Quaint quintic

$x+y|f(x,y)\\ \implies g(x,y)=(x+y)\\\implies h(x,y)=\frac{f(x+y)}{g(x+y)}\\ g(2,2)=4, f(2,2) = 64\implies h(2,2) = 16\\\implies |g(2,2)|+|h(2,2)|=\boxed{20}$

Consider $f(x) = x^5+3y^2-3x^2+y^5 ~~~~~~~~~ \therefore f(-y)=0$

By the factor theorem, we know that $(x+y)$ is a factor of $f(x)$ ...

Further factorizing $f(x)$ , we get... $f(x,y)=(x+y) (x^4-x^3 y+x^2 y^2-x y^3+y^4-3x+3 y)$

Since, this factorization is irreducible, hence, $g(x,y)$ and $h(x,y)$ must be $(x+y)$ and $(x^4-x^3 y+x^2 y^2-x y^3+y^4-3x+3 y)$ in some order. The order doesn't matter since the problem requires the answer for same values of $x$ and $y$ . Simply plugging in $x=2$ and $y=2$ in both equation, we get...

$|g(2,2)|=|4|=4$ and $|h(2,2)|=|16|=16$

Therefore, $|g(2,2)|+|h(2,2)|=4+16=\fbox{20}$

De $x^n + y^n = (x + y)(x^{n - 1} \cdot y^0 - x^{n - 2} \cdot y^1 + x^{n - 3} \cdot y^3 - ... - x^1 \cdot y^{n - 2} + x^0 \cdot y^{n - 1})$ ,

Temos que $\boxed{x^5 + y^5 = (x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4)}$

Segue que,

$f(x, y) = y^5 + x^5 + 3y^2 - 3x^2 \\\\ f(x, y) = (y^5 + x^5) + 3(y^2 - x^2) \\\\ f(x, y) = (y + x)(y^4 - y^3x + y^2x^2 - yx^3 + x^4) + 3(y + x)(y - x) \\\\ f(x, y) = (y + x)(y^4 - y^3x + y^2x^2 - yx^3 + x^4 + 3y - 3x) \\\\ \begin{cases} g(x, y) = x + y \\ h(x, y) = y^4 - y^3x + y^2x^2 - yx^3 + x^4 + 3y - 3x \end{cases}$

Por conseguinte,

$g(x, y) = x + y \\ g(2, 2) = 2 + 2 \\ g(2, 2) = 4 \\ \boxed{|g(2, 2)| = 4}$

E,

$h(x, y) = y^4 - y^3x + y^2x^2 - yx^3 + x^4 + 3y - 3x \\ h(2, 2) = 16 - 16 + 16 - 16 + 16 + 6 - 6 \\ h(2, 2) = 16 \\ \boxed{|h(2, 2)| = 16}$

Por fim, $\boxed{\boxed{|g(2, 2)| + |h(2, 2)| = 20}}$

Factoring out $x^5+y^5$ we have:

$x^5+y^5=(x+y)(x^4-x^{3}y+x^{2}y^2-xy^{3}+y^4)$

$x^5+y^5+3y^2-3x^2 = (x+y)(x^4-x^{3}y+x^{2}y^2-xy^{3}+y^4) + 3(y-x)(x+y)$

Factoring for $(y+x)$

$x^5+y^5+3y^2-3x^2 =(x+y)(x^4-x^{3}y+x^{2}y^2-xy^{3}+y^4+3(y-x)$

Thus we have

$f(x,y)=x+y$ and $g(x,y)=x^4-x^{3}y+x^{2}y^2-xy^{3}+y^4+3(y-x)$

$f(2,2)+g(2,2)=4+16=20$

Notice that we can write

$\begin{aligned} f(x,y) &= x^5 + 3y^2 - 3x^2 + y^5 \\ &= (x^5+y^5) - (3x^2-3y^2) \\ &= (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4) \\ &\qquad \qquad\qquad\qquad - (x+y)(3x-3y) \\ &= (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4+3x-3y) \end{aligned}$

[It remains to prove that the second factor cannot be factorized further.]

Therefore, $g(x,y) = x+y$ and $h(x,y) = x^4-x^3y+x^2y^2-xy^3+y^4+3x-3y,$ so we have $|g(2,2)| + |h(2,2)| = 4 + 16 = \boxed{20}.$

hint : root of f(x,y) is 0,0 then, f(x,y) is divisible by xy. g(x,y) = xy h(x,y) = f(x,y)/g(x,y)

evaluate the question at point (2,2)

Rewrite $f(x,y)$ as

$(x^5 + y^5) + (3y^2 - 3x^2)$

$(x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) + 3(y-x)(y+x)$

$(x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4 + 3y-3x))$

Therefore $g(x,y) = (x+y)$ and $h(x,y) = (x^4 - x^3y + x^2y^2 - xy^3 + y^4 + 3y - 3x)$

And $|g(2,2)| + |h(2,2)| = |(2+2)| + |(2^4 - 2^3*2 + 2^2*2^2 -2*2^3 + 2^4 + 3*2 - 3*2)|$

$= |4| + |16| = 4+16 = \boxed{20}$

Input this into Wolfram Alpha:
* $x^{5} - 3x^{2} + 3y^{2} + y^{5}$ *

In alternative forms, you should see this:

$(x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}-3x+y^{4}+3y)$

Replace x and y with 2, as shown below,

$(2+2)(2^{4}-2^{3}2+2^{2}2^2-2\times2^{3}- 3 \times2+2^{4}+3\times2)$

then solve $|4|+|16-16+16-16-6+16+6|$

Notice the 6's cancel out, as do all but one 16

Now add the remaining numbers $|4|+|16|$

The answer is 20

x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4) 3(y^2-x^2)=3(y-x)(y+x)

on observing the eq., it is clear that x=-y is one sol. hence a factor: g(x,y)=(x+y). now, g(2,2)=4 therefore, h(2,2)=f(2,2)/g(2,2)=16 so that h(2,2)+g(2,2)=20

$f(x,y)=x^5+y^5-3(x^2-y^2)$
$f(x,y)=(x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4) - 3(x+y)(x-y)$
$f(x,y) = (x+y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4 - 3x + 3y)$
Set $g(x,y)=x+y$ and $h(x,y)=x^4 - x^3y + x^2y^2 - xy^3 + y^4 - 3x + 3y$
$g(2,2) = 2+2=4$
$h(2,2) = 2^4 - 2^3\times 2 + 2^2 \times 2^2 - 2 \times 2^3 + 2^4 - 3(2)+3(2)$
$h(2,2) = 16$
$|g(2,2)|+|h(2,2)|=4+16=20$

Note that $f(k,-k) = 0$ , hence $x + y$ is a factor by the factor theorem. By the polynomial division we get $\frac{x^5 + 3y^2 - 3x^2 + y^5}{x + y} = x^4-x^3 y+x^2 y^2-x y^3-3 x+y^4+3 y$ Hence $f(x,y) = (x + y)(x^4-x^3 y+x^2 y^2-x y^3-3 x+y^4+3 y)$ So $|g(2,2)| + |h(2,2)| = \boxed{20}$