Question: Given that $-8i$ is a zero, factor the following polynomial function completely. Use the Conjugate…

Given that $-8i$ is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable.

$$f(x) = x^4 - 8x^3 + 79x^2 - 512x + 960$$

Answer:

$$f(x) =$$

Solution

Given that \(-8i\) is a zero, we can use the Conjugate Roots Theorem, which states that if a polynomial has real coefficients, then any non-real complex roots must occur in conjugate pairs. Since \(-8i\) is a zero, its conjugate \(8i\) is also a zero. Therefore, \((x + 8i)(x - 8i) = x^2 + 64\) is a factor. Now, we’ll use polynomial division to divide \(f(x)\) by \(x^2 + 64\). First, set up the division: \[ f(x) = x^4 - 8x^3 + 79x^2 - 512x + 960 \] Divide by \(x^2 + 64\): 1. Divide the leading term \(x^4\) by \(x^2\): \(x^2\) 2. Multiply \(x^2\) by \(x^2 + 64\): \(x^4 + 64x^2\) 3. Subtract from the original polynomial: \[ (x^4 - 8x^3 + 79x^2 - 512x + 960) - (x^4 + 64x^2) = -8x^3 + 15x^2 - 512x + 960 \] 4. Divide the leading term \(-8x^3\) by \(x^2\): \(-8x\) 5. Multiply \(-8x\) by \(x^2 + 64\): \(-8x^3 - 512x\) 6. Subtract: \[ (-8x^3 + 15x^2 - 512x + 960) - (-8x^3 - 512x) = 15x^2 + 960 \] 7. Divide the leading term \(15x^2\) by \(x^2\): \(15\) 8. Multiply \(15\) by \(x^2 + 64\): \(15x^2 + 960\) 9. Subtract: \[ (15x^2 + 960) - (15x^2 + 960) = 0 \] The quotient is \(x^2 - 8x + 15\). Now, factor \(x^2 - 8x + 15\): The factors of 15 that add to \(-8\) are \(-3\) and \(-5\). Thus, \(x^2 - 8x + 15 = (x - 3)(x - 5)\). Combining all factors, we have: \[ f(x) = (x^2 + 64)(x - 3)(x - 5) \]