Attachment 25610

I let (x+1)=x and (x-3)=y and I got $\displaystyle \frac{2x^4y^2(3x+y)}{x^{10}}$

where do I go from here?

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- Nov 9th 2012, 09:49 PMkingsolomonsgraveHow to simplify this expression
Attachment 25610

I let (x+1)=x and (x-3)=y and I got $\displaystyle \frac{2x^4y^2(3x+y)}{x^{10}}$

where do I go from here? - Nov 9th 2012, 10:02 PMMarkFLRe: How to simplify this expression
Rewrite the denominator as $\displaystyle x^4 \cdot x^6$ and divide out the factor common to the numerator and denominator.

- Nov 10th 2012, 06:39 AMSorobanRe: How to simplify this expression
Hello, kingsolomonsgrave!

Quote:

$\displaystyle \text{Simplify: }\:\frac{6(x+1)^5(x-3)^2 + 2(x+1)^4(x-3)^3}{(x+1)^{10}}$

I let (x+1)=x and (x-3)=y. . Don't do that!

Use two "new" variables: .$\displaystyle {\color{blue}x + 1 = u,\;x-3 = v}$

and I got $\displaystyle \frac{2x^4y^2(3x+y)}{x^{10}}$

where do I go from here?

You can cancel the x's . . . and back-substitute.

We have: .$\displaystyle \frac{6(x+1)^5(x-3)^2 + 2(x+1)^4(x-3)^3}{(x+1)^{10}}$

Factor: . $\displaystyle \frac{2(x+1)^4(x-3)^2\big[3(x+1) + (x-3)\big]}{(x+1)^{10}} $

Reduce: . $\displaystyle \frac{2(x-3)^2\big[3x+3 + x - 3\big]}{(x+1)^6} $

Simplify: .$\displaystyle \frac{2(x-3)^2\cdot4x}{(x+1)^6} \;=\;\frac{8x(x-3)^2}{(x+1)^6} $

Where did this problem come from?

If it came from an application of the Quotient Rule,

. . something is drastically wrong.