a binomial thm problem

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- Jan 10th 2008, 08:09 PMafeasfaerw23231233a binomial thm problem
a binomial thm problem

- Jan 10th 2008, 08:26 PMJhevon
- Jan 10th 2008, 09:28 PMtopsquark
$\displaystyle (1 + 2x)^4(1- x)^3$

The most clever thing I can come up with is to define two new variables:

$\displaystyle y = 1 + \frac{x}{2}$

and

$\displaystyle z = \frac{3x}{2}$

Then

$\displaystyle y + z = 1 + 2x$

and

$\displaystyle y - z = 1 - x$

Thus

$\displaystyle (1 + 2x)^4(1- x)^3 = (y + z)^4(y - z)^3$

$\displaystyle = [(y + z)^3(y - z)^3](y + z)$

$\displaystyle = (y^2 - z^2)^3(y + z)$

$\displaystyle = ([1 + 2x]^2 - [1 - x]^2)^3(1 + 2x)$

etc.

But I doubt this actually cuts down on the work load.

-Dan