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How to simplify $6x^2 - 5xy - 6y^2$ ?

$(2x-3y)(3x+2y)$

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    So what's the question then? – 2017-02-09
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    I don't know how to do it :( @GoodDeeds – 2017-02-09
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    @Bacon common let's make maths great again,I wish there is some way… – 2017-02-09
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    You can factor it by grouping. Think if it as $6x^2 - (5y)x - (6y^2)$ so it "looks" quadratic in $x$, but with some $y$'s in the coefficients. – 2017-02-09
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    Use the quadratic formula on x for example and treat y as a constant. – 2017-02-09
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    Have you learn't the [**cross method**](http://www.mathsteacher.com.au/year9/ch08_factors/07_cross/mult.htm)? Some sorts of trial-and-error, inspection or mental calculation should be built up by doing plenty of exercises. Look, $6=2\times 3$ and $-5=(2)(2)-(3)(3)$. So $(3x-2y)(2x-3y)$ or $(3x+2y)(2x-3y)$, etc. Personally, I don't rely on quadratic formula which involves square root. Sometime, it's really hard to find the correct factors when the square terms have many factors which means many combinations need to be considered. – 2017-02-09
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    @NgChungTak I didn't thought $(2)(2)-(3)(3)=-5 $, I was just thing $2+3=5 $:P thanks btw – 2017-02-09
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    "Simplify" and "factor" are generally regarded as opposite processes. The expression is already simplified, you really want to factor it. Just sayin'. – 2017-02-09

6 Answers 6

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Once you see that $4*6*6=12^2$, the discriminat is $\sqrt{5^2+12^2}=13$ from the well known 5-12-13 right triangle.

This means that the quadratic will facror and you will get the result that others have given.

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You do something called the middle term factorisation.

So, from that linked algorithm, we see that you can rewrite things as $$6x^2-5xy-6y^2$$ $$=6x^2+(-9xy+4xy)-6y^2$$ $$=3x(2x-3y)+2y(2x-3y)$$ $$=(3x+2y)(2x-3y)$$

Hope this helps you.

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Another approach

Do you know $$\large\color{red} {ax^2+bx+c=a(x-x_1)(x-x_2)}$$ ? if you see $6x^2 - 5xy - 6y^2 $ like that form ,you will have $$6(x^2) + (-5y)x +(- 6y^2)=0\\$$ solve for $x$ $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\\ \frac{+5y\pm\sqrt{25y^2-4.6.(-6y^2)}}{2.6}=\\\dfrac{5y\pm13y}{12}\to x_1=\dfrac{5y+13y}{12}=\dfrac{3y}{2}\\ x_2=\dfrac{5y-13y}{12}=\dfrac{-2y}{3}\\$$ now $$6(x^2) + (-5y)x +(- 6y^2)=6(x-\dfrac{3y}{2})(x-\dfrac{-2y}{3})=\\2.3(x-\dfrac{3y}{2})(x+\dfrac{2y}{3})=\\(2x-3y)(3x+2y)$$

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    Nice information , thanks :) – 2017-02-09
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One way is write

$$6x^2-5xy-6y^2=(ax+by)(cx+dy)$$

What give us:

$$ac=6\quad (1)\\ ad+bc=-5\quad (2)\\ bd=-6\quad (3)$$

Now try to find integer solutions (that is something that happens in many cases, including this one).

We can start with $ac=6$ and the possibilities are: $(a,c)\in \{(\pm 1,\pm 6),(\pm 2,\pm 3),(\pm 3,\pm2),(\pm6,\pm1)\}$

Pick up each one and replace in equation $(2)$ and then you can find $b,d$ using $(2)$ and $(3)$.

You can spend some time doing that but it is a very usefull way when you don't have any idea how to proced.

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    I liked this way the most ^ ;) – 2017-02-09
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$$\begin{align}6x^2-5xy-6y^2&=6x^2-9xy+4xy-6y^2\\ &=3x(2x-3y)+2y(2x-3y)\\ &=(3x+2y)(2x-3y)\end{align}$$

Given $ax^2+bx+c$, you try to find $\alpha$ and $\beta$ such that $$\alpha+\beta=-b$$ $$\alpha\beta=c$$

This can be derived by completing the square in $ax^2+bx+c$. If finding $\alpha,\beta$ is not easy in a given case, completing the square also results in a formula to find the factors:

$$\alpha,\beta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

and the factorization is then $$ax^2+bx+c=(x-\alpha)(x-\beta)$$

Here, you could use $a=6,b=-5y,c=-6y^2$.

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Splitting the middle term,

$$6x^2 - 9xy + 4xy - 6y^2$$

$$3x(2x - 3y) + 2y(2x - 3y)$$

$$(3x + 2y)(2x - 3y)$$

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    Why don't you split middle term to$ -2x - 3x$ ? Any certain procedure? – 2017-02-09
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    If any doubt please ask. – 2017-02-09
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    Yes let me explain you. – 2017-02-09
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    First skip y from second and third term we have $6x^2 - 5x - 6$, then splitting middle term we have $6x^2 - 9x + 5x - 6$. Now add y with middle terms and $y^2$ with last term $6x^2 - 9xy + 5xy - 6y^2$ then take common terms from pairs. – 2017-02-09
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    I will never think of including 5 because it's prime and not having any factor with 6 – 2017-02-09
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    Sorry my mistake in comment. It is $6x^2 - 9xy + 4xy - 6y^2$ – 2017-02-09
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    Now you can take common terms out. – 2017-02-09
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    Yes,thank you very much – 2017-02-09