1. ## Polynomials...

I just can't seem to get my head around this problem...

Suppose that a polynomial function of degree 4 with rational coefficiets has the given numbers as zeros. Find the othe zero(s).

17. -1, (the square root of)3, 11/3

This one has really got me stumped and the book is of no help.

Thanks for any help anyone can grant me...

2. Originally Posted by The Sky Patrol
I just can't seem to get my head around this problem...

Suppose that a polynomial function of degree 4 with rational coefficiets has the given numbers as zeros. Find the othe zero(s).
17. -1, (the square root of)3, 11/3

This one has really got me stumped and the book is of no help.

Thanks for any help anyone can grant me...
Hello,

if a olynomial function has zeros it can be described by liear factors containing the zeros. So your function can be written as:
$f(x)=(x+1)(x-\sqrt{3})(x-\frac{11}{3})(x-a)$

Because you said that the coefficient must be rational you have to look for a zero which will transform the root value into a rational number. This will happen only if $a = -\sqrt{3}$, because $(x-\sqrt{3})(x+\sqrt{3})=x^2-3$

Your function is therefore: $f(x)=(x+1)(x-\sqrt{3})(x-\frac{11}{3})(x+\sqrt{3})$
$f(x)=x^4-\frac{8}{3} x^3-\frac{29}{3}x^2+8x+11$

Greetings

EB

3. Got it!

Thanks for the fast and enlightening response.

Much, much appreciated!!!

4. I hate to be a bother, but here is another problem that bugged me in the homework.

Given that the polynomial function has the given zero, find the other zeros.

45. f(x) = x^4 - 5x^3 + 7x^2 - 5x + 6; -i

I am using the synthetic division method, but am having trouble because it includes the -i as the zero.

5. Yeah, I guess I wasn't sure which forum to post in. I chose College Algebra because it is a college Algebra class. Sorry... but thanks for moving my thread to it's appropriate place.

6. Originally Posted by The Sky Patrol
Yeah, I guess I wasn't sure which forum to post in. I chose College Algebra because it is a college Algebra class. Sorry... but thanks for moving my thread to it's appropriate place.
No this is called 'elementary algebra'. The algebra posted in that forum is totally different from the one you know called 'abstract algebra'.