Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Find a polynomial with integer coefficients that satisfy the given condition. R has degree 4 and zeros 1-2i and 1, with a zero of multiplicity 2.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- mathstudent55

A zero of multiplicity 2 means there is one zero that appears twice.

- mathstudent55

Also, if a polynomial with integer coefficients has complex roots, then those roots must appear in pairs of complex conjugate roots.

- mathstudent55

For example, if the polynomial has the root 3 + 5i, then it must also have 3 - 5i as a root.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

i got everything it's just that i don't know how to multiply [x-(1+2i)] with [x-(1-2i)]

- mathstudent55

This is one way of doing it:
First, simplify the parentheses inside each expression.
Then multiply them together using polynomial multiplication.
That is, multiply every term of the first polynomial by every term of the second polynomial, then collect like terms.

- mathstudent55

The other way of doing it is to turn the product into the product of a sum and a difference and end up with the difference of two squares. Then you simplify.

- mathstudent55

Are you ok with this problem now?
Do you need more help?

- mathstudent55

\(\large y = [x-(1+2i)][x-(1-2i)](x - 1)^2\)

- campbell_st

looking at your complex zero you know
\[x = 1 \pm 2i\]
then rewriting you get
\[x- 1 = \pm 2i\]
square both sides of the equation
\[(x -1)^2 = 4i^2\]
or
\[(x -1)^2 = -4\]
then you can find the quadratic factor

- mathstudent55

Method 1:
\(\large y = \color{red}{[x-(1+2i)][x-(1-2i)]}(x - 1)^2\)
\(\large y = \color{red}{(x-1- 2i)(x- 1+2i)}(x - 1)^2\)
\(\large y = \color{red}{(x^2-x + 2xi -x +1 -2i -2xi + 2i -4i^2)}(x - 1)^2\)
\(\large y = \color{red}{(x^2-2x +1+4)}(x - 1)^2\)
\(\large y = \color{red}{(x^2-2x +5)}(x - 1)^2\)
Method 2:
\(\large y = \color{red}{[x-(1+2i)][x-(1-2i)]}(x - 1)^2\)
\(\large y = \color{red}{[(x-1)-2i][(x-1)+2i]}(x - 1)^2\)
\(\large y = \color{red}{[(x-1)^2-(2i)^2]}(x - 1)^2\)
\(\large y = \color{red}{[x^2 - 2x + 1 - (-4)]}(x - 1)^2\)
\(\large y = \color{red}{(x^2 - 2x + 5)}(x - 1)^2\)
After doing the step above with either method 1 or method 2 or with @campbell_st 's method, you still need to square x - 1, and then multiply it all together.

Looking for something else?

Not the answer you are looking for? Search for more explanations.