College Algebra Review Chapter.

H

Hollis

New member
Joined
Nov 19, 2007
Messages
12
Which of the following polynomials is prime when factoring over the integers?
A: x^3+1
B: x^3-1
C: x^2-1
D: x^2+1
E: x^4-1

I know that the answer is not E, and I do not believe it is C. I tried working them out individually, but this is one of the last problems and I think I'm just frustrated and making stupid mistakes.
 


You're right. (E) is of the form known as a difference of squares; it's one of the special factoring patterns that comes up often in math.

The following are only three of the basic special factoring patterns that we should know.

A Difference of Squares: (A^2 - B^2) = (A + B)(A - B)

A Difference of Cubes: (A^3 - B^3) = (A - B)(A^2 + AB + B^2)

A Sum of Cubes: (A^3 + B^3) = (A + B)(A^2 - AB + B^2)

A Sum of Squares (A^2 + B^2) is Prime. In the Real number system, a sum of squares cannot be factored.

(A) x^3 + 1 is a sum of cubes because it equals x^3 + 1^3 .

(B) x^3 - 1 is a difference of cubes because it equals x^3 - 1^3 .

(C) x^2 - 1 is a difference of squares because it equals x^2 - 1^2 .

(D) x^2 + 1 is a sum of squares because it equals x^2 + 1^2 .

(E) x^4 - 1 is a diference of squares because it equals (x^2)^2 - 1^2 .

8-)

 
Hollis said:
Which of the following polynomials is prime when factoring over the integers?
A: x^3+1
B: x^3-1
C: x^2-1
D: x^2+1
E: x^4-1

I know that the answer is not E, and I do not believe it is C. I tried working them out individually, but this is one of the last problems and I think I'm just frustrated and making stupid mistakes.
Click to expand...

I don't understand the problem. Is it supposed to be for all 'x'? What if x is odd

x = (2n+1)

x[sup:252dvaud]2[/sup:252dvaud] = (2n+1)[sup:252dvaud]2[/sup:252dvaud] = 4n[sup:252dvaud]2[/sup:252dvaud] + 4n + 1

x[sup:252dvaud]2[/sup:252dvaud] + 1 = 2 * (2n[sup:252dvaud]2[/sup:252dvaud] + 2n + 1)

None of those are prime for arbitrary 'x'.
 
A prime element of a unique factorization domain (UFD) is an element that cannot be written as a product of two elements of the ring except for itself times the ring's unity (or its additive inverse).

He is not referring to the ring of integers, but instead to the ring Z[x] where "numbers" are polynomials.

So, just like 3 is prime in Z, x^2-2 is prime in Z[x], but not in R[x].
 
You must log in or register to reply here.
Top