Respuesta :

lesliekolleroxrkcf
if 1+i is a root then 1-i must be a root because complex roots come in conjugate pairs, so
(x-8)² * (x-1-i) * (x-1+i) = 0
(x-8)² * (x²-x+xi-x+1-i-xi+i+1)=0
(x-8)² * (x²-2x+2) = 0
(x²-16x+64)(x²-2x+2)=0
x⁴-2x³+2x²-16x³+32x²-32x+64x²-128x+128=0
f(x)=x⁴-18x³+98x²-160x+128

if I didn't lose track... that was tough on a phone with me paper to scratch on!

Using the Factor Theorem, the polynomial function is given by:

[tex]p(x) = a(x^4 - 18x^3 + 96x^2 - 160x + 128)[/tex]

Many correct answers according to the value of the leading coefficient a.

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The Factor Theorem states that a polynomial function with zeros [tex]x_1, x_2,...,x_n[/tex] is defined by:

[tex]p(x) = a(x - x_1)(x - x_2)...(x - x_n)[/tex]

  • In which a is the leading coefficient.

In this question:

  • 8 is a zero with multiplicity 2, thus [tex]x_1 = x_2 = 8[/tex]
  • If 1 + i is a zero, it's conjugate also is, thus [tex]x_3 = 1 + i, x_4 = 1 - i[/tex]

Then, the polynomial is:

[tex]p(x) = a(x - 8)(x - 8)(x - 1 - i)(x - 1 + i)[/tex]

[tex]p(x) = a(x^2 - 16x + 64)(x^2 - 2x + 1 - i^2)[/tex]

Since [tex]i^2 = -1[/tex]

[tex]p(x) =a(x^2 - 16x + 64)(x^2 - 2x + 2)[/tex]

[tex]p(x) = a(x^4 - 18x^3 + 96x^2 - 160x + 128)[/tex]

Many correct answers according to the value of the leading coefficient a.

A similar problem is given at https://brainly.com/question/24380382

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