View Full Version : Finding the Remainder

Parker

07-24-2009, 02:15 PM

Hi, I think this is in the right place...sorry if it's not...anyways, I'm having trouble with this one; it's one of those things that I'm terminally bad at...any help would be loved. Thanks so much.

What is the remainder when x^3-5x^2+3x+4 is divided by x-1 ?

Subhotosh Khan

07-24-2009, 03:07 PM

Hi, I think this is in the right place...sorry if it's not...anyways, I'm having trouble with this one; it's one of those things that I'm terminally bad at...any help would be loved. Thanks so much.

What is the remainder when x^3-5x^2+3x+4 is divided by x-1 ?

What methods have you been taught/ synthetic division? Long division?

Review synthetic division at:

http://www.purplemath.com/modules/synthdiv.htm

Review long division at:

http://www.purplemath.com/modules/polydiv2.htm

soroban

07-26-2009, 03:24 PM

Hello, Parker!

What is the remainder when x^3-5x^2+3x+4 is divided by x-1 ?

If you know the Remainder Theorem, it's easy.

. . \text{When a polynomial }f(x)\text{ is divided by }(x-a)\text{, the remainder is: }\:f(a)

\text{Dividing by }(x-1)\text{, we have: }\:a = 1

\text{Hence, the remainder is: }\:f(1) \:=\:1^3 - 5(1^2) + 3(1) + 4 \;=\;\boxed{3}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The long division looks like this:

. . . \begin{array}{ccccccccc} &&&& x^2 & - & 4x & - & 1 \\ & & -- & -- & -- & -- & -- & -- \\ x-1 & | & x^3 & - & 5x^2 & + & 3x & + & 4 \\ & & x^3 & - & x^2 \\ &&-- & -- & -- \\ &&& - & 4x^2 & + & 3x \\ &&& - & 4x^2 & + & 4x \\ &&& -- & -- & -- & -- \\ &&&&& - & x & + & 4 \\ &&&&& - & x & + & 1 \\ &&&&& -- & -- & -- & -- \\ &&&&&& && \boxed{3} \end{array}

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