Question

Find all other zeros of P(x)=`x^3-9x^2+28x-30`, given that 3+i is a Zero.?

`x^3-9x^2+28x-30`, given that 3+i is a Zero.

Answer 1

The zeros are `x = 3+i, 3-i, and 3`

Explanation:

If `3+i` is a zero, then, because complex zeros always exist with a conjugate pair, then `3-i` must be a zero and `(x-3-i)` and `(x-3+i)` must be factors.

Multiply these two factors:

`(x-3-i)(x-3+i) = x(x-3+i) -3(x-3+i) - i(x-3+i)`

`(x-3-i)(x-3+i) = x^2-3x+ix -3x+9-3i - ix+3i-i^2`

`(x-3-i)(x-3+i) = x^2-3x -3x+9-i^2`

`(x-3-i)(x-3+i) = x^2-6x+10`

The above trinomial must evenly divide the original polynomial into a binomial. We do not need to perform the long division; we only need to observe that `x^3/x^2 = x` and `-30/10 = -3`, therefore:

`(x^3-9x^2+28x-30)/(x^2-6x+10) = x-3`

The last zero occurs when the above binomial is equal to zero:

`x - 3 = 0`

`x = 3`

In summary, the zeros are `x = 3+i, 3-i, and 3`