Factoring

pilivicky

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Sep 13, 2005
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I have two problems that I cannot solve for the life of me!!! Actually, there are a few of them, but I figured that if I can learn how to do these two, I should have an example on how to do the other four:

4x^-1 – 3x^-2 + 5x^-3

and the other one is (3x+1)^5/3 - 2(x+4)(3x+1)^2/3

Please help!!!! All I need is to know is what process to use. I tried using the quadratic equation, tried getting rid of the negative exponent, but to no avail.

Thank you for any help!
 
4x^-1 – 3x^-2 + 5x^-3 = y

x^-1(4 -3x^-1 +5x^-2) =y

x^-2 (4x -3 +5x^-1) =y

x^-3 (4x^2 -3x +5) =y

now you have a quadratic to solve....
does this help?
 
Thank you for your prompt reply! I am still running into the same problem: I would end up with an imaginary number (the square root of a negative number!) unless I have forgotten how to calculate the quadratice equation. (9+-√9-80)/8 so that would be (9 +-√-79)/8 right?

Were you able to figure out the second one?

Thank you again!
 
pilivicky said:
Thank you for your prompt reply! I am still running into the same problem: I would end up with an imaginary number (the square root of a negative number!) unless I have forgotten how to calculate the quadratice equation. (9+-√9-80)/8 so that would be (9 +-√-79)/8 right?
Correct; except for typo: 9 - 80 = -71, not -79

"4x^-1 – 3x^-2 + 5x^-3"
Another way to do that one is:
4/x - 3/x^2 + 5/x^3 ; multiply through by x^3:
4x^2 - 3x + 5


"and the other one is (3x+1)^5/3 - 2(x+4)(3x+1)^2/3"

(3x + 1)^(2/3)[3x + 1 - 2(x + 4)] = 0
(3x + 1)^(2/3)[3x + 1 - 2x - 8)] = 0
(3x + 1)^(2/3)[x - 7] = 0

so (3x + 1)^(2/3 = 0 or x - 7 = 0
x = -1/3 or x = 7

...surprisingly easy; and I'm fairly sure it's correct...apm??
 
factor & simplify

Thanks for the answer. Yes, when you put it that way solving for x seems easy. I guess I thought that if I found what the values of x were, I could just plug it in. The problem asks to factor and simplify, but I have never factored negative rational numbers.
 
Re: factor & simplify

pilivicky said:
Thanks for the answer. Yes, when you put it that way solving for x seems easy. I guess I thought that if I found what the values of x were, I could just plug it in. The problem asks to factor and simplify, but I have never factored negative rational numbers.
You've got the buggy ahead of the horse; you "plug in" AFTER you solve, to ckeck if your answer is correct.
 
The problem asks to factor and simplify

I think we might have gotten off track here. We've been given an expression, NOT an equation; therefore, we should not be trying to "solve" it -- just "factor and simplify."

APM has already provided three different factored forms:

x^-1(4 -3x^-1 +5x^-2)

x^-2 (4x -3 +5x^-1)

x^-3 (4x^2 -3x +5)

I'm not sure what more they're specifically looking for in this problem.
 
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