Find the prime factorization for the integer

121

this is what I think as far as I go
11x11
am I right?

Right, but we usually write it as 11².

Simplify the expression. Assume all variables represent

nonzero real numbers.

-10b^5c^9/2b^5c^3

Since it is a fraction, we look for like factors in the numerator and denominator and simplify accordingly.

Start with the coefficients:
(-10)/2 = -5
b5/b5 = 1
c9/c3 = c9-3 = c6
So putting it altogether:
-10b^5c^9/2b^5c^3
= -5c6

Yes, you are correct! The prime factorization of 121 is indeed 11 * 11.

To find the prime factorization of an integer, you can start by dividing it by the smallest prime number, which is 2. If the number is divisible by 2, continue dividing it by 2 until it is no longer divisible. Then move on to the next prime number, which is 3, and repeat the process until the number cannot be divided any further.

In the case of 121, 2 is not a factor of 121, so we move on to 3. 121 is also not divisible by 3. Next, we try 5, but again, 121 is not divisible by 5. Finally, we try 7, but 121 is not divisible by 7 either.

So, the remaining prime factors are 11 and 11, which gives us the prime factorization of 121 as 11 * 11.