Factor x^4 + 2x^3 y + 3x^2 y^2 + 2x y^3 + y^4.

Step-by-step solution

1. 1

   Read the coefficient pattern. Listing the coefficients in order gives

   $1,\;2,\;3,\;2,\;1$

   a palindrome, and the expression is homogeneous of degree $4$ — every term has total degree $4$. A degree-$4$ homogeneous palindrome is a strong hint that it is the square of a degree-$2$ homogeneous expression.

2. 2

   Set up the square of a general trinomial. Try $\left(x^2+kxy+y^2\right)^2$ and expand using $(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$ with $a=x^2$, $b=kxy$, $c=y^2$:

   $\left(x^2+kxy+y^2\right)^2=x^4+2kx^3y+\left(k^2+2\right)x^2y^2+2kxy^3+y^4$

3. 3

   Match coefficients. Comparing with the target:

   $2k=2\;\Longrightarrow\;k=1,\qquad k^2+2=1+2=3\;\checkmark$

   Both conditions are satisfied by the same $k$, which is exactly why the factorization exists. Had the $x^2y^2$ coefficient been anything but $3$, no single $k$ would have worked.

4. 4

   Write the factorization.

   $x^4+2x^3y+3x^2y^2+2xy^3+y^4=\left(x^2+xy+y^2\right)^2$

5. 5

   Confirm numerically and note irreducibility. At $x=2$, $y=1$: the original is $16+16+12+4+1=49$, and $\left(4+2+1\right)^2=7^2=49\;\checkmark$. The inner factor $x^2+xy+y^2$ does not factor further over the reals — as a quadratic in $x$ its discriminant is $y^2-4y^2=-3y^2<0$ — so this is the complete factorization.