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

1 Answer
Ratnaker Mehta
Jun 12, 2016

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#.

Related questions
See all questions in Zeros
Impact of this question
2000 views around the world
You can reuse this answer
Creative Commons License