Polynomial: $x^3 + 6yx^2 + 12xy^2 + 8y^3$
The simplest factorization is $(x+2y)^3$. How does one factor it into this form?
2026-03-29 13:51:23.1774792283
On
On
On
Factor $x^3 + 6 y x^2 + 12 x y^2 + 8y^3$
269 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
5
There are 5 best solutions below
0
On
Alt. hint: using the homogeneity of the expression, look for factors of $\,z^3 + 6z^2+12z+8\,$ where $\,z = x/y\,$. The rational root theorem finds them fairly quickly.
0
On
Because this is a cubic and has leading term $x^3$, try the form $(x + a y)^3$, expand and solve for $a$ and find $a = 2$.
0
On
If I had to simplify this, I'd start by trying the easier problem you get when you plug in $y=1$. So I'd try to factor $x^3 +6x^2+12x+8$. By the rational root theorem, any rational root has to divide $8$, so I try to divide the polynomial by $x-a$ where $a = \pm 1, \pm2, \pm4, \pm 8$. I would discover that $x+2$ divides the polynomial and so it factors as $(x+2)(x^2+4x+4)$. Then I could factor the last quadratic to get $(x+2)(x+2)(x+2)$. Then it would be obvious that I just need to stick the $y$ back in. $(x+2y)^3$.
If you already know the answer to the "simplification", it's sometimes easier to start there and work your way back. Observe, \begin{align} (x + 2y)^3 &= (x + 2y)(x + 2y)(x + 2y)\\ &= (x^2 + 4xy + 4y^2)(x + 2y)\\ &= x^3 + 2x^2 y + 4x^2 y + 8x y^2 + 4y^2 x + 8y^3\\ &= x^3 + 6x^2 y + 12xy^2 + 8y^3 \end{align} So expanding it to verify is tedious, but it's easy. Going backwards, however, is pretty challenging. It's always easy to expand just polynomials, but factoring will be an eternal struggle. In fact, this idea of a mathematical operation being easy to carry out one way, but hard to reverse is the basis of many security systems.. As you gain more experience, you'll know what to guess.
If you know the biomial theorem, you might see the expanded form and recognize the 3rd row of Pascal's triangle \begin{matrix} &&&&&1\\ &&&&1&&1\\ &&&1&&2&&1\\ &&1&&3&&3&&1\\ \end{matrix} Except it's been multiplied by $2^0, 2^1, 2^2$, and $2^3$ to get $1, 6, 12, 8$ instead of $1, 3, 3, 1$. This might lead you to guess that the factored form of $x^3 + 6x^2y + 12xy^3 + 8y^3$ is $(x + 2y)^3$. Again, there's no "formula" that can really help you guess this apart from the Binomial Theorem and a whole lot of experience.