Mathematics, 13.07.2019 02:30 qiuedhg

Using the given zero, find one other zero of f(x). explain the process you used to find your solution.
2 - 3i is a zero of f(x) = x4 - 4x3 + 14x2 - 4x + 13.
*can someone show the work i have the answers

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Answer from: kholman

The zeros of a function are the points where the function cross the x-axis.

One other zero of $\mathbf{f(x) = x^4 - 4x^3 + 14x^2 - 4x + 13}$ is 2 + 3i.

The zero of $\mathbf{f(x) = x^4 - 4x^3 + 14x^2 - 4x + 13}$ is given as:

$\mathbf{Zero = 2 - 3i}$

The above number is a complex number.

If a complex number a + bi is the zero of a function f(x), then the conjugate a - bi is also the zero of f(x).

This means that, one other zero of $\mathbf{f(x) = x^4 - 4x^3 + 14x^2 - 4x + 13}$ is 2 + 3i.

Read more about zeros of functions at:

link

Answer from: iBrain

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)

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

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

Answer from: Quest

I'll say b) log(- that if not let me know what i can do
Asap 25 pts + brainliest to right/best answer

Answer from: Quest

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9

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