SOLUTION: Use the given zero to find the remaining zeros of the polynomial function P(x) =x^4-6x^3+19x^2-6x+18; 3-3i

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Question 1201674: Use the given zero to find the remaining zeros of the polynomial function
P(x) =x^4-6x^3+19x^2-6x+18; 3-3i
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

The coefficients of the polynomial are real numbers, so the complex roots come in conjugate pairs. So if 3-3i is a root, 3+3i is another root.

You can use Vieta's Theorem to find the quadratic polynomial with roots 3-3i and 3+3i:
coefficient of linear term: -(sum of the two roots) = -((3-3i)+(3+3i)) = -6
constant term: product of the two roots: (3-3i)(3+3i) = 9-9i^3 = 9+9 = 18

The quadratic polynomial with roots 3-3i and 3+3i is x^2-6x+18. So

We know the leading coefficient of the second quadratic factor is 1, because the leading coefficient of the product is 1. And we know the constant term of the second quadratic factor is 1, because the constant term of the product is 18. So we find the constant "a" by looking at the linear term in the product.

On the left, the linear term is ; on the right it is . So a is 0; the other quadratic factor is . The zeros of that quadratic are i and -i.

ANSWER: given the one root 3-3i, the other roots are 3+3i, i, and -i.

Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

The term "root" is another way of saying "zero of a function"
The root is visually represented by the x intercept, assuming the root is a real number.

In the case of 3-3i being a root, we unfortunately cannot represent this as an x intercept.
It has a conjugate pair of 3+3i

We'll take the root x = 3-3i and work backwards, so to speak, to arrive at a quadratic polynomial.
x = 3-3i
x-3 = -3i
(x-3)^2 = (-3i)^2
(x-3)^2 = 9i^2
(x-3)^2 = 9(-1)
(x-3)^2 = -9
(x-3)^2+9 = 0
x^2-6x+9+9 = 0
x^2-6x+18 = 0
Similar, if not almost identical, steps will apply for the root x = 3+3i.

If you were to use the quadratic formula to solve x^2-6x+18 = 0, then you should get the complex roots x = 3+3i and x = 3-3i.
You can use WolframAlpha or the CAS feature in GeoGebra to check your work.

Because 3-3i is a root of P(x)=x^4-6x^3+19x^2-6x+18, we know that x^2-6x+18 is a factor of P(x).

P(x) = (x^2-6x+18)*Q(x)
where Q is some quotient polynomial.
The remainder is 0 because x^2-6x+18 is a factor.

Isolating Q means we divide both sides by (x^2-6x+18)
Q(x) = P(x)/(x^2-6x+18)
Q(x) = (x^4-6x^3+19x^2-6x+18)/(x^2-6x+18)

From here we use polynomial long division to determine Q.
Synthetic division will not work because the denominator x^2-6x+18 is not linear.

Here is what the polynomial long division will look like.

We therefore find that
Q(x) = x^2+1
P(x) = Q(x)*(x^2-6x+18)
P(x) = (x^2+1)(x^2-6x+18)

In other words,
(x^2+1)(x^2-6x+18) = x^4-6x^3+19x^2-6x+18
is an identity. The equation is true for all values of x.

We can confirm this by use of the box method
We'll have the terms of x^2+1 along the left side and the terms of x^2-6x+18 along the top

x^2 -6x 18
x^2
1

Then to fill out each inner cell, multiply the left and top headers.
Example: x^2 times x^2 = x^4 in the upper left corner

x^2 -6x 18
x^2 x^4 -6x^3 18x^2
1 x^2 -6x 18

Add up the terms inside those cells we filled out
x^4-6x^3+18x^2+x^2-6x+18 = x^4-6x^3+19x^2-6x+18
We have confirmed that (x^2+1)(x^2-6x+18) expands to x^4-6x^3+19x^2-6x+18
In other words, we have confirmed x^4-6x^3+19x^2-6x+18 factors to (x^2+1)(x^2-6x+18)

As stated earlier, the two roots x = 3+3i and x = 3-3i were from solving x^2-6x+18=0.

To determine the other two missing other roots, we solve x^2+1 = 0 to get x = i or x = -i
You could use the square root method.
Or the quadratic formula could be applied (even if it may be a bit overkill).

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The four roots of P(x) are:
x = 3+3i
x = 3-3i
x = i
x = -i
where i = sqrt(-1)

Confirmation through use of WolframAlpha
https://www.wolframalpha.com/input/?i=x%5E4-6x%5E3%2B19x%5E2-6x%2B18%3D0

GeoGebra also can be used to confirm the answer.
You need to be in the CAS mode and you'd use the function called CSolve.
The "C" in CSolve stands for "complex"
The regular "solve" function will return an empty set to indicate "no real solutions", which is why you need to avoid using it.
More info found here
https://wiki.geogebra.org/en/CSolve_Command