escNotice how the imaginary terms cancel out, leaving only realOne more thing to note when working with the quadratic formula and a negative discriminant: the i can only cowhen the discriminant is negative, we get two complex roots whose imaginary part only differs by + or -as feare called conjugates.In our example above, we saw the conjugates: 1+ 3i and 1-3i.!19aPracticeGive an expression for f(x) of degree 3 with roots = 8,x=-2-1, and x = -2+i.. Your final answer should not contain any i's.f(x) = is a polynomial of degree 3 with the given roots.←Preview My AnswersYou have attempted this problem 1 time.Your overall recorded score is 0%.You have unlimited attempts remaining.Ask For Help@2WS#m3Submit AnswersedCс$4f%5tA66y&7Ohu* 008j9

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Answered: Give an expression for f(x) of degree 3…

Give an expression for f(x) of degree 3 with roots x = 8,x=-2-i, and x = -2+i. • Your final answer should not contain any i's. f(x) = is a polynomial of degree 3 with the given roots..

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter8: Complex Numbers And Polarcoordinates

Section8.4: Roots Of A Complex Number

Problem 41PS: Recall from the introduction to Section 8.2 that Jerome Cardans solutions to the equation x3=15x+4...

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Question

esc
Notice how the imaginary terms cancel out, leaving only real
One more thing to note when working with the quadratic formula and a negative discriminant: the i can only co
when the discriminant is negative, we get two complex roots whose imaginary part only differs by + or -as fe
are called conjugates.
In our example above, we saw the conjugates: 1+ 3i and 1-3i.
!
1
9
a
Practice
Give an expression for f(x) of degree 3 with roots = 8,x=-2-1, and x = -2+i.
. Your final answer should not contain any i's.
f(x) = is a polynomial of degree 3 with the given roots.
←
Preview My Answers
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For the complex number 3+4i real part is ______________, and the imaginary part is _______________.
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Solve: 6a2+9a=3a .
Recall from the introduction to Section 8.2 that Jerome Cardans solutions to the equation x3=15x+4 could be written as x=2+11i3+211i3 Lets assume that the two cube roots are complex conjugates. If they are, then we can simplify our work by noticing that x=2+11i3+211i3=a+bi+abi=2a which means that we simply double the real part of each cube root of 2+11i to find the solutions to x3=15x+4. Now, to end our work with Cardan, find the three cube roots of 2+11i. Then, noting the discussion above, use the three cube roots to solve the equation x3=15x+4. Write your answers accurate to the nearest thousandth.