Q. \(x^3 + 8\)

Q: Factor x^3+8.

Use sum of cubes: x^3+8=x^3+2^3=(x+2)(x^2-2x+4).

Q: Solve x^3+8=0.

Set (x+2)(x^2-2x+4)=0. Then x=-2 and x^2-2x+4=0 has discriminant -12, so no other real solutions.

Q: What are the complex roots of x^2-2x+4=0?

\Delta=(-2)^2-16=-12. So x=\frac{2\pm\sqrt{-12}}{2}=1\pm i\sqrt{3}.

Q: Expand (x+2)(x^2-2x+4).

Multiply to get x(x^2-2x+4)+2(x^2-2x+4)=x^3-2x^2+4x+2x^2-4x+8=x^3+8.

Q: Determine whether x^3+8 is divisible by x+2.

Yes, since 0^? check: (-2)^3+8=-8+8=0, so x+2 is a factor.

Answer

\(x^3+8\) is a sum of cubes since \(8=2^3\). Using the identity \(a^3+b^3=(a+b)(a^2-ab+b^2)\) with \(a=x\) and \(b=2\):

\[
x^3+8 = x^3+2^3 = (x+2)\bigl(x^2-2x+4\bigr).
\]

Detailed Explanation

We want to simplify or factor the expression \(x^3 + 8\).

Step 1: Recognize a special pattern

Notice that \(8\) is a perfect cube because \(8 = 2^3\). So we can rewrite the expression as

\[
x^3 + 8 = x^3 + 2^3.
\]

Step 2: Use the sum of cubes formula

The sum of cubes formula is

\[
a^3 + b^3 = (a+b)(a^2 – ab + b^2).
\]

Here, \(a = x\) and \(b = 2\).

Step 3: Substitute \(a = x\) and \(b = 2\)

Substitute into the formula:

\[
x^3 + 2^3 = (x+2)(x^2 – x\cdot 2 + 2^2).
\]

Step 4: Simplify inside the parentheses

Compute each term in \(x^2 – 2x + 2^2\):

\[
2^2 = 4,
\]
so the expression becomes

\[
(x+2)(x^2 – 2x + 4).
\]

Final answer

\[
x^3 + 8 = (x+2)(x^2 – 2x + 4).
\]

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Algebra FAQ

Factor \(x^3+8\).

Use sum of cubes: \(x^3+8=x^3+2^3=(x+2)(x^2-2x+4)\).

Solve \(x^3+8=0\).

Set \((x+2)(x^2-2x+4)=0\). Then \(x=-2\) and \(x^2-2x+4=0\) has discriminant \(-12\), so no other real solutions.

Find real roots of \(x^3+8=0\).

Only real root is \(x=-2\).

What are the complex roots of \(x^2-2x+4=0\)?

\(\Delta=(-2)^2-16=-12\). So \(x=\frac{2\pm\sqrt{-12}}{2}=1\pm i\sqrt{3}\).

Expand \((x+2)(x^2-2x+4)\).

Multiply to get \(x(x^2-2x+4)+2(x^2-2x+4)=x^3-2x^2+4x+2x^2-4x+8=x^3+8\).

Determine whether \(x^3+8\) is divisible by \(x+2\).

Yes, since \(0^? \) check: \((-2)^3+8=-8+8=0\), so \(x+2\) is a factor.

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