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Let f(x) be a polynomial such that f(0)=4, f(1)=5, and f(2)=10. Find the remainder when f(x) is divided by x(x-1)(x-2).

----Thanks!

 Mar 11, 2020

Best Answer 

 #2
avatar+33661 
+4

The remainder will, in general, be a polynomial of order less than that of the divisor (otherwise it would still be divisible by the divisor). So

 

remainder = u*x2 + v*x + w 

 

The non-remainder part vanishes when x=0, 1 or 2, so 

 

f(0) = 4 = u*0 +v*0 + w, or w = 4

f(1) = 5 = u*1 + v*1 + 4, or u + v = 1

f(2) = 10 = u*4 + v*2 + 4 or 4u + 2v = 6

From these last two equations we get u = 2 and v = -1

 

Remainder = 2x2 - x + 4

 Mar 12, 2020
edited by Alan  Mar 12, 2020
edited by Alan  Mar 12, 2020
edited by Alan  Mar 12, 2020
 #1
avatar+118687 
0

I would like to see someone knowledgable do this one too.   frown

 Mar 12, 2020
 #2
avatar+33661 
+4
Best Answer

The remainder will, in general, be a polynomial of order less than that of the divisor (otherwise it would still be divisible by the divisor). So

 

remainder = u*x2 + v*x + w 

 

The non-remainder part vanishes when x=0, 1 or 2, so 

 

f(0) = 4 = u*0 +v*0 + w, or w = 4

f(1) = 5 = u*1 + v*1 + 4, or u + v = 1

f(2) = 10 = u*4 + v*2 + 4 or 4u + 2v = 6

From these last two equations we get u = 2 and v = -1

 

Remainder = 2x2 - x + 4

Alan Mar 12, 2020
edited by Alan  Mar 12, 2020
edited by Alan  Mar 12, 2020
edited by Alan  Mar 12, 2020
 #4
avatar+118687 
0

Thanks Alan,

I am still not understanding this statement

"The non-remainder part vanishes when x=0, 1 or 2, "

 

Could you expand upon that please?

Melody  Mar 13, 2020
 #5
avatar+33661 
+3

In general we would have f(x) = m(x)*x*(x-1)*(x-2) + u*x2 + v*x + w

 

where m(x)*x*(x-1)*(x-2) is what I meant by the non-remainder part.  Clearly this part will be zero, when x = 0, 1 or 2.

Alan  Mar 13, 2020
 #6
avatar+118687 
0

I finally get it.

Thanks Alan.   cool

Melody  Mar 14, 2020
 #3
avatar+288 
+1

Thanks for the help! laugh

 Mar 12, 2020

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