# Thread: Resolution of polynomial 2

1. Factor $30x^4-41x^3y-129x^2y^2+100xy^3+150y^4$.
Please help me get started. I tried grouping the terms but still cant see any factorization that is familiar to me.

Thanks.

2. Originally Posted by paulmdrdo
Factor $30x^4-41x^3y-129x^2y^2+100xy^3+150y^4$.
Consider $$P_y(x)=30x^4+(-41y)x^3+(-129y^2)x^2+(100y^3)x+150y^4$$ as a polynomial in $x$ and $y$ as a parameter Suppose there are roots of the form $x=ky$. Then, $$P_y(ky)=\left(30k^4-41k^3-129k^2+100k+150\right)y^4=0.$$ Consider the equation $$30k^4-41k^3-129k^2+100k+150=0.$$ Using the theorem of the rational roots we get $k=-3/2$ and $k=5/3$, so $x=-(3/2)y$ and $x=(5/3)y$ are roots of $P_y(x)$ for all $y$, so: $$P_y(x)=\left(x+\frac{3}{2}y\right)\left(x-\frac{5}{3}y\right)q(x,y).$$ Could you continue?

I'm still not sure how to continue from there. Do I need to use rational roots theorem?

And What do you call that method? Does this always work on the polynomials that has the same form I posted?

4. Originally Posted by paulmdrdo
I'm still not sure how to continue from there. Do I need to use rational roots theorem?
Using the Ruffini Algorithm for the root $x=-\dfrac{3}{2}y$
$$\begin{array}{r|rrrrr} & 30 & -41y & -129y^2 & 100y^3 & 150y^4 \\ -\dfrac{3}{2}y & & -45y & 129y^2 & 0 & -150y^4 \\ \hline & 30 & -86y & 0 & 100y^3 & 0\end{array}$$
This implies $$p(x,y)=\left(x+\frac{3}{2}y\right)(\underbrace{30x^3-86yx^2+100y^3}_{Q(x,y)}).$$
Now, decompose $Q(x,y)$ in the same way with the root $x=\dfrac{5}{3}y$.

Originally Posted by paulmdrdo
And What do you call that method? Does this always work on the polynomials that has the same form I posted?
No name, this is a strategy in this particular case, based in the fact that the polynomial is homogeneous, testing linear factors, and all of this with the "hope" that the equation in $k$ has rational roots.

5. I,ve decided to complete the solution:

Applying the Ruffini algorithm to $Q_1(x,y)$: $$\begin{array}{r|rrrrr} & 30 & -86y & 0 & 100y^3 \\ \dfrac{5}{3}y & & 50y & -60y^2 & -100y^3 \\ \hline & 30 & -36y & -60y^2 & 0 \end{array}$$ $$\Rightarrow P(x,y)=\left(x+\frac{3}{2}y\right)(\underbrace{30x^3-86yx^2+100y^3}_{Q_1(x,y)})$$ $$=\left(x+\frac{3}{2}y\right)\left(x-\frac{5}{3}y\right)\left(30x^2-36yx-60y^2\right)$$ $$=\dfrac{1}{6}(2x+3y)(3x-5y)\left(30x^2-36yx-60y^2\right)$$ $$=(2x+3y)(3x-5y)\left(5x^2-6yx-10y^2\right).$$
Let us now factorice $Q_2(x,y)=5x^2-6yx-10y^2$. Solving the quadratic equation on $x$: $$5x^2-6yx-10y^2=0,\quad x=\dfrac{6y\pm \sqrt{236y^2}}{10}=\dfrac{6y\pm 2\sqrt{59}y}{10},$$ $$x=\dfrac{3+\sqrt{59}}{5}y,\quad x=\dfrac{3-\sqrt{59}}{5}y$$ So, we can write:
$$\boxed{\; P(x,y)=5(2x+3y)(3x-5y)\left(x-\dfrac{3+\sqrt{59}}{5}y\right)\left(x-\dfrac{3-\sqrt{59}}{5}y\right)\;}$$

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