Question

find all of the zeros of P(x) = x^3-8x+32, given that 2 + 2i is a zero. ( if there is more than one zero, separate them with commas.)

Answer 1

The zeros of the polyomial P(x) = x³ - 8x + 32 are -4, 2 + 2i and 2 - 2i

How to determine all the zeros of the polyomial

From the question, we have the following parameters that can be used in our computation:

P(x) = x³ - 8x + 32

Given that 2 + 2i is a zero, then 2 - 2i is also a zero

So, we have

Divisor = (x - [2 + 2i]) * (x - [2 - 2i])

Evaluate

Divisor = x² -2x + 2ix - 2x - 2ix + 8

Divisor = x² - 4x + 8

To calculate the last zero, we use the long division method of quotient

So, we have

x + 4

x² - 4x + 8 | x³ - 8x + 32

x³ - 4x² + 8x

---------------------

4x² - 16x + 32

4x² - 16x + 32

---------------------

0

So, we have

x + 4 = 0

Evaluate

x = -4

Hence, the zeros of the polyomial are -4, 2 + 2i and 2 - 2i

Answer 2

According to the Conjugate Roots Theorem, the complex zeroes of a polynomial always occur in pairs with its conjugate. This means that is a+ib is the root of the equation then its conjugate a-ib must also be a root of the same polynomial.

Given that x=2+2i is a root, then x=2-2i will also be a root of the given cubic polynomial.

Note that since the degree of polynomial is 3, there can be maximum 3 zeroes. Two of these are known complex zeroes, there is only one zero remaining i.e. not occurring in pair. So it must be a real number.

Let the third root be 'a', this means (x-a) ia a factor of the polynomial.

`\begin{gathered} (x-2-2i)(x-2+2i)(x-a)=x^3-8x+32 \\ ((x-2)^2-(2i)^2)(x-a)=x^3-8x+32 \\ (x^2-4x+4+4)(x-a)=x^3-8x+32 \\ (x^2-4x+8)(x-a)=x^3-8x+32 \\ (x-a)=(x^3-8x+32)/(x^2-4x+8) \end{gathered}`

Apply the Long Division,

It is found that,

`\begin{gathered} x-a=x-4 \\ a=4 \end{gathered}`

Thus, the third root of the given cubic polynomial is 4.