Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$

Question

This problem is from my Math Challenge II Algebra class, and it's really confusing. How can you factor something like this? Here's the question again: Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$.

Answer 1

Basically by educated guesses. The factorization, if it exists, looks like $(ax+by+cz)(dx+ey+fz)$. Concentrate first on the $x^2$, $xy$, and $y^2$ terms and try to factor that. That should give you $a$, $b$, $c$, and $d$. Then look at the $z$ terms and figure out what $c$ and $f$ are.

Answer 2

First, split up the expression into pieces.

\begin{align*} F(x,y,z) &= 3x^2-11xy+6y^2-xz-4yz-2z^2 \\ &= 3x^2\underbrace{-11xy-xz}_{G(x,y,z)}+\underbrace{6y^2-4yz-2z^2}_{H(y,z)} \end{align*}

Start with the easy one : $H(y,z)$. \begin{align*} H(y,z) &= 6y^2-4yz-2z^2 \\ &= 6y^2-6yz+2yz-2z^2 \\ &= 6y(y-z)+2z(y-z) \\ &= 2(y-z)(3y+z) \end{align*}

Now, the tuff one : $G(x,y,z)$. Remember, we want to find some factors like those in $H(x,y)$. So, we have to use our imagination...

\begin{align*} G(x,y,z) &= -11xy-xz \\ &= -3x(3y+z)-2xy+2xz \\ &= -3x(3y+z)-2x(y-z) \\ \end{align*}

Finish, with all the pieces together. \begin{align*} F(x,y,z) &= 3x^2 + G(x,y,z) + H(y,z) \\ &= 3x^2-3x(3y+z)-2x(y-z)+2(y-z)(3y+z) \\ &= (3y+z)[2(y-z)-3x]-x[2(y-z)-3x] \\ &= [2(y-z)-3x][(3y+z)-x] \\ &= (3x-2y+2z)(x-3y-z) \end{align*}

Answer 3

The most systematic way is to complete the squares and then factor as a difference of squares, but I dont have the energy to do that right now, maybe someone else would like to post such a solution.

Another way is treat it like ordinary factorisation, say in variable $x$,

$$3x^2-(11y+z)x+6y^2 -4yz+2z^2$$ and if we factor the non $x$ terms we have $$6y^2 -4yz+2z^2=2(z-y)(z+3y)$$

then a little trial ane error gives

$$[3x+2(z-y)][x-(z+3y)]$$