Approaches to factor $12xy - 4xz + 4x^2 + 9y^2 + 6yz$

Question

What approaches would you take to factor the following?

$$12xy - 4xz + 4x^2 + 9y^2 + 6yz$$

There is no common factor, and there seems to be no common relationship between every term that seems reasonable.

Answer 1

By observing the combination of the variables, we can infer that the factorized form, if any, should be $(ax+by+cz)(dx+ey)$.

By the method of matching coefficients, we have the following: $$\begin{cases}ad = 4\ldots(1)\\be = 9\ldots(2)\\cd = -4\ldots(3)\\ce = 6\ldots(4)\\ae+bd=12\ldots(5)\end{cases}$$

In addition, it is required that $$GCD(a, b, c) = GCD(d, e) = 1$$ or it would contradict to the fact that there are no common factors for all the terms.

Therefore, from (3) & (4), it can be deduced that $c=2$ or $c=-2$.

Then, $(c, d, e) = (2, -2, 3) \text{ or } (-2, 2, -3)$.

Going to (1) & (2), we obtain $(a, b) = (-2, 3) \text{ or } (2, -3)$.

Substituting the two possibilities to (5), we obtain $ae+bd=-12\neq 12$.

Hence, it is shown that the expression cannot be factorized.

Answer 2

a ternary quadratic form factors (over the complex numbers possibly) if and only if the discriminant is zero, that being the determinant of the Hessian matrix.

This determinant is negative, so the factoring asked for in the original question is impossible.

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 3 }{ 2 } & - 6 & 1 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 36 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & - 6 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 6 & - 2 \\ 6 & 9 & 3 \\ - 2 & 3 & 0 \\ \end{array} \right) $$

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One suggestion is to negate the coefficient 12, and that works:

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 2 } & 0 & 1 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & - 6 & - 2 \\ - 6 & 9 & 3 \\ - 2 & 3 & 0 \\ \end{array} \right) $$

This recipe says $$ 4 \left(x - \frac{3}{2} y - \frac{1}{2} z \right)^2 - z^2 , $$ or $$ \left(2x - 3 y - z \right)^2 - z^2 , $$ or (difference of squares) $$ \left(2x - 3 y -z +z \right)\left(2x - 3 y -z - z \right) $$ $$ \left(2x - 3 y \right)\left(2x - 3 y -2z \right) $$

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One I made up, semidefinite positive:

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 8 }{ 3 } & 1 & 0 \\ 6 & 3 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 5 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 8 }{ 3 } & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 8 & 18 \\ 8 & 23 & 53 \\ 18 & 53 & 123 \\ \end{array} \right) $$

or $$ 3 x^2 + 23 y^2 + 123 z^2 + 106 yx + 36 z x + 16 xy $$

So

This recipe says $$ 3 \left(x + \frac{8}{3} y +6 z \right)^2 + \frac{5}{3} (y+3z)^2 $$ or $$ \frac{1}{3} \left( 9 \left(x + \frac{8}{3} y +6 z \right)^2 + 5 (y+3z)^2 \right) $$ or $$ \frac{1}{3} \left( \left(3x + 8 y +18 z \right)^2 + 5 (y+3z)^2 \right) $$ $$ \frac{1}{3} \left( \left(3x + 8 y +18 z \right)^2 - (-5) (y+3z)^2 \right) $$ or (difference of squares) $$ \frac{1}{3} \left( \left(3x + 8 y +18 z + i \sqrt 5 (y+3z)\right) \left(3x + 8 y +18 z - i \sqrt 5 (y+3z)\right) \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 3

This cannot be parsed, but you can do this : $$12xy - 4xz + 4x^2 + 9y^2 + 6yz=(2x+3y-z)^2-z^2+12yz=(2x+3y)(2x+3y-2z)+12yz$$

Answer 4

We observe that, multiplier polynomials must be linear .

The given polynomial is a quadratic polynomial respect to $x$ . We want to divide both side of $P(x,y)=0$ by $y^2, \thinspace y≠0$ :

$$ \small {\begin{align}12xy-4xz+4x^2+9y^2+6yz=0\\ \underbrace{4a^2+2a(6-2b)+(6b+9)}_{\mathrm {Q(a,b)}}=0\end{align}} $$

where $a=\dfrac xy$ and $b=\dfrac zy$ .

We know that, if $Q(a,b)$ is a factorable polynomial, then $P(x,y)$ is also a factorable polynomial. Then, we find the discriminant $\Delta_a$ :

$$\Delta_a=4b(b-12)$$

We see that if factorization is possible, then $b(b-12)=(b-6)^2-36$ must always be a perfect square . A contradiction .