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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:

enter image description here

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.

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  • $\begingroup$ The "x" in the diamond diagram should be a 1* this does not change my problem $\endgroup$ – Sphygmomanometer Mar 19 '18 at 0:35
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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:$$R(x)=2x-3, \text{ where } x\ne-\dfrac{7}{4}$$

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  • $\begingroup$ 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 $\endgroup$ – Sphygmomanometer Mar 19 '18 at 0:53
  • $\begingroup$ Did you mean $2(4x^2+x-21/2)$, but that's really of no use @Sphygmomanometer $\endgroup$ – user535339 Mar 19 '18 at 0:55
  • $\begingroup$ 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) $\endgroup$ – Sphygmomanometer Mar 19 '18 at 0:57
  • $\begingroup$ 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 $\endgroup$ – user535339 Mar 19 '18 at 0:58
  • $\begingroup$ I did change the expression, is $$8x^2+2x-21= 2(4x^2+x-21/2)$$ not correct? $\endgroup$ – Sphygmomanometer Mar 19 '18 at 1:02

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