Degree 5; zeros: -8; -i; -5+i , enter the polynomial

Question

please help

Answer 1

First off, if x=-8 is a root, then it came from x+8 = 0 which makes (x+8) a factor of your polynomial.

Secondly, complex roots always come in conjugate pairs, so if -i is a root, then so is i and likewise if -5+i is a root then 

-5-i is also a root.

So if x=-i is an answer, then x+i = 0 and (x+i) is a factor.

    if x=i is an answer, then x-i=0 and (x-i) is a factor.

    if x=-5+i is a root, then x+5-i =0 and (x+5-i) is a factor, and

    if x=-5-i is a root, then x+5+i =0 and (x+5+i) is a factor.

 

So just multiply these all out.

(x+8)(x-i)(x+i)(x+5+i)(x+5-i) = 0

 

multiplying the conjugates makes life a little easier rather than multiplying in order. So,

(x+8)( x²+1)(x²+10x+26) =0

(x+8)(x⁴+10x³+27x²+10x+26)=0 and lastly,

x⁵ + 18x⁴+107x³+226x²+106x+208=0

 

Hope this helped.

Answer 2

If the coefficients are real, then for any complex zero, the complex conjugate must also be a zero.  You will then have 5 zeros:

 

-8

-i

+i

-5 + i

-5 - i

 

The polynomial will be 

 

P(x) = A(x - r₁)(x - r₂)...(x - r₅)

 

where A ≠ 0 and the r_i are the zeros.

 

P(x) = A(x + 8)(x + i)(x - i)(x + 5 - i)(x + 5 + i)

 

You can expand and simplify from here.