# form a 5th degree polynomial f(x) with real coeffcients having the given zeros: 4;-i;4+i must show work donot enter the polynomial in factored form, multiply it out

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### 1 Answer

If the polynomial has real coefficients, then either the zeros are real, or they come in complex conjugate pairs. So, for example, if 4+i is a root, then 4-i must be a root as well. Thus, our polynomial must have the following roots:

4, -i, i, 4+i, 4-i

We can find a polynomial with these roots by multiplying out:

`(x-4)(x+i)(x-i)(x-4-i)(x-4+i)`

It will be easier to do this in stages:

`(x+i)(x-i)=x^2-ix+ix-i^2=x^2+1`

`(x-4-i)(x-4+i)=x^2-4x+ix-4x+16-4i-ix+4i-i^2`

`=x^2-8x+16+1=x^2-8x+17`

Thus we want to multiply:

`(x-4)(x^2+1)(x^2-8x+17)`

`=(x^3+x-4x^2-4)(x^2-8x+17)`

`=x^5-8x^4+17x^3+x^3-8x^2+17x-4x^4+32x^3-68x^2-4x^2+32x-68`

`=x^5-12x^4+50x^3-80x^2+49x-68`