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Given equation

$$ x^2 + Bxy + Cy^2 = 0. $$

I want to factorize it in the form

$$ (x + my)(x + ny) = 0. $$

What are the values of $m$ and $n$ in terms of $B$ and $C$?


I tried writing the equation as

$$ x^2 + Bxy + Cy^2 = (x + kCy)(x + \dfrac{1}{k}y) = 0. $$

In this case the following equation must be satisfied:

$$ kC + \dfrac{1}{k} = B $$

I solved it for $k$ to find

$$ k = \dfrac{B \mp \sqrt{B^2 - 4C}}{2C}. $$

That makes

$$ x^2 + Bxy + Cy^2 = (x + \dfrac{B \mp \sqrt{B^2 - 4C}}{2}y)(x + \dfrac{2C}{B \mp \sqrt{B^2 - 4C}}y) = 0. $$

Which implies that

$$ \begin{array}{rcl} m &=& \dfrac{B \mp \sqrt{B^2 - 4C}}{2} \\ n &=& \dfrac{2C}{B \mp \sqrt{B^2 - 4C}}, \end{array} $$

and there are at most four $(m,n)$ pairs.

Is my solution correct? Are there any (other) solution methods for this factorization?

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2 Answers 2

up vote 3 down vote accepted

Your approach is quite satisfactory. It just skips a formal step by assuming the solution form rather than deriving it.

$$ (x+my)(x+ny) = x^2 +(m+n)xy + mny^2 $$ Therefore $$ B = m+n\\ C = mn $$ From the equation for $C$, we can write $$ n = \frac{C}m $$ which turns the first equation into $$ B = m + \frac{C}m $$ Now, if you let $k=\frac1m$, then you can write these as $$ n = Ck\\ m = \frac1k\\ B = \frac1k + Ck $$ And this is the same as the assumed form that you started with. As you can see, this derivation requires no additional assumptions (except, perhaps, that $m\neq 0$, which should be fine because $m=0$ would make $C=0$ anyway), and so there are no other solutions than the one you provided.

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Nice solution! Please note that Glen O has answered your question very well, and what comes below does not. Its purpose is to give a slightly different view of the matter, and to connect it with other things that you know.

Let us factor $4x^2+4Bxy+4Cy^2$. Completing the square, we get $(2x+By)^2-(B^2-4C)y^2$. If $B^2-4C\lt 0$, there is no factorization over the reals. But if we are willing to work with complex non-real numbers, we can go on. If we are unwilling, assume that $B^2-4C\ge 0$.

Let $D=B^2-4C$. We are looking at $(2x+By)^2-(\sqrt{D}y)^2$. This is a difference of squares, and factors as $$(2x+By+\sqrt{D}y)(2x+By-\sqrt{D}y).$$ But we had multiplied by $4$, so let's undo that by dividing each term by $2$. We get $$\left(x-\frac{B+\sqrt{D}}{2}y\right)\left(x+\frac{B+\sqrt{D}}{2}y\right).$$

Remark: Note that in particular we have derived the Quadratic Formula. For an alternate approach, let's use the Quadratic Formula. Divide our quadratic by $y^2$, not worrying about possible division by $0$, and temporarily let $w=\frac{x}{y}$. Then our quadratic is $y^2(w^2+Bw+C)$. The quadratic equation $w^2+Bw+C=0$ has the roots $\frac{-B\pm\sqrt{B^2-4C}}{2}$, and therefore the polynomial $w^2+Bw+C$ factors as $$w^2+Bw+C=\left(w-\frac{-B+\sqrt{B^2-4C}}{2}\right) \left(w-\frac{-B-\sqrt{B^2-4C}}{2}\right) .$$ Now replace $w$ by $\frac{x}{y}$, and multiply each term by $y$.

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