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anonymous

  • one year ago

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 4, -2, and -1 + 2i How would I solve this?

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  1. campbell_st
    • one year ago
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    so the polynomial has 4 roots.. the real roots are x = 4 and x = -2 the complex roots are \[x = -1 \pm i\]

  2. campbell_st
    • one year ago
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    so to find the complex roots start with \[x = -1 \pm i\] add 1 to both sides \[x + 1 = \pm i\] square both sides of the equation \[(x + 1)^2 = i^2\] remember i^2 = -1 then \[x^2 + 2x + 1 = -1\] add 1 to both sides of the equation and you will have the quadratic factor of the polynomial... so the polynomial is \[P(x) = (x^2 + 2x +1)(x -4)(x+1)\] just distribute for the equation

  3. anonymous
    • one year ago
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    Okay.

  4. anonymous
    • one year ago
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    2i, 5, -5?

  5. campbell_st
    • one year ago
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    is that another question...?

  6. anonymous
    • one year ago
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    No, is that the correct answers @campbell_st

  7. campbell_st
    • one year ago
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    well the polynomial is \[P(x) = (x^2 + 2x + 2)(x -4)(x +2)\] which becomes \[P(x) = (x^2+ 2x +2)(x^2 -2x - 8)\] you just need to multiply the 2 quadratic factors together

  8. anonymous
    • one year ago
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    The answer I come up with is not one of the options. X^4-10x^2-20x-16

  9. Mertsj
    • one year ago
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    f(x)=(x-4)(x+2)[x-(-1+2i)][x-(-1-2i)]

  10. Mertsj
    • one year ago
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    Multiply the four factors and that will be your answer.

  11. anonymous
    • one year ago
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    Okay I got x^4+6.5x^2-26x

  12. anonymous
    • one year ago
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    Is that correct?

  13. Mertsj
    • one year ago
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    f(x)=(x^2-2x-8)(x^2+5)=x^4-2x^3-8x^2+5x^2-10x-40=x^4-2x^3-3x^2-10x-40

  14. Mertsj
    • one year ago
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    That's what I get from multiplying those 4 factors. Is that one of your choices?

  15. anonymous
    • one year ago
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    Yes, I must have done it wrong. Thank you.

  16. anonymous
    • one year ago
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    Wait, no.

  17. anonymous
    • one year ago
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    what the hell

  18. Mertsj
    • one year ago
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    If 4is a root, then x-4 is a factor. If -2 is a root then x-(-2)=x+2 is a factor If -1+2i is a root then -1 - 2i is also a root If -1+2i is a root then x-(-1+2i) is a factor If -1-2i is a root then x - (-1-2i) is a factor

  19. Mertsj
    • one year ago
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    So now we have 4 factors of the polynomial: (x-4)(x+2)[x-(-1+2i)][x-(-1-2i)] (x-4)(x+2)=x^2-2x-8 [x-(-1+2i)][x-(-1-2i)]=x^2-x(-1-2i)-x(-1+2i)+(-1+2i)(-1-2i)= x^2+x+2xi+x-2xi+5=x^2+2x+5

  20. Mertsj
    • one year ago
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    So now we must multiply (x^2-2x+8)(x^2+2x+5)= x^4+2x^3+5x^2-2x^3-4x^2-10x+8x^2+16x+40= x^4+9x^2+6x+40

  21. Mertsj
    • one year ago
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    Is that one of your choices?

  22. anonymous
    • one year ago
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    No.

  23. anonymous
    • one year ago
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    @Mertsj

  24. Mertsj
    • one year ago
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    What are the answer choices?

  25. anonymous
    • one year ago
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    a. f(x) = x4 - 7x2 - 26x - 40 b. f(x) = x4 - 3x3 - 8x2 - 13x - 40 c. f(x) = x4 + 6.5x2 - 26x - 40 d. f(x) = x4 - 3x3 + 8x2 + 13x + 40

  26. Mertsj
    • one year ago
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    I made a typo when I multiplied: (x^2-2x-8)(x^2+2x+5)=x^4+2x^3+5x^2-2x^3-4x^2-10x-8x^2-16x-40= x^4-7x^2-26x-40

  27. Mertsj
    • one year ago
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    Before I used x^2-2x+8 instead of x^2-2x-8

  28. anonymous
    • one year ago
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    Ah I see.

  29. anonymous
    • one year ago
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    Thanks.

  30. Mertsj
    • one year ago
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    yw

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