P(x)=x^4-6x^3+7x^2-6x+6
and given that i is a zero, use THIS zero to determine the exact(not decimal) but simplified remaining zeros?

Question

P(x)=x^4-6x^3+7x^2-6x+6
and given that i is a zero, use THIS zero to determine the exact(not decimal) but simplified remaining zeros?

anonymous · Answer

Since they tell you i is one of the zeros, and all the coefficients of the polynomial are real, the complex conjugate of i (which is -i)  must be one of the other roots.  So we know that:$$(x-i)(x+i)=(x^2+1)$$is a factor of that polynomial.  If you can do the division:$$\frac{x^4-6x^3+7x^2-6x+6}{x^2+1}$$your answer should be of degree 2, a quadratic.  From there, use the quadratic formula to get the remaining two zeros.

directrix · Answer

Continuing from above, the division will give a quotient of (x^2 - 6x + 6) which should be set to 0 as the root hunt continues.

(x^2 - 6x + 6) = 0 appears not to factor so run the Quadratic Formula on it. You'll find the other two roots to go with i and -i.

anonymous · Answer

GRRRREAT.... thank you guys very much... :D

anonymous · Answer

bith of those bring me to a decimal, do i just round it?

directrix · Answer

@vanessam --> I myself have an affinity for the exact quantities. So, I would go with i, -i, 3 - √3, and 3 + √3. Look at the instructions for the problem to determine if you should approximate.

anonymous · Answer

yay.. thanxxxx :)))))

directrix · Answer

We're happy to help.