Please help me figure this out and pls explain to me how to do this I’m so confused
1 answer:
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Suppose that the polynomial x^2+y^2-1 is not irreducible.
It means that there exists two fractions such that :
We found that q divide t and q divide s, but s and t are supposed to be prime each other. It is a contradiction.
The polynomial x^2+y^2-1 is then irreducible in q[x,y]
It means that there exists two fractions such that :
We found that q divide t and q divide s, but s and t are supposed to be prime each other. It is a contradiction.
The polynomial x^2+y^2-1 is then irreducible in q[x,y]
Answer:
x = 2
and
x = 3
Step-by-step explanation:
The function given is:
In this fraction, the bottom part is the denominator. We need to set it equal to 0 and solve for 2 values of x.
This is middle term factorization process. Hence the 2 solutions (roots) of the denominator is:
x = 2, 3