Q. \(x^4-16\)

Q: Factor x^4-16 completely?

Use difference of squares: x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4).

Q: Solve x^4-16=0?

x^4=16\Rightarrow x=\pm2 (real solutions). Over complexes: x=\pm2,\ \pm2i.

Q: What are the zeros of x^4-16?

From (x-2)(x+2)(x^2+4)=0: zeros are x=2,\ x=-2,\ x=2i,\ x=-2i.

Q: Is x^4-16 divisible by x-2?

Yes. Since f(2)=16-16=0, x-2 is a factor. The quotient is (x+2)(x^2+4) after full factorization.

Q: Find the roots using substitution y=x^2?

Let y=x^2. Then y^2-16=0\Rightarrow y=\pm4. So x^2=4\Rightarrow x=\pm2, and x^2=-4\Rightarrow x=\pm2i.

Q: Expand (x-2)(x+2)(x^2+4) to verify?

First (x-2)(x+2)=x^2-4. Then (x^2-4)(x^2+4)=x^4-16.

Answer

We factor the expression using the difference of squares.

\[
x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)
\]

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

\[
x^2+4 \text{ is irreducible over the reals.}
\]

\[
\boxed{x^4-16=(x-2)(x+2)(x^2+4)}
\]

Detailed Explanation

We are asked to work with the expression \(x^4 – 16\). A common first step is to recognize that both terms are powers with a shared pattern that can be factored.

Step 1: Rewrite \(x^4 – 16\) using a difference of squares idea

Notice that \(x^4\) can be written as \((x^2)^2\), and \(16\) can be written as \((4)^2\). That means

\[
x^4 – 16 = (x^2)^2 – 4^2.
\]

Step 2: Use the difference of squares formula

The difference of squares formula is

\[
a^2 – b^2 = (a – b)(a + b).
\]

Here, \(a = x^2\) and \(b = 4\). Substitute into the formula:

\[
(x^2)^2 – 4^2 = (x^2 – 4)(x^2 + 4).
\]

Step 3: Factor \(x^2 – 4\) further as a difference of squares

Again, \(x^2 – 4\) is a difference of squares because \(x^2 = (x)^2\) and \(4 = (2)^2\). So use the same formula with \(a = x\) and \(b = 2\):

\[
x^2 – 4 = x^2 – 2^2 = (x – 2)(x + 2).
\]

Step 4: Check whether \(x^2 + 4\) factors further over the integers

\(x^2 + 4\) does not factor into real or integer linear factors over the integers. (It would require complex numbers.) In integer factorization form, we leave it as \(x^2 + 4\).

Final Answer (fully factored over the integers)

\[
x^4 – 16 = (x – 2)(x + 2)(x^2 + 4).
\]

See full solution

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Algebra FAQ

Factor \(x^4-16\) completely?

Use difference of squares: \(x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)\).

Solve \(x^4-16=0\)?

\(x^4=16\Rightarrow x=\pm2\) (real solutions). Over complexes: \(x=\pm2,\ \pm2i\).

What are the zeros of \(x^4-16\)?

From \((x-2)(x+2)(x^2+4)=0\): zeros are \(x=2,\ x=-2,\ x=2i,\ x=-2i\).

Is \(x^4-16\) divisible by \(x-2\)?

Yes. Since \(f(2)=16-16=0\), \(x-2\) is a factor. The quotient is \((x+2)(x^2+4)\) after full factorization.

Find the roots using substitution \(y=x^2\)?

Let \(y=x^2\). Then \(y^2-16=0\Rightarrow y=\pm4\). So \(x^2=4\Rightarrow x=\pm2\), and \(x^2=-4\Rightarrow x=\pm2i\).

Expand \((x-2)(x+2)(x^2+4)\) to verify?

First \((x-2)(x+2)=x^2-4\). Then \((x^2-4)(x^2+4)=x^4-16\).

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