SOLUTION: Can someone please help me factor this problem completely? x^10y^3 - 4x^9y^2 - 21x^8y????? Here is what I have as my attempt, but am not sure I am even correct with what I have

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Can someone please help me factor this problem completely? x^10y^3 - 4x^9y^2 - 21x^8y????? Here is what I have as my attempt, but am not sure I am even correct with what I have      Log On


   

Question 101511This question is from textbook Intermediate Algebra
: Can someone please help me factor this problem completely?
x^10y^3 - 4x^9y^2 - 21x^8y?????
Here is what I have as my attempt, but am not sure I am even correct with what I have..
Simplify:
x^10y^3 - 36xy^2 - 168xy
x^10y^3 - 72xy - 168xy
Combine like terms:
72xy + 168xy = 240xy
240xy* x^10*y^3=?
This problem was given to me by my college Professor, but it is not from our textbook. Thanks. Below are additional directions from Professor:
**Make sure to superscript all exponents. If you believe that a polynomial is not factorable, label it as prime. Show all stages of factoring separately (i.e., show the result of factoring out a GCF before taking the resulting polynomial and factoring it again).**
This question is from textbook Intermediate Algebra

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
x%5E10y%5E3+-+4x%5E9y%5E2+-+21x%5E8y
.
First look at the numbers that multiply each of the terms to see if there is are common
factors. The first term is multiplied by 1 (understood), the second term by -4, and the
third term by -21. These three numbers have no common factors, so we don't change them.
.
Next look at the x part of each of the terms. The first term contains x%5E10, the second
term x%5E9, and the third term x%5E8. x%5E8 is common to each term because
x%5E10+=+x%5E2%2Ax%5E8 and x%5E9+=+x%2Ax%5E8. So we can pull an x%5E8 from each term and
we get:
x%5E10y%5E3+-+4x%5E9y%5E2+-+21x%5E8y+=+%28x%5E8%29%2A%28x%5E2%2Ay%5E3+-+4x%2Ay%5E2+-21y%29
.
Next notice that you have a "y" common to all terms. So pull a y out as a multiplier and
you have:
.

.
Notice that we can multiply out %28x%5E8%29%2A%28y%29%28x%5E2y%5E2+-4xy-21%29 and the result should be
the original given expression. This is just a "check".
.
So we are down to %28x%5E8%29%2A%28y%29%28x%5E2y%5E2+-4xy-21%29. We now need to see if we can factor the
x%5E2y%5E2+-+4xy+-21 Yes we can. It might be a little easier to see if we wrote
x%5E2y%5E2+ in an equivalent form of %28xy%29%5E2 to change the expression to:
.
%28xy%29%5E2+-+4xy+-+21 and we might see it a little more clearly if we let xy = A to get:
.
A%5E2+-+4A+-+21
.
We can try to factor this into (A + ___)*(A + ___). The underscores have to be two numbers
that are factors of -21. These numbers could be 21 and 1 or 7 and 3. With their appropriate signs
they must sum to -4. -7 and + 3 do that. If they are multiplied they give -21 and if they are
added they give -4 which is the multiplier of the center term that contains just A.
.
So we can write that A%5E2+-+4A+-+21 factors to %28A+-+7%29%2A%28A%2B+3%29. But also recall
that we said A was equal to xy. So now we can substitute xy into the two factors and get:
.
%28A-7%29%2A%28A%2B3%29+=+%28xy+-+7%29%2A%28xy+%2B+3%29
.
And so we can say that x%5E2y%5E2+-+4xy+-+21 factors to %28xy+-+7%29%2A%28xy%2B3%29 and we can
carry this result back into this equation that we got earlier:
.
%28x%5E8%29%2A%28y%29%28x%5E2y%5E2+-4xy-21%29
.
In this equation, replace %28x%5E2y%5E2+-4xy-21%29 with %28xy+-+7%29%2A%28xy%2B3%29 and the result is:
.
%28x%5E8%29%2A%28y%29%2A%28xy+-+7%29%2A%28xy%2B3%29
.
and that's the answer to your problem.
.
I know this is confusing, but I hope that you can see your way through it. If you can
understand all the maneuvering in this you will have made a lot of progress.