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 lalex0710 September 23rd, 2012 12:01 PM

rational expression

x^3-8 / 3x^2-x - 10

 soroban September 23rd, 2012 12:36 PM

Re: rational expression

Hello, lalex0710!

Quote:
 $\frac{x^3\,-\,8}{3x^2\,-x\,-\,10}$

A wild guess . . . you want this simplified?

Do you know anything about Factoring?

$\frac{x^3\,-\,8}{3x^2\,-\,x\,-\,10} \;=\;\frac{(x\,-\,2)(x^2\,+\,2x\,+\,4)}{(x\,-\,2)(3x\,+\,5)} \;=\; \frac{\cancel{(x\,-\,2)}(x^2\,+\,2x\,+\,4)}{\cancel{(x\,-\,2)}(3x\,+\,5)} \;=\; \frac{x^2\,+\,2x\,+\,4}{3x\,+\,5}$

 lalex0710 September 23rd, 2012 12:44 PM

Re: rational expression

Yes, I do know how to factor, but I am unfamiliar with exponents higher than power of 2. Anyways, thanks for the answer, and I have one more question lol.

x^2-4 / x^3+4x^2-4x-16

 lalex0710 September 23rd, 2012 12:46 PM

Re: rational expression

For the previous question, could you also elaborate on the answer. I don't quite understand how you pulled out the (x-2) as common factor, and factored the rest.

 jks September 23rd, 2012 03:10 PM

Re: rational expression

Hi lalex0710,

For the second problem, $\frac{x^2-4}{x^3+4x^2-4x-16}$, here is how I would work it:

First, find the x values of the zeroes of the numerator and plug then into the denominator to see which ones result in 0. You then can divide through by that term (or terms) using long division to factor.

$x^2-4$ is 0 at $x=-2,2$ so $(x+2)$ and $(x-2)$ are factors.

The denominator at $x=-2$ is $(-2)^3+4(-2)^2-4(-2)-16=0$ and at $x=2$ the denominator is $2^3+4(2)^2-4(2)-16=0$

So both $x+2$ and $x-2$ divide the denominator. Therefore $x^2-4$ divides the denominator. Use long division to factor out $x^2-4$ giving:

$\frac{x^2-4}{x^3+4x^2-4x-16}=\frac{(x^2-4)}{(x^2-4)(x+4)}=\frac{1}{(x+4)}$

Note that this is also $\frac{x^2-4}{x^3+4x^2-4x-16}=\frac{(x+2)(x-2)}{(x+2)(x-2)(x+4)}=\frac{1}{(x+4)}$

Also note that at $x=-4$, the denominator is $(-4)^3+4(4)^2-4(-4)-16=0$ as expected. Finally, check all calculations by picking $x=7$ (not a zero, and I like to use a number not likely to coincidentally give an equality):

$\frac{7^2-4}{7^3+4(7)^2-4(7)-16}=\frac{45}{495}=\frac{1}{11}=\frac{1}{(7+4)}$

Carrying this back to the first problem, note that $x^3-8=0$ at $x=2$ so $(x-2)$ is a factor. At $x=2$ the denominator is $3(2)^2-2-10=0$ so $(x-2)$ is a factor of the denominator as well. Long division may be used to find the results presented. Finally, there are no real roots of $x^2+2x+4$ so the expression cannot be simplified any further.

 greg1313 September 23rd, 2012 03:17 PM

Re: rational expression

Quote:
 Originally Posted by lalex0710 Yes, I do know how to factor, but I am unfamiliar with exponents higher than power of 2. Anyways, thanks for the answer, and I have one more question lol. x^2-4 / x^3+4x^2-4x-16
The numerator factors as $^{(x\,-\,2)(x\,+\,2).}$ To factor the denominator we may group the terms and factor, like so:

$x^3\,+\,4x^2\,-\,4x\,-\,16\,=\,x(x^2\,-\,4)\,+\,4(x^2\,-\,4)\,=\,(x\,+\,4)(x^2\,-\,4)\,=(x\,+\,4)(x\,+\,2)(x\,-\,2)$

Quote:
 Originally Posted by lalex0710 For the previous question, could you also elaborate on the answer. I don't quite understand how you pulled out the (x-2) as common factor, and factored the rest.
It's a difference of cubes: $^{a^3\,-\,b^3\,=\,(a\,-\,b)(a^2\,+\,ab\,+\,b^2)}$

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