Use the given zero to find all the zeros of the function. Include the given zero in your answer. Function: g(x) = 5x^3 + 39x^2 + 77x − 17 Zero: −4 + i

Question

Function: g(x) = 5x³ + 39x² + 77x − 17

    Zero: −4 + i

Answer 1

Given zero: x = -4 + i → x - (-4 + i) = 0

This comes in the conjugate pair.

x = -4 - i → x - (-4 - i) = 0

Distribute x and (-4 + i) to each term in the bracket:

[x - (-4 + i)]*[x - (-4 - i)] = x² - x(-4 - i) - x(-4 + i) + [(-4 + i)*(-4 - i)]

= x² + 4x - xi + 4x + xi + (16 + 4i - 4i - i²)

= x² + 8x + (16 - i²)

= x² + 8x + (16 + 1), where i² = -1

= x² + 8x + 17

By long division, we get:

(5x³ + 39x² + 77x - 17) ÷ (x² + 8x + 17) = 5x - 1

Zeros: x = 1/5, -4 + i, -4 - i

Answer 2

By Conjugate Zeros Theorem (- 4 - i) also zero of g(x). By Factor Theorem (x + 4 - i)(x + 4 + i) is factor of g(x).

So g(x) has factor (x + 4)² - i² = x² + 8x + 17. If we use long division g(x) by (x² + 8x + 17) we will get

5x - 1 quotient. By setting 5x - 1 = 0, x = 1/5 is zero of g(x): {- 4 - i, - 4 + i, 1/5}