Factorise the equation

Question

I tried a lot to factorise $6x^2+7xy+2y^2+11xz+7yz+3z^2$.

I couldn't, Please help if it's possible. I tried to reduce it to $(a+b+c)^2$ format but I couldn't.

Answer 1

First factorize the $x/y$ terms (by solving a quadratic equation)

$$6x^2+7xy+2y^2=(3x+2y)(2x+y).$$

The remaining terms are

$$11xz+7yz+3z^2$$ which you can identify with

$$(3x+2y+az)(2x+y+bz)-6x^2+7xy+2y^2.$$

This yields the system

$$\begin{cases}2a+3b&=11,\\a+2b&=7,\\ab&=3.\end{cases}$$ that has the solution $a=1,b=3$.

Answer 2

Here is a better way than trying "a lot". Since the polynomial is homogeneous of degree $2$ in $x,y,z$ one could try $$ (ax+by+cz)(a'x+b'y+c'z)=6x^2+7xy+2y^2+11xz+7yz+3z^2. $$ Comparing coefficients gives a solution, namely $(2x+y+3z)(3x+2y+z)$.

Answer 3

Let $6x^2+7xy+2y^2+11xz+7yz+3z^2=(ax+by+cz)(dx+ey+fz)$. Then

$$\begin{cases}ad=6\\ be=2 \\ cf=3 \\ ae+bd=7 \\ af+cd=11 \\ bf+ce=7 \end{cases}$$

$(a,b,c,d,e,f)$ is not unique, as it can be replaced by $\displaystyle \left(ka,kb,kc,\frac{d}{k},\frac{e}{k},\frac{f}{k}\right)$ for non-zero constant $k$.

Arbitrarily take $a=2$. Then $d=3$ and

$$\begin{cases} be=2 \\ cf=3 \\ 2e+3b=7 \\ 2f+3c=11 \\ bf+ce=7 \end{cases}$$

So $\displaystyle e=\frac{2}{b}$ and hence

\begin{align} 2\left(\frac{2}{b}\right)+3b&=7\\ 3b^2-7b+4&=0\\ b&=1\quad\text{or}\quad\frac{4}{3} \end{align}

If $b=1$, then $e=2$ and hence

$$\begin{cases} cf=3 \\ 2f+3c=11 \\ f+2c=7 \end{cases}$$

Solving, $c=3$ and $f=1$.

$(a,b,c,d,e,f)=(2,1,3,3,2,1)$ is a possible solution.

$6x^2+7xy+2y^2+11xz+7yz+3z^2=(2x+y+3z)(3x+2y+z)$.

Answer 4

Hint:  consider it as a quadratic in $\,x\,$: $$6x^2+(7y+11z)x+2y^2+7yz+3z^2$$

Its discriminant is:

$$ (7y+11z)^2-4\cdot 6 \cdot (2y^2+7yz+3z^2) = y^2 - 14 y z + 49 z^2 = (y-7z)^2 $$

Therefore the roots in $\,x\,$ are $x_{1,2}=\frac{1}{12}\big(-(7y+3z)\pm(y-7z)\big)\,$, and so the polynomial factors as $\,6(x-x_1)(x-x_2)\,$.