# Writing a polynomial as a linear combination of other polynomials

I'm currently working on writing $3(x)_4 - 12(x)_3 + 4(x)_1 - 17$ as a linear combination of $(x)_4,\ldots,(x)_0$ and am having difficulty understanding where the conversion comes from. I have the answer as $3x^4 - 30x^3 + 69x^2 -38x - 17$ and need help with how to work these problems. Or how to work it the other way such as writing $(x)_4 - 2(x)_2$ as a linear combination of the power $x$?

Those numbers following the $x$'s should be falling factorials

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It is difficult to understand your question or what you have tried? – copper.hat May 6 '13 at 17:13
imgur.com/uDgj6Gl I've attached a picture to better show the formatting since I'm new to formatting on stackexchange. @copper.hat – aelsedoudi May 6 '13 at 17:23
Have you heard of divided differences? – J. M. May 6 '13 at 18:00

It will help a little to write out the falling factorials in a more familiar way:

\begin{align*} (x)_4 &= x^4 - 6x^3 + 11x^2 -6 x\\ (x)_3 &= x^3-3 x^2+2 x\\ (x)_2 &= x^2 - x\\ (x)_1 &= x\\ (x)_0 &= 1 \end{align*}

To write $x^4 - 2x$ as a linear combination of these, we must start with $(x)_4$, since this is the only way to get a term of $x^4$. This term brings with it, however, an unwanted $-6x^3$. This can only be cancelled by adding $6(x)_3$. Continuing in this way, we end up with $$x^4 - 2x = (x)_4 + 6(x)_3 +7(x)_2 - (x)_1.$$

For the other direction, just refer to our chart to get $$(x)_4 - 2(x)_2 = x^4 - 6x^3 + 9x^2 -8x.$$

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