Question

given the function f (x) = x³ + 8, a real zero is located at x = select choice.

Explanation:

Step1: Recall the formula for sum of cubes

The sum of cubes formula is \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\). For the function \(f(x)=x^3 + 8\), we can rewrite \(8\) as \(2^3\), so \(f(x)=x^3+2^3\).

Step2: Factor the function

Using the sum of cubes formula, \(x^3 + 2^3=(x + 2)(x^2-2x + 4)\).

Step3: Find the real zero

To find the real zero, we set \(f(x) = 0\), so \((x + 2)(x^2-2x + 4)=0\). The quadratic factor \(x^2-2x + 4\) has a discriminant \(\Delta=(-2)^2-4\times1\times4=4 - 16=- 12<0\), so it has no real roots. The real root comes from \(x + 2=0\), which gives \(x=-2\).

Answer:

\(-2\)