SOLUTION: Suppose that a polynomial function P(x) of degree 3 with rational coefficiants has 2 and -3i as zeroes. what are the following a. find the other zeroes b. write P(x) as a product

Algebra ->  Rational-functions -> SOLUTION: Suppose that a polynomial function P(x) of degree 3 with rational coefficiants has 2 and -3i as zeroes. what are the following a. find the other zeroes b. write P(x) as a product      Log On


   

Question 858768: Suppose that a polynomial function P(x) of degree 3 with rational coefficiants has 2 and -3i as zeroes. what are the following
a. find the other zeroes
b. write P(x) as a product of its linear factors
c. write P(x) in polynomial form
Found 3 solutions by Fombitz, josgarithmetic, solver91311:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since -3i is a root then so is 3i
a) 2, 3i, -3i
b) P%28x%29=%28x-2%29%28x-3i%29%28x%2B3i%29
c) P%28x%29=%28x-2%29%28x%5E2%2B9%29
P%28x%29=x%5E3%2B9x-2x%5E2-18
P%28x%29=x%5E3-2x%5E2%2B9x-18

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Complex roots with imaginary components come as conjugate pairs.
The roots include 3i.

P%28x%29=%28x-2%29%28x-%283i%29%29%28x-%28-3i%29%29
%28x-2%29%28x-3i%29%28x%2B3i%29-------Factored form with all linear factors.
%28x-2%29%28x%5E2-%283i%29%5E2%29
%28x-2%29%28x%5E2-%28-1%29%289%29%29
%28x-2%29%28x%5E2%2B9%29-------Factored form with Real coefficients and constant terms.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Complex zeros always come in conjugate pairs, that is, if is a zero of a polynomial then is also a zero of that polynomial.

Since your given zeros are 2 and (which is to say ), the third and final zero must be , or simply

If is a zero of a polynomial, then is a linear factor of the polynomial. Hence the linear factors of the desired polynomial are , or more simply:

You can multiply the factors to derive the standard form polynomial yourself. Hint: do the two complex factors first, remembering that the product of a pair of conjugates is the difference of two squares and that

John

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