Using the Given Zero, Find All Other Zeros of f(x) = x^4 – 32x^2 + 144

To find all the zeros of the polynomial function f(x) = x^4 – 32x^2 + 144, we start by utilizing the given zero, 2i. Since polynomial functions with real coefficients have complex roots that come in conjugate pairs, if 2i is a zero, then its conjugate -2i must also be a zero.

Next, let’s use these zeros to factor the polynomial. We know that both 2i and -2i correspond to the factor:

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

Now we need to divide the original polynomial f(x) by x^2 + 4 to find the remaining factors. We can perform polynomial long division.

         x^2 - 32
        _______________________
x^2 + 4 | x^4 + 0x^3 - 32x^2 + 0x + 144
         - (x^4 + 4x^2)
         _______________________
               -36x^2 + 144
              - (-36x^2 - 144)
         _______________________
                         288

This division shows that:

f(x) = (x^2 + 4)(x^2 - 36)

The second factor, x^2 – 36, can be factored further:

x^2 - 36 = (x - 6)(x + 6)

Thus, we can express the complete factorization of f(x) as:

f(x) = (x - 6)(x + 6)(x^2 + 4)

The zeros of f(x) can now be found:

• x – 6 = 0 => x = 6
• x + 6 = 0 => x = -6
• x^2 + 4 = 0 => x = 2i and x = -2i

In conclusion, the complete set of zeros of f(x) includes:

• 6
• -6
• 2i
• -2i