Two questions - factoring and exponents

[superdave]

These aren't homework questions. They were on a test I had today.

The first was how do I factor 2x^3-3x^2-16x-5. I'm fairly certain that's right. It was actually part of a larger problem, involving finding a slant asymptote. I just used my graphing utility to find the answer, but I'd like to know how to actually factor that. I'm pretty sure you get (x+1) and a quadratic equation, just not sure what that equation is.

Second question is about exponents. I think I got this one wrong.

It was 1.7^(-3/5). Now, I got .787 I think. But what I was wondering, shouldn't the answer be equal to 1.7^-3 - 1.7^5? When I tested it, it wasn't. Isn't that one of the properties of exponents? x^(a/b) = x^a-x^b?

 

[finchie_88]

shouldn't the answer be equal to 1.7^-3 - 1.7^5?

No, what it should be is $$\frac{x^a}{x^b} = x^{a-b}$$.

Also, I hate to break it to you, but I don't think that x+1 is a factor of that cubic (unless my mental maths is wrong - which is quite likely). To factor that, you might be able to find a root, by substituting numbers in and seeing if you get 0. Then, use polynomial division (assuming you can do that), to reduce the degree by 1, and then use the quadratic formula to get the other two roots.

You could always look at the co-efficients of the terms, e.g. the product of the three roots will equal 5/2, that might help.

 

[superdave]

err, I meant x-1.

 

[topsquark]

err, I meant x-1.

Still doesn't work.

-Dan

 

[ksinclair13]

Are you sure that that is what you need to factor? I ask this because this doesn't have nice roots. The roots are approximately -1.947985, -.3389005, and 3.7868854. So the factored form would look something like this:

f(x) = (x+1.948)(x+.3389)(x-3.787).

 

[superdave]

No. I could be remembering wrong. I didn't actually have to factor it. The question was about finding a slant asymptote. I assumed there was factoring. But it was too hard, so I just used my TI 84 to graph it, then found the asymptote through trial and error. (It was a multiple choice question. I just put in each option and looked to see if it worked with the other function.)

When we get our tests back Monday, I'll put the problem up. Cause I would like to know if there was an easy way. The final exam is not multiple choice, so it won't work for that.

 

[moose]

Do you realize that there is no slant asymptote on that graph?

Are you sure it wasn't dealing with a different graph or something?

 

[ksinclair13]

Maybe that was the numerator of a rational expression. In this case, you would divide it out and then the oblique asymptote is the line y = [the answer excluding the remainder].