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# *Clever* way to factor (x^5 + y^5) = (x +y) * f(x,y) without long division. That is...

shoshanna6e23 | eNotes Newbie

Posted April 1, 2010 at 1:31 AM via web

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*Clever* way to factor (x^5 + y^5) = (x +y) * f(x,y) without long division. That is determine f(x.y) using "out of the box" thinking.

My daughter e-mailed me this one from college. Here is what I get: both (x^5 + y^5) & (x+y) are symmetric in x & y therefore f(x.y) must be as well.

Since both x & y are first degree and the answer is fifth degree then we can confine ourselves to f(x,y) of 4th degree only - is that a correct assumption?????

f(x,y) must look like: (x^4 +ax^3y + bx^2y^2 + axy^3 + y^4) to maintain the symmetry f(x.y) = f(y,x)

a little elementary analysis shows a=-b but no constraints on the answer a or b.

a little poking around shows:

(x^4  -x^3y + bx^2y^2 - axy^3 + y^4) works.

but poking around is not "clever" enough.

Is there another constraint I can impose like sum of coefficients or something?

neela | High School Teacher | Valedictorian

Posted April 1, 2010 at 2:23 AM (Answer #1)

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We know that:

(x+y)/(x+y = 1

(x^3+y^3)/(x+y) = x^2-xy+y^2.

(x^5+y5)/(x+y) = x^4-x^3y+x^2y^2-x^1y^3+y^4 is a homogeneous equation of 4 th degree.

Similarly

((x^2n+1)+y(2n+1))/(x+y) = x^(2n)-x^(2n-1)y+x^(2n-2)y^2+.....+y^2n = f(x,y)

So,

((x^(2n+1)+y^(2n+1)) = (x+y)f(x,y), where f(x,y) is a homogeneous function of 2n degree  and the coefficient of  x^r*(y^(2-r) is (-1)^r

giorgiana1976 | College Teacher | Valedictorian

Posted April 1, 2010 at 5:59 PM (Answer #2)

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Hi!

f(x,y)= (x^4 +ax^3*y + bx^2*y^2 + ax*y^3 + y^4) is perfect.

All it has to be done is to multiply (x +y) * f(x,y).

The result, after multiplying is:

x^5+y^5+y*x^4(a+1)+y^2*x^3(a+b)+y^3*x^2(c+b)+y^4*x(c+1)=x^5 + y^5

Now it comes the important part!

In order to respect the identity, meaning x^5 + y^5=(x +y) * f(x,y), you have to impose the constraint that the corresponding coefficients of each term from both sides, to be equal!

As you can see, we have x^5 + y^5 and no other terms like y*x^4, y^2*x^3 and so on. But it doesn't mean that they do not exist, it means that their coefficients are zero!

So, you'll obtain the system:

a+1=0, it results that a=-1

a+b=0,  it results that a=-b, but a=-1, so b=1

c+b=0, it results that c=-b, but b=1, so c=-1

c+1=0, c=-1, true!

f=x^5+y^5+0*y*x^4+0*y^2*x^3+0*y^3*x^2+0*y^4*x

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