Factoring a given quartic form

Question

Description

Below is our polynomial: $$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4$$ I've gone to almost end of it but,it seems there's a small -or conversely huge- fault in my solving which made $\mathbf{A}\:and\: \mathbf{B}$ imaginary numbers as I didn't find any two numbers which Could have the sum of $-8xy$ and product of $24x^2y^2$ OR in another sense , I can not find $\mathbf{A}\:and\: \mathbf{B}$.

My work

$$\begin{align} & 3x^4-8x^3y+14x^2y^2-8xy^3+3y^4 \\ & = 3(x^4+y^4)-8xy(x^2+y^2)+14x^2y^2 \\ & = 3(x^4+y^4+2x^2y^2)-8xy(x^2+y^2)+8x^2y^2\\ & = 3(x^2+y^2)^2-8xy(x^2+y^2)+8x^2y^2\\ & = \frac{1}{3}\times[3^2(x^2+y^2)^2- 3\times8xy(x^2+y^2)+24x^2y^2] \\ & =\frac{1}{3}(3(x^2+y^2)+\mathbf{A})(3(x^2+y^2)+\mathbf{B}) \end{align}$$

Answer 1

Assume by symmetry

$$(ax^2+bxy+cy^2)(cx^2+bxy+ay^2)=\\=acx^4+(ab+bc)x^3y+(a^2+b^2+c^2)x^2y^2+(ab+bc)xy^3+(ac)y^4$$

then we need

• $ac=3 \implies a=3\, c=1$
• $ab+bc=-8 \implies3b+b=-8\implies b=-2$
• $a^2+b^2+c^2=14 \implies 9+4+1=14$

thus

$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4=(3x^2-2xy+y^2)(x^2-2xy+3y^2)$$