\) using a common factor?

\(x^2-y^2=(x-y)(x+y)\) or \(x^2-y^2=(x-y)(x+y)\) (no single monomial factor unless you factor further).

What is \(x^2-y^2\) expanded when \((x-y)(x+y)\) is multiplied?

\((x-y)(x+y)=x(x+y)-y(x+y)=x^2+xy-xy-y^2=x^2-y^2\).

What are zeros of \(x^2-y^2\) if \(x\) and \(y\) are real?

\(x^2-y^2=0\Rightarrow x^2=y^2\Rightarrow x=y\) or \(x=-y\).

How do I compute \(\frac{x^2-y^2}{x-y}\) (assuming \(x\neq y\))?

\(\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=x+y\).

How do I compute \(\frac{x^2-y^2}{x+y}\) (assuming \(x\neq -y\))?

\(\frac{x^2-y^2}{x+y}=\frac{(x-y)(x+y)}{x+y}=x-y\).

Can \(x^2-y^2\) be written as a difference of two squares?

Yes: it is already \(x^2-y^2=(x)^2-(y)^2\).

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