Anonymous

A degree 4 polynomial with integer coefficients has zeros -4-2 i and 1, with 1 a zero of multiplicity 2. If the coefficient of x^4 is 1,

then the polynomial is ___________

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• Anonymous

The complex roots come in conjugate pairs. So the roots are

-4-2i and -4+2i.

The other roots are 1 and 1.

So the polynomial would looks like

(x-(-4-2i)) * (x-(-4+2i)) * (x-1)^2

=x^4+6*x^3+5*x^2-32*x+20

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• 4

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• Well, since the coefficients of the polynomial are real numbers, then the root -4 - 2i, has its conjugate pair as a root also, i.e.

-4 + 2i

((x - 1)^2)(x + 4+2i)(x + 4 - 2i)

Multiply this out, and you will get a fourth degree polynomial with real integer coefficients.

x^4 + 6x^3 + 5x^2 - 32x + 20

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• if -4-2i is zero then -4+2i is a zero

if x=1 is a double zero

then

(x-1)^2*((x+4)^2+4)=

(x^2-2x+1)(x^2+8x+20)=

x^4+6x^3+5x^2-32x+20

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• Anonymous

First, you find all of the zeroes.

(complex numbers have conjugates)

(-4-2i), (-4+2i), (+1), (+1)

Then, you make it into the factored form of the polynomial.

(x+4-2i)(x+4+2i)(x-1)(x-1)=0

Then you multiply it out.

(x^2+8x+20)(x^2-2x+1)=0

x^4+6x^3+5x^2-32x+20=0

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• 4 isn't a polynomial

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• #'s are everlasting they never end and a polynomial is in the 4th power an on & on & on 00000000000000000

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