Question

How do you find all the zeros of `P(x) = 6x^3 − 13x^2 + 14x − 2` with 1+i as a root?

Answer 1

All the zeros of the given polynomial `P(x)=6*x^3-13*x^2+14*x-2`
are `1/6,1+i,1-i`.

Explanation:

Let us remember that the complex roots occur in conjugate pairs. Now one such root is given to be`1+i`, so, the root has to be `1-i`.

Since `P(x)` is a cubic polynomial (poly.), `P(x)` must have 3 roots, out which 2 roots are now known to us, namely, `1+-i`.

Let the 3rd root be `alpha`.

Now, by Vieta's Rule, `alpha`+`(1+i)`+`(1-i)` = `-(-13/6)` = `13/6.`
`alpha` `(1+i)`+`(1+i)` `(1-i)`+`alpha` `(1-i)` = `14/6.`
`alpha` `(1+i)` `(1-i)`=`-(-2/6)` = `2/6.`

From the 1st equation (eqn.), we get `alpha` = `1/6.`
We verify that this value of `alpha` satisfy the remaining 2 eqns.
Hence the roots `P(x)` are `1/6, 1+i, 1-i`.