Question

Question #30ee9

Answer 1

The three zeros are `x=1+i, 1-i`, and `-4`

Explanation:

We start with some knowledge of the nature of cubic polynomials, in that there are only 3 possible cases for its roots:

1. the equation has three distinct real roots
2. the equation has a multiple root and all its roots are real
3. the equation has one real root and two non-real complex conjugate roots

The question has determined for us that situation 3 applies since the first root is complex: `1+i`. Therefore, we also know another root which is the complex conjugate of the first: `1-i`. Now there is only one root left to find. Let's find the answer by assuming the form of the solution where our unknown root is `a`:

`0=(x-a)(x-1-i)(x-1+i)`

lets multiply it out and set it equal to our original polynomial

`x^3-(a+2)x^2+2(a+1)x-2a = x^3+2x^2−6x+8`

By setting the coefficient of each power of `x` on either side of the equation equal to the other, we get

`-(a+2)=2`
`2(a+1)=-6`
`-2a=8`

therefore `a=-4`

(in all three cases which means it is consistent)