# 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/

durbanville | High School Teacher | (Level 2) Educator Emeritus

Posted on

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`

Sources:

user4825270 | Student | (Level 3) eNoter

Posted on

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