Q. \(\)

Q: . Factor x^4-81.

. Use difference of squares: x^4-81=(x^4-9^2)=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9).

Q: . Solve x^4-81=0.

. From (x^2-9)(x^2+9)=0: x^2-9=0 \Rightarrow x=\pm3. Also x^2+9=0 \Rightarrow x=\pm3i.

Q: . What are the real zeros of x^4-81?

. Real zeros come from x^2-9=0, so x=-3 and x=3. The factor x^2+9 gives only complex roots.

Q: . Does x^4-81 have quadratic factors over integers?

. Yes: x^4-81=(x^2-9)(x^2+9). These are monic quadratics with integer coefficients.

Q: . Find the roots of x^4-81 using substitution y=x^2.

. Let y=x^2. Then y^2-81=0\Rightarrow y=\pm9. So x^2=9\Rightarrow x=\pm3, and x^2=-9\Rightarrow x=\pm3i.

Q: . Determine the sign of x^4-81 for real x.

. Since x^4=81 at x=\pm3, and x^4-81 is even, it’s positive for |x|>3 and negative for |x|<3. At x=\pm3, it equals 0.

Answer

We factor the polynomial using difference of squares.

\[
x^4-81=(x^2)^2-9^2
\]
\[
x^4-81=(x^2-9)(x^2+9)
\]
\[
x^2-9=(x-3)(x+3)
\]

\[
x^4-81=(x-3)(x+3)(x^2+9)
\]

Final result: \((x-3)(x+3)(x^2+9)\).

Detailed Explanation

Problem: Solve and simplify the expression \(x^4 – 81\).

We are asked to simplify \(x^4 – 81\). This is a difference of squares after rewriting \(x^4\) appropriately.

Step 1: Rewrite \(x^4\) as a perfect square.

Notice that \(x^4\) can be written as \((x^2)^2\):

\[
x^4 – 81 = (x^2)^2 – 81
\]

Step 2: Rewrite \(81\) as a perfect square.

Note that \(81 = 9^2\). So the expression becomes:

\[
(x^2)^2 – 9^2
\]

Step 3: Use the difference of squares formula.

Recall the identity:

\[
a^2 – b^2 = (a-b)(a+b)
\]

Here, let \(a = x^2\) and \(b = 9\). Substitute into the formula:

\[
(x^2)^2 – 9^2 = (x^2 – 9)(x^2 + 9)
\]

Step 4: Factor \(x^2 – 9\) further.

Again, \(x^2 – 9\) is a difference of squares because \(9 = 3^2\):

\[
x^2 – 9 = x^2 – 3^2 = (x-3)(x+3)
\]

Step 5: Leave \(x^2 + 9\) as it is.

The expression \(x^2 + 9\) is a sum of squares and does not factor into real linear factors.

Final Answer: The fully factored form of \(x^4 – 81\) is:

\[
x^4 – 81 = (x-3)(x+3)(x^2+9)
\]

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Algebra FAQ

. Factor \(x^4-81\).

. Use difference of squares: \(x^4-81=(x^4-9^2)=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9)\).

. Solve \(x^4-81=0\).

. From \( (x^2-9)(x^2+9)=0\): \(x^2-9=0 \Rightarrow x=\pm3\). Also \(x^2+9=0 \Rightarrow x=\pm3i\).

. What are the real zeros of \(x^4-81\)?

. Real zeros come from \(x^2-9=0\), so \(x=-3\) and \(x=3\). The factor \(x^2+9\) gives only complex roots.

. Does \(x^4-81\) have quadratic factors over integers?

. Yes: \(x^4-81=(x^2-9)(x^2+9)\). These are monic quadratics with integer coefficients.

. Find the roots of \(x^4-81\) using substitution \(y=x^2\).

. Let \(y=x^2\). Then \(y^2-81=0\Rightarrow y=\pm9\). So \(x^2=9\Rightarrow x=\pm3\), and \(x^2=-9\Rightarrow x=\pm3i\).

. Determine the sign of \(x^4-81\) for real \(x\).

. Since \(x^4=81\) at \(x=\pm3\), and \(x^4-81\) is even, it’s positive for \(|x|>3\) and negative for \(|x|<3\). At \(x=\pm3\), it equals \(0\).

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