Horner Scheme

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Exactly what I was looking for, great job, thank you!

CroArtVandelay
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@intuit13 if we apply this process for the rest divisors of the last term (-6), we shall finally get that the polynomial can be written (x-1)(x-3)(x+2)(x-1). The result will be the same if we apply the process successively for the quotients we get when remainder equals to 0. Graphically speaking, for x=3 and x=-2, x-axis intersects function's graph, and x-axis is tangent at x=1. This is why factor (x-1) takes part twice. I hope that this is what you were looking for. :)

mathm
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c'est très bien expliqué merci .thank you.

dounia
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thanks ;) didn't pay attention in class and you just learned it to me in 1.54 :p

SpankY
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switch to full screen, press "pause" a bit before the end and wait 2 seconds without moving the mouse. YouTube bar will be automatically hidden and you' ll be able to see the last line of text.

mathm
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@qq75970138 my intention was to make a clear presentation of how this scheme works, not to research if Horner was the first one that introduced it. Worldwide is known as such. I am pretty sure that in your rich mathematical tradition, of which you should be proud, there should be that algorithm, probably described a bit different, however earlier than other people knew that.

mathm
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thanks, but small question how were the degrees of the final polynomial found ?
 x^3 - x^2 - 5^x + 1, you are just giving each entity(to each X)  the degree starting from 0 and increase it until you get the last entity ?

Scotty
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@datHulkinZ if I had put my music, it would have been part of the presentation and probably you wouldn't like it :) You can play any music you like at your media player while watching that. You choose ... It's better this way ...

mathm
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Bah, I was looking for Horner's Method for breaking down polynomials.... This is just synthetic division. :(

intuit
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@mathm09
you are watching this clip in order to obtain some knowledge, not to entertain yourself :)

Invernf
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thxx a lot but i cann't see the last step

braveryindraneel