How do you factor these kinds of equations?

2008-03-09 3:54 pm
I'm sorry I know this is my second question but my sister isn't at home to help me with Math because I really don't get it. Can someone explain to me? If only you have time...

117x³y + 21x²y + 57xy?

Sorry for taking up your time, but I really need to learn this and I can't understand it..
更新1:

Umm, that's a 3 and a 2, if it's too small..

回答 (7)

2008-03-11 8:08 pm
✔ 最佳答案
(3 x y) ( 39 x ² + 7 x + 19 )
2008-03-09 11:01 pm
take out the common number from these: maybe 3?

3xy(39x^2 + 7x + 19)

ther u go...("^" indicates to the power of)

you take out the common number first, then look for the similar x's and y's like ive done and u shud be rite, sorry, good luck
參考: experience
2008-03-09 11:00 pm
I think it would be

xy(117x²+21x+57)

You have to find the common factor, which in this case is xy. The common factor is the largest number you can pull out of each term. Then you times is by what is left in each term.
2008-03-09 11:37 pm
117x^3y + 21x^2y + 57xy
= 3xy(39x^2 + 7x + 19)
2008-03-09 11:23 pm
117x³y + 21x²y + 57xy= 3xy(39x²+7x+19)
2008-03-09 11:14 pm
This is an expression, not an equation. We are usually asked to "solve" equations but "factor" expressions (if possible). You need to refer back to the section in your textbook which discusses Greatest Common Factor. It would also be beneficial to know exactly what is meant by the word "factor."

Each term in your expression appears to have a common factor of 3, x and y. We can write the following which may help us see something interesting.

117x³y + 21x²y + 57xy

= 39x²•3xy + 7x•3xy + 19•3xy

Since a factor is one of two or more "things" multiplied together, then 3xy is a common factor to all three terms. Actually, 3xy is the largest factor that is common to all three terms. Therefore, 3xy is the largest common factor. Oops, I meant Greatest Common Factor.

117x³y + 21x²y + 57xy

= 3xy(39x² + 7x + 19)

Try working some examples with less difficulty than this one. Once you "see" what I am talking about with the simple problems, factoring GCF's becomes very simple even on the longer expressions.
2008-03-09 11:10 pm
In my honest opinion, I hate math now that college is done, but as for your question hun, I would have to say that you need to
117x^3 y + 21x^ 2 y + 57xy =
Seeing that you have X^3 X^2 X and y y y ,it would be like looking at it like this
x + x + x+ x + x + x= X^6
y + y + y = Y^3
So then 117 + 21 + 57 = 195
Don't forget to readd the letters back in so it would look like this 117x^3 y + 21x^ 2 y + 57xy = 195x^6 Y^3

If I am wrong please forgive me and re ask your sister


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