How can I factor these polynomials?

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The discussion focuses on the process of factoring polynomials, specifically finding the greatest common factor (GCF) among terms. Participants explain that to find the GCF, one must identify the smallest power of each variable and break down numbers into their prime factors. Examples provided include finding the GCF of monomials and polynomials, emphasizing that the method remains consistent across different types of expressions. The conversation highlights that while specific examples can be addressed, a comprehensive understanding of polynomial factorization is beyond the scope of the forum. The thread concludes with a reminder that the book contains worked examples for further study.
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I'm in desperate need of help with factoring. Basically, how do you do it?

Below is an example of what I'm up against.

54c^2d^5e^3; 81d^3e^2

It wants me to find the greatest common factor.

http://www.rempub.com/80-activities-to-make-basic-algebra-easier

That is the book I'm working out of.
 
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Notice that d and e are factors in common?
BTW: just checking - is that an actual example from the book or one you made up?
 
a^3=a\times a^2
So if you wanted to find the greatest common factor of a^2; a^3 then it would be a^2. Basically, if you have any a^n for n greater than 2, then the common factor will always be the one with the smallest power index, in this case 2.

m^3n^5; n^7m^2
In this case, the smallest index of the m's is 2, and the smallest of the n's is 5, hence the greatest common factor is m^2n^5.

For numbers, you need to break them up into their prime factors. For example,
12=4\times 3 = 2^2\times 3
and
90 = 9\times 10 = 3^2\times 2\times 5

This is the same problem as earlier with the variables, but now we just have numbers where variables would've otherwise been. It's the exact same procedure though, as long as you break the number into its prime factors.
So to find the great common factor of 12 and 90, we notice that the highest power of the 2's is 1 (hence just 2^1=2), the highest power of the 3's is 1, and 5 is only present in one of the numbers so no 5 common factor. Hence the answer is 2^1\times 3^1=2\times 3 = 6.
 
Simon Bridge said:
Notice that d and e are factors in common?
BTW: just checking - is that an actual example from the book or one you made up?
It's straight from the book.
 
How do I factor monomials, binomials, trinomials and quadratic equations?

Here are some more examples from the book.

24x^3y^2-20x^2y^2+16xy^2

x^2+5x+6

3y^2+8y+4
 
cjackson said:
How do I factor monomials, binomials, trinomials and quadratic equations?
General factorization of polynomials is too involved for us to attempt to answer it here in an online forum. Your book must show worked examples of each type. If you have a question about a particular step in a single factorization, we can help you with that, but our role here is not to teach you a broad topic in algebra or other area of mathematics.

Thread closed.
cjackson said:
Here are some more examples from the book.

24x^3y^2-20x^2y^2+16xy^2

x^2+5x+6

3y^2+8y+4
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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