I wouldn’t use the term polynomials for those expressions, but what you do is factor them to the simplest factors and see which of those they have in common.

54c^2d^5e^3 = 2333ccdddddee*e

81d^3e^2 = 3333dddee

I wouldn’t normaly expand the exponentials, but I don’t know where you’re struggling.

For any of these factors see if the other number has a corresponding factor and keep only those. 333ddde*e = 3^3 d^3 e^2

And that’s the greatest common factor.

The reason is that there’s just one unique way to factor a number into primes (and unknowns in this case).

Suppose you’re told to find the greatest common factor of 3,696,694,848 and 88,058,880.

Now, you start dividing by small primes and discover that 3,696,694,848 happens to equal (2^6) times (3^7) times (7^4) times (11^1) and 88,058,880 equals (2^10) times (3^3) times (5^1) times (7^2) times (13^1).

Each of those numbers is divisible by 2^6. Each is divisible by 3^3. Each is divisible by 7^2. (2^6) times (2^3) times (7^2) equals 84,672. That’s the greatest common factor.

You do the same thing in your problem, except that you can use variables for factors in addition to primes. Each of the two terms is divisible by 3^3, d^3, and e^2. So the greatest common factor is (3^3) times (d^3) times (e^2).

I would say the two things that they asked you to find the greatest common factor of are terms rather than polynomials. Yeah, they are also polynomials, but they are polynomials with just one term. Calling them polynomials made us think that this problem was about factoring polynomials, and that’s a more complicated problem.

Let us assume that we understand c,d,e as independent variables, so you are in the ring of polynomials in 3 variables. This has unique prime factorization, but the primes are very complicated. The three variables are prime, but so are such expressions as c + d, 2c + e, etc. And integer primes are also prime here. However, any divisor of a monomial is monomial (because you cannot multiply a non-monomial by anything and get a monomial). Now suppose you were asked to find the gcd (greatest common divisor) of (545^2)(7^5)(11^3) and (817^3)(11^2). This is actually 23^35^27^511^3 and 3^47^311^2, which is obviously 3^37^3*11^2. Similarly, the answer to your question is 27d^3c^2. The point is that primes are primes and c,d,e act just like 5,7,11.

It helps to insert * in the numerical example, but true exponential notation would help more.

The proof that polynomial rings have unique prime factorization is well less than obvious and was one of David Hilbert’s early successes.

You left us to assume that c,d,e are all equal to or greater than 1,

Also they aren’t factors of 54 or 81 …

the common factor can be divided from both sides.

Factors of 54, 6 * 9, well thats 233*3

Factors of 81, 9 * 9 = 333*3

So you can take the 333*3 out, of both expressions leaving 2 on the left ,3 on the right .
Can’t divide any c’s from the right hand side, so the c^2 stays at the left.
divide d^3 from both sides
divide e^2 from both sides.

So the common factor is 27d^3e^2,
Well, if d and e were equal, or they were 2 or 3, or multiples of 2 or 3, then there could be a larger common factor, so we have to assume that d and e do not have such similarities.

For the GCF of numbers, there is a second method I teach my students, in addition to the prime factorization method. It’s rather haphazard, but my students seem to grasp it easily.

Let’s say we want the GCF of 36 and 96. The largest possibility is the smallest of the numbers, 36. But 36 doesn’t divide into 96 (without a remainder), so 36 isn’t the GCF.

The next biggest possibility is 36/2=18. But 18 doesn’t divide into 96 either, so no dice.

The next biggest possibility is 36/3=12. 12 does divide into 96. Found it! The GCF of 36 and 96 is 12.

I wouldn’t suggest trying this on huge numbers like in Wendell’s example, but for smallish numbers this method can be easily done in your head.

Factoring and finding greatest common divisors are not the same problem. There is no need to factor in order to find common divisors.

For integers, there’s good old Euclid’s Algorithm. Very simple and efficient. So good, it’s the standard beginner’s algorithm for 2300 years. (And if you’re not teaching Euclid’s algorithm, you’re doing it wrong.)

For single variable polynomials, it easily extends to GCF of polynomials using synthetic division. But, with multi-variate polynomials, there’s some more stuff going on. Start here.

Do not think of or try to factor. That is a horrible, slow, frequently impossible task.

ftg, if you’re addressing the OP, cjackson specifically asked about factoring: that’s what he needs to know how to do. See his later post (#6) for clarification of this.

In some cases (such as the first example given in that post) factoring a polynomial involves finding the greatest common factor/divisor of its terms. But, as long as the numbers involved are fairly small (as they are in that example), I think it’s easier to find the GCF by inspection rather than by any systematic algorithm.

When I was about 40, a community college math teacher showed me a method for factoring trinomials that I had never seen before, which certainly wasn’t what I ever saw in any algebra text or class. It’s called the AC method. Has anyone else here ever seen this?

Take this example, from OP in Post #6:
3y[sup]2[/sup] + 8y + 4
First, we note that it’s much easier to read if you use real superscripts for exponents, and add some white space.

To start, multiply the a and c numbers: 3 * 4 = 12 (This is where the name AC method comes from.)

Next, factor that completely:
1 | 12
2 | 6
3 | 4

Next, scan down that list of factor pairs, hoping to find a pair whose sum is the middle term:
1 | 12 2 | 6 <— !!!
3 | 4

Found one! (If no such pair, then the trinomial doesn’t factor.)

Next re-write the original trinomial, splitting the middle term, using the two coefficients thus found:
3y[sup]2[/sup] + 6y + 2y + 4

Okay, that looks like a lot of steps. But it’s less work than the way we were traditionally taught (which involves factoring the a and c numbers separately, and trying all combinations of those until you get the middle term right). This method has less trial-and-error to it.

For the second and third, the quadratic formula is your friend and I will leave you to learn and use it. The first is actually 4xy^2(6x^2 - 5x + 4) and again the quadratic formula gives the (complex) factors of the quadratic.

The thing to recall is that there is a close connection between factors of a polynomial and its roots. There are (rather complicated) formulas for finding roots of polynomials of degrees 3 and 4, but there are not, nor can there be, formulas for any higher degree.