Polynomdivison?

Hallo,

wie wendet man hier die Polynomdivison an?
Ich habe den ersten Schritt gemacht, aber bei mir kommt direkt nach (27/125)x^2 irgendwie null raus. Die Lösungen sollen ja 5, (-5) und 5 sein. Nur der Punkt 5 ist gegeben, weshalb ich durch (x-5) geteilt habe.

Aufgabe:

0= ((27/125x^3)-(27/25)x^2)-5,4x+27)):(x-5)

LG

(1 votes)
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eterneladam
8 months ago

Multiply the thing once with 125/27 to get rid of the annoying fractures, the factor can be replicated afterwards,

(x^3 – 5 x^2 – 25 x + 125 : (x – 5) = x^2 – 25

x^3 – 5 x ^2

————–

0 – 25 x + 125

-25 x + 125

————–

0

Halbrecht
8 months ago
Reply to  eterneladam

what such tasks are to do ( /25 and .4 ), is a mystery to me

evtldocha
8 months ago

First you multiply the whole equation with 125/27 and write 5.4 as 54/10 and get the equation:

Or after a briefing:

The polynomial division then supplies:

(  x³ - 5x² - 25x + 125 ) : (x - 5)  = x²
 - x² + 5x²  
 -----------------------
            - 25x + 125   : (x - 5) = -25
            + 25x - 125
 -----------------------           
 
 Von oben nach unten: x² - 25   also: 
 (  x³ - 5x² - 25x + 125 ) : (x - 5) = x² - 25

For the remainder term, the 3rd binary formula obviously applies:
x2 – 25 = (x + 5)·(x – 5)

Tannibi
8 months ago

Put the whole job down. It may be,
that you are already in the factors with 3.
Two. Potence has failed.

DerRoll
8 months ago
Reply to  Tannibi

At least the solution of the zero-point task is correct, as

https://www.arndt-bruenner.de/mathe/scripts/polynome.htm

betrayed. But we don’t know what the questioner expected.