Q. Convert the question \(x^2+4\) factored.

Q: What are the factors of x^2+4 over the integers?

x^2+4 does not factor over the integers because it has no rational number roots. Its discriminant is negative: b^2-4ac=-16.

Q: Can x^2+4 be factored over the real numbers?

Yes. Solve x^2+4=(x-2i)(x+2i) over complex numbers. Over reals, it stays irreducible as a product of real linear factors.

Q: How do you derive the factorization using the quadratic formula?

For x^2+4=0, x=\frac{-0\pm\sqrt{0-16}}{2}=\pm 2i. So x^2+4=(x-2i)(x+2i).

Q: Is there a way to factor x^2+4 like a difference of squares?

Not directly, since x^2+4=x^2-(-4). You can write \[x^2+4=(x)^2-(2i)^2=(x-2i)(x+2i).\]

Answer

We want to factor \(x^2+4\) into a product of polynomials. Since \(x^2+4 = x^2+2^2\), use the sum of squares fact:

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

Final result: \( (x-2i)(x+2i) \).

Detailed Explanation

We want to factor the expression \(x^2+4\).

Step 1: Recognize the form

The expression \(x^2+4\) looks like a sum of squares, because \(4\) is a perfect square: \(4=2^2\). So we can rewrite it as \(x^2+(2)^2\).

Step 2: Use the difference of squares / sum of squares structure

Over the real numbers, factoring \(x^2+4\) is the same as factoring \(x^2+2^2\), which does not factor nicely into real linear factors. But it does factor over the complex numbers using the identity:

\[\,a^2+b^2=(a+bi)(a-bi)\,.\]

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

Step 3: Substitute into the identity

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

Final answer

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

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Algebra FAQ

What are the factors of \(x^2+4\) over the integers?

\(x^2+4\) does not factor over the integers because it has no rational number roots. Its discriminant is negative: \(b^2-4ac=-16\).

Can \(x^2+4\) be factored over the real numbers?

Yes. Solve \(x^2+4=(x-2i)(x+2i)\) over complex numbers. Over reals, it stays irreducible as a product of real linear factors.

What is the factorization of \(x^2+4\) over complex numbers?

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

How do you derive the factorization using the quadratic formula?

For \(x^2+4=0\), \(x=\frac{-0\pm\sqrt{0-16}}{2}=\pm 2i\). So \(x^2+4=(x-2i)(x+2i)\).

Is there a way to factor \(x^2+4\) like a difference of squares?

Not directly, since \(x^2+4=x^2-(-4)\). You can write \[x^2+4=(x)^2-(2i)^2=(x-2i)(x+2i).\]

What are the roots of \(x^2+4\)?

\[x^2+4=0 \Rightarrow x=\pm 2i.\]

Why doesn’t \(x^2+4\) factor into real linear factors?

Real linear factors require real roots. Since the discriminant of \(x^2+4\) is negative \((-16)\), the roots are complex, so no real linear factorization exists.

Use this to factor \(x^2+4\).
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