Question

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

Answer 1

Answer:

One other zero is 2+3i

Step-by-step explanation:

If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.

The conjugate of (a+b) is (a-b).

The conjugate of (a-b) is (a+b).

The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.

Ok so we have two zeros 2-3i and 2+3i.

This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.

I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).

(x-(2-3i))(x-(2+3i))

Foil!

First: x(x)=x^2

Outer: x*-(2+3i)=-x(2+3i)

Inner:  -(2-3i)(x)=-x(2-3i)

Last:  (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)

---------------------------------Add together:

x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)

Simplifying:

x^2-2x-3ix-2x+3ix+4+9  (since i^2=-1)

x^2-4x+13                     (since -3ix+3ix=0)

So x^2-4x+13 is a factor of the given polynomial.

I'm going to do long division to find another factor.

Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.

                x^2+1

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

x^2-4x+13|  x^4-4x^3+14x^2-4x+13                    

              -( x^4-4x^3+ 13x^2)

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

                                 x^2-4x+13

                               -(x^2-4x+13)

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

                                    0

So the other factor is x^2+1.

To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.

$x^2+1=0$

$x^2=-1$

$x=\pm \sqrt{-1}$

$x=\pm i$

So the zeros are i, -i , 2-3i , 2+3i