anonymous
  • 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathstudent55
  • mathstudent55
A zero of multiplicity 2 means there is one zero that appears twice.
mathstudent55
  • mathstudent55
Also, if a polynomial with integer coefficients has complex roots, then those roots must appear in pairs of complex conjugate roots.
mathstudent55
  • mathstudent55
For example, if the polynomial has the root 3 + 5i, then it must also have 3 - 5i as a root.

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More answers

anonymous
  • anonymous
i got everything it's just that i don't know how to multiply [x-(1+2i)] with [x-(1-2i)]
mathstudent55
  • 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
  • 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
  • mathstudent55
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mathstudent55
  • mathstudent55
\(\large y = [x-(1+2i)][x-(1-2i)](x - 1)^2\)
campbell_st
  • 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
  • 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.

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