Just a quickie...

is

$\displaystyle (-\frac{2}{4}u^{-4}-\frac{7}{5}u^{-5}) = -\frac{1}{20}u^{-4}(10+28u^{-1})$

correct?

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- Apr 21st 2010, 02:53 AMdarksupernovatake a power of -1 out
Just a quickie...

is

$\displaystyle (-\frac{2}{4}u^{-4}-\frac{7}{5}u^{-5}) = -\frac{1}{20}u^{-4}(10+28u^{-1})$

correct? - Apr 21st 2010, 03:12 AMADARSH
yes.

- Apr 21st 2010, 03:13 AMTekken
yes your correct due to rules of indices i.e. $\displaystyle u^{-4}.u^{-1} = u^{-4+(-1)} = u^{-5} $

- Apr 21st 2010, 03:20 AMSoroban
Hello, darksupernova!

Quote:

$\displaystyle \left(-\frac{2}{4}u^{-4}-\frac{7}{5}u^{-5}\right) \;=\; -\frac{1}{20}u^{-4}(10+28u^{-1})$

Correct?

Assuming that the instructions are: "Factor*completely*" . . .

When factoring out a variable, we factor out the "least exponent".

Example: .$\displaystyle 4x^5 + 7x^3 \:=\:x^3(4x^2 + 7)$

. . . . . . . Otherwise, we could claim that:

. . . . . . . $\displaystyle x(4x^4 + 7x^2)\,\text{ and }\,x^2(4x^3 + 6x)$ are both correct.

Also, the LCD of $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{7}{5}$ is 10.

Preferred answer: .$\displaystyle -\frac{1}{10}u^{-5}(5u + 14)$

- Apr 21st 2010, 03:53 AMdarksupernova
thank you very much :)