Q. \(x^4-1\)

Q: Factor x^4-1 completely?

x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1).

Q: Solve x^4-1=0 over the reals?

x^4-1=0\Rightarrow x=\pm1. Since x^2+1\neq0 for real x, no other real solutions.

Q: Solve x^4-1=0 over the complex numbers?

x-1=0\Rightarrow x=1, x+1=0\Rightarrow x=-1, x^2+1=0\Rightarrow x=\pm i.

Q: What is the difference of squares factorization?

Use x^4-1=(x^2)^2-1^2=(x^2-1)(x^2+1).

Q: Can x^4-1 be written using conjugates?

x^4-1=(x^2-1)(x^2+1), and x^2-1=(x-1)(x+1). This comes from conjugate-like splitting into products of quadratics/linears.

Q: Find the roots and their multiplicities?

The roots are x=1,-1,i,-i. Each factor is linear or irreducible quadratic with distinct roots, so each root has multiplicity 1.

Answer

Factor \(x^4-1\) using difference of squares.

\[x^4-1=(x^2)^2-1^2=(x^2-1)(x^2+1)\]
\[x^2-1=(x-1)(x+1)\]

So the full factorization is

\[x^4-1=(x-1)(x+1)(x^2+1)\]

Detailed Explanation

We want to factor the polynomial \(x^4 – 1\).

Step 1: Recognize a difference of squares.

Notice that \(x^4\) can be written as \((x^2)^2\). Also, \(1\) can be written as \(1^2\). So we rewrite the expression as:

\[
x^4 – 1 = (x^2)^2 – 1^2
\]

Step 2: Use the difference of squares formula.

The identity is:

\[
a^2 – b^2 = (a-b)(a+b)
\]

Here we identify \(a = x^2\) and \(b = 1\). Substitute into the formula:

\[
(x^2)^2 – 1^2 = (x^2 – 1)(x^2 + 1)
\]

Step 3: Factor \(x^2 – 1\) again (difference of squares).

We rewrite \(x^2 – 1\) as \(x^2 – 1^2\). Use the same formula \(a^2 – b^2 = (a-b)(a+b)\) with \(a = x\) and \(b = 1\):

\[
x^2 – 1 = x^2 – 1^2 = (x-1)(x+1)
\]

Step 4: Combine the factors.

We still have the factor \((x^2 + 1)\) from Step 2, so the full factorization is:

\[
x^4 – 1 = (x-1)(x+1)(x^2+1)
\]

Final Answer:

\[
x^4 – 1 = (x-1)(x+1)(x^2+1)
\]

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Algebra FAQ

Factor \(x^4-1\) completely?

\(x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\).

Solve \(x^4-1=0\) over the reals?

\(x^4-1=0\Rightarrow x=\pm1\). Since \(x^2+1\neq0\) for real \(x\), no other real solutions.

Solve \(x^4-1=0\) over the complex numbers?

\(x-1=0\Rightarrow x=1\), \(x+1=0\Rightarrow x=-1\), \(x^2+1=0\Rightarrow x=\pm i\).

What is the difference of squares factorization?

Use \(x^4-1=(x^2)^2-1^2=(x^2-1)(x^2+1)\).

Can \(x^4-1\) be written using conjugates?

\(x^4-1=(x^2-1)(x^2+1)\), and \(x^2-1=(x-1)(x+1)\). This comes from conjugate-like splitting into products of quadratics/linears.

Find the roots and their multiplicities?

The roots are \(x=1,-1,i,-i\). Each factor is linear or irreducible quadratic with distinct roots, so each root has multiplicity \(1\).

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