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I am looking for a method that I can use to factorize the following equation $$3{y}^{2}+8yx-3x^{2}$$

Into this: $$(3y-x)(y+3x)$$

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5 Answers

up vote 9 down vote accepted

Hint: Write $3y^2+8yx-3x^2 = x^2 (3z^2 + 8z -3)$, where $z=y/x$. The second factor is a quadratic polynomial in just one variable, and this you perhaps know how to factorize?

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For what it's worth:

You might try a simpler problem for warm up; say the same problem, but with $x=2$: $$\tag{1} 3y^2+8\cdot 2 y -3\cdot 2^2 $$ To factor this, you could guess and check. The first step would be to guess that the factorization of $(1)$ is $$\tag{2} (3y + a)(y+b); $$ since, if you multiplied this out, you would at least get the correct $3y^2$ term in $(1)$.

Now we need to find $a$ and $b$. To get the correct constant term, $-3\cdot2^2$, in $(1)$, make a guess; say, $a=\mp3$ and $b=\pm2^2$. Note here $a$ and $b$ have to have different signs and that we are simply making a guess with $ab=-3\cdot2^2$.

But, would this guess give the right factorization? It would be if the "cross terms" resulting from multiplying $(2)$ out combined to give you $8\cdot2y=16y$. Let's check if that's possible. The cross terms of the product $(2)$ are shown in maroon and green below:

$$ (\overbrace{\color{maroon}{3y} \pm \underbrace{\color{darkgreen}3)(\color{darkgreen}y}_{\color{darkgreen}{\mp3y}}\pm\color{maroon}{2^2}}^{\color{maroon}{\pm3y\cdot2^2\ =\ \pm 12y}}) $$ But we can't get $16y$ from $\color{maroon}{\pm12y}\mp\color{darkgreen}{3y} $. You'd either have $12y-3y=9y$ or $-12y+3y=-9y$. So our guess isn't correct.

That's ok; try a different guess: say $a=\mp2$, $b=\pm3\cdot2$.

Then you'd have $$ (\overbrace{\color{maroon}{3y} \pm \underbrace{\color{darkgreen}2)(\color{darkgreen}y}_{\color{darkgreen}{\mp2y}}\pm\color{maroon}{3\cdot2}}^{\color{maroon}{\pm3y\cdot3\cdot2\ =\ \pm 18y}}) $$ Here, we can combine the cross terms to get $16y$, yay! Indeed $16y=\color{maroon}{+18y}+\color{darkgreen}{(-)2y}$. So we need to take $a=-2$ and $b=+3\cdot2$; and the factorization of $(1)$ is $$ (3y-2)(y+3\cdot 2). $$

Now back to your problem: factor $$ 3y^2+ 8xy-3x^2. $$

From our warm up, where $x=2$, you might suspect that the factorization is $$ (3y-x)(y+3\cdot x) $$ (this is why I didn't do all the arithmetic in the preceding; that would have made this supposition harder to see).

Checking this guess will reveal that this is indeed the factorization.

Of coure, you could have just used this guess and check procedure on the original expression.

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Write $8yx$ as $9yx-yx$ to see the factorisation.

$$\begin{align} 3y^2+8yx-3x^2&=3y^2+9yx-yx-3x^2\\&=3y(y+3x)-x(y+3x)\\&=(3y-x)(y+3x)\end{align}$$

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If you know that all the factors will have integer coefficients there's another, more ad hoc approach you can take; the advantage is that it's sometimes easier to apply 'on-the-fly' (for instance, during a test). Write your product as $(ay+bx)(cy+dx)$; then you know that $a\cdot c = 3$ and $b\cdot d=-3$. Since $a$ and $c$ must be factors of 3, either $a=1,c=3$ or $a=3,c=1$ (there are negative versions of these cases, but the negative signs can be absorbed into the $b$ and $d$ factors); likewise, either $b=1, d=-3$, $b=3, d=-1$, $b=-1, d=3$, or $b=-3, d=1$; it's now just a matter of checking $bc+ad$ for the different assignments to see which one gives the correct central coefficient $8$.

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Hint $\ $ By the AC method $\rm\ 3\:\!(3x^2 - 8yx - 3y^2)\: =\: y^2\:\! (X^2 - 8\:\!X - 9),\:$ for $\rm\: X = 3x/y $

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