Question

Solve the equation x^16 – 2x^15 – x^14 + 4x^13 – x^12 – 2x^11 + x^10 = 0 in the real number system.

a) x = 0 with multiplicity 10, x = –1 with multiplicity 4, x = 1 with multiplicity 2

b) x = 0 with multiplicity 10, x = –1 with multiplicity 3, x = 1 with multiplicity 3

c) x = 0 with multiplicity 10, x = –1 with multiplicity 2, x = 1 with multiplicity 4

d) x = –1 with multiplicity 2, x = 1 with multiplicity 4

Answer 1

C:x = 0 with multiplicity 10, x = –1 with multiplicity 2, x = 1 with multiplicity 4

Answer 2

Start by taking out a common factor.
x^10
x^10(x^6 - 2x^5 - x^4 + 4x^3 - x^2 - 2x + 1) = 0
So far we know that D won't work.
x - 1 is a factor because putting 1 in for the xs in the expression inside the brackets gives 0 Now you need to do a division. I'm going to assume you can do that.
Using division, I get
x^5 - x^4 - 2x^3 + 2x^2 + x - 1 Now divide x - 1 into this mess again. You get
x^4 - 2x^2 + 1 which factors into
(x^2 - 1)(x^2 - 1) which factors into
(x - 1)(x + 1)(x - 1)(x + 1) = 0

The first two divisions add (x -1)(x - 1)
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So we have (x - 1) with a multiplicity of 4 and (x + 1) with a multiplicity of 2
or x = 1 with a multiplicity of 4 and x = -1 with a multiplicity of 2 and x = 0
with a multiplicity of 10
That's the answer.
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C <<<< ===== Answer.