SOLUTION: If x+y=a and xy=b, find the value of {{{x^2y+xy^2+x^2+y^2}}} I know that you have to square root {{{x^2+y^2}}} and {{{xy^2}}} to find b and a, but I'm not really sure what else

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: If x+y=a and xy=b, find the value of {{{x^2y+xy^2+x^2+y^2}}} I know that you have to square root {{{x^2+y^2}}} and {{{xy^2}}} to find b and a, but I'm not really sure what else       Log On


   



Question 888735: If x+y=a and xy=b, find the value of x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2
I know that you have to square root x%5E2%2By%5E2 and xy%5E2 to find b and a, but I'm not really sure what else to do. Thanks!

Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2=xy%28x%2By%29%2Bx%5E2%2By%5E2
x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2=xy%28x%2By%29%2B%28x%5E2%2B2xy%2By%5E2%29-2xy
x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2=xy%28x%2By%29%2B%28x%2By%29%5E2-2xy
Now substitute,
x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2=b%28a%29%2Ba%5E2-2b
highlight%28x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2=a%5E2%2Bab-2b%29

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
If x+y=a and xy=b, find the value of x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2
We have to get that expression in terms of a and b,
Since a is the sum of x and y and b is the product of x and y,
we look for ways to get the sum a=x+y and the product b=xy.

Looking at this:

x%5E2y%2Bxy%5E2%2Bx%5E2%2By%5E2

We notice that the first two terms factor as xy(x+y)

xy%28x%2By%29%2Bx%5E2%2By%5E2

which immediately becomes ba 

ba%2Bx%5E2%2By%5E2

We look at the last two terms.

Since %28x%2By%29%5E2=x%5E2%2B2xy%2By%5E2, we notice that those last two 
terms will be %28x%2By%29%5E2 if we add 2xy between them, but 
if we do that we'll have to subtract 2xy to offset, so we 
do that:

ba%2Bx%5E2%2B2xy%2By%5E2-2xy

Factor the middle three terms as %28x%2By%29%5E2.

ba%2B%28x%2By%29%5E2-2xy

Finally, we substitute a for x+y and b for xy, and we end up with

ba%2Ba%5E2-2b

Edwin