# Does factoring by x-method always work?

Given the following function $R(x)$ [in pic] find the vertical, horizontal, oblique asymptotes. In order to find the asymptotes you need to reduce the function, so I did the following:

This is incorrect, the solution is $(2x-3)$; I noticed that the x method/ diamond method did not work here since the factors $(2x-3)(4x+7)$ did not multiply to $4x^2+x-21/2$ but rather $8x^2 +2x-21$, how this happens I do not know. Now, I understand there are many other ways of factoring this problem but I seem to have either made a mistake somewhere or have found a situation where the criss cross method does not work? I would like help finding out exactly what I did wrong, to help strengthen my math foundation.

• The "x" in the diamond diagram should be a 1* this does not change my problem – Sphygmomanometer Mar 19 '18 at 0:35

You are incorrect in the following part:

$$8x^2+2x-21\ne 2(4x+7)(2x-3)$$

There is another part where you without reason, just add another $2$ also.

The correct factoring should be:$$R(x)=2x-3, \text{ where } x\ne-\dfrac{7}{4}$$

• No, I wrote: $$8x^2+2x-21= 2((4x+7)(2x-3))$$ the reason another 2 comes up is because the 2 is distributed in both terms; the main issue I cant wrap my head around is the x method not working – Sphygmomanometer Mar 19 '18 at 0:53
• Did you mean $2(4x^2+x-21/2)$, but that's really of no use @Sphygmomanometer – user535339 Mar 19 '18 at 0:55
• I figured that I would factor out a 2 and then factor the smaller expression using the criss-cross method (which apparently doesnt work for this or I did something wrong) – Sphygmomanometer Mar 19 '18 at 0:57
• I figured that I would factor out a 2. Then you also have to change the expression it is being factored out of. A very basic example would be saying that $2x$ isn't equal to $2(2x)$ @Sphygmomanometer – user535339 Mar 19 '18 at 0:58
• I did change the expression, is $$8x^2+2x-21= 2(4x^2+x-21/2)$$ not correct? – Sphygmomanometer Mar 19 '18 at 1:02