SOLUTION: Q. Show that 2 - i is a root of z^4 - 8z^3 + 28z^2 - 48z + 35. Hence fond all the roots

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Question 135973: Q. Show that 2 - i is a root of z^4 - 8z^3 + 28z^2 - 48z + 35. Hence fond all the roots
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Actually, technically speaking, and we should always speak technically when talking about this stuff, z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35 doesn't have any roots. z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35+=+red%280%29 has roots, and those roots are related to the factors of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35 as we shall see shortly.

Assume that 2-i is a root of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0

Then 2%2Bi must also be a root of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0 because complex roots always occur in conjugate pairs.

a is a root of a polynomial equation if and only if x-a is a factor of the polynomial. That means that both z-%282-i%29 and z-%282%2Bi%29 are factors of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35. Furthermore, the product of those two factors must also be a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35.

The product of z-%282-i%29 and z-%282%2Bi%29, %28z-%282-i%29%29%28z-%282%2Bi%29%29=%28z-2%2Bi%29%28z-2-i%29 works out to z%5E2-2z-zi-2z%2B4%2B2i%2Bzi-2i-i%5E2=z%5E2-4z%2B5 (recalling that i%5E2=-1 so -%28i%5E2%29=-%28-1%29=%2B1)

Since z%5E2-4z%2B5 is a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35, the result of polynomial long division of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35 by z%5E2-4z%2B5 will be a quotient polynomial with a zero remainder. If this is true, then our original assumption is true, i.e., 2-i is a root of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0 because we will have proven that z-%282-i%29 is a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35

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Polynomial Long Division Process
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z%5E2 goes into z%5E4 z%5E2 times. green%28z%5E2%29 is the first term of the quotient.

Multiply z%5E2 times z%5E2-4z%2B5 to get z%5E4-4z%5E3%2B5z%5E2 and subtract from z%5E4+-+8z%5E3+%2B+28z%5E2 resulting in -4z%5E3%2B23z%5E2. Bring down the -48z to form -4z%5E3%2B23z%5E2-48z

z%5E2 goes into -4z%5E3 -4z times. green%28-4z%29 is the second term of the quotient.

Multiply -4z times z%5E2-4z%2B5 to get -4z%5E3%2B16z%5E2-20z and subtract from -4z%5E3%2B23z%5E2-48z resulting in 7z%5E2-28z. Bring down the 35 to form 7z%5E2-28z%2B35

z%5E2 goes into 7z%5E2 7 times. green%287%29 is the third term of the quotient.

Multiply 7 times z%5E2-4z%2B5 to get 7z%5E2-28z%2B35 and subtract from 7z%5E2-28z%2B35 resulting in a remainder of zero. The quotient is the sum of the three parts noted above, namely green%28z%5E2-4z%2B7%29

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End of Polynomial Long Division Process
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That means that z%5E2-4z%2B5 is a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35, and therefore z-%282-i%29 must also be a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35, and finally z=2-i (and its conjugate, z=2%2Bi)must be a root of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0.

Since z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0 is a quartic (degree 4) polynomial equation, there must be 4 roots. We have found two of them. Since the quotient derived by the polynomial long division process just performed must also be a factor of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35, the factors of that quotient must also be factors of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35.

Therefore solving z%5E2-4z%2B7=0 will give the remaining two roots of z%5E4+-+8z%5E3+%2B+28z%5E2+-+48z+%2B+35=0. I'll leave that one to you. Hint: The quadratic does NOT factor over the integers (or the reals, for that matter), so either complete the square or use the quadratic formula. I completed the square and that worked out rather tidily. Write back and tell me your results.

The following graph of f%28x%29=x%5E4+-+8x%5E3+%2B+28x%5E2+-+48x+%2B+35 supports the notion that there are no real roots to your given equation because the graph does not intersect the x-axis, so f%28x%29%3C%3E0 for all real x.