# Determine the remainder using the remainder theorem: `3x^2+2x-7` `divide (2x-1)` What is the answer for the 2nd equation on question 1) ? http://postimg.org/image/behx7ffhr/

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Using remainder theorem, the answer is `-21/4`

Use the mantra for long division

D (divide) then M(multiply) then S(subtract) then B(bring down):

In this case `2x-1 | 3x^2 +2x-7` renders `3/2x+7/4` rem `-21/4`

First divide `3x^2` by `2x = 3/2 x` then using the above mantra multiply `3/2 x` to get the `3x^2` for subtracting and also multiply the `-1` by `3/2 x` = `-3/2 x` then minus this new expression:

Thus `3/2 x +7/4` rem `-21/4`

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`2x-1|3x^2+2x-7`

`-(3x^2 - 3/2x)`

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`0` `7/2x -7` NB: `2x-(-3/2 x) = 7/2 x`

`-(7/2 x -7/4)` NB:`7/4` comes from `7/2 x divide 2x`

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`0` `-21/4` NB: `-7 -(-7/4) =` `-21/4`

**Ans: **

**`therefore 3x^2+2x-7 = (2x-1)(3/2 x+7/4) -21/4` **

**remainder is `- 21/4` **

let f(x)=3x^2+2x-7

using the remainder theorem, when f(x) is divided by 2x-1,the remainder= f(1/2)

=3(1/2)^2+2(1/2)-7

=3(1/4)+1-7

=3/4-6

=-21/4