Question

factor the polynomial, then determine the real zeros of the function: f(x) = x⁴ - 16. options: 2i, -2i, 2, -2; 0; 2, -2; 2

Explanation:

Step1: Recognize as difference of squares

The polynomial \( f(x) = x^4 - 16 \) can be written as a difference of squares: \( (x^2)^2 - 4^2 \). Using the difference of squares formula \( a^2 - b^2=(a - b)(a + b) \), we get \( (x^2 - 4)(x^2 + 4) \).

Step2: Factor \( x^2 - 4 \) further

\( x^2 - 4 \) is also a difference of squares: \( x^2 - 2^2=(x - 2)(x + 2) \). So now \( f(x)=(x - 2)(x + 2)(x^2 + 4) \).

Step3: Find real zeros

To find the real zeros, set \( f(x) = 0 \). So \( (x - 2)(x + 2)(x^2 + 4)=0 \). The factor \( x^2 + 4 = 0 \) gives \( x^2=-4 \), which has no real solutions (only complex solutions \( \pm 2i \)). The factors \( x - 2 = 0 \) gives \( x = 2 \) and \( x + 2 = 0 \) gives \( x=-2 \). So the real zeros are \( 2 \) and \( -2 \).

Answer:

C. 2, -2 (assuming the third option is labeled C as per the options: first option: 2i, -2i, 2, -2; second: 0; third: 2, -2; fourth: 2)