# Polynomial Division Problem

Polynomial Division Problem

Calculate the following and state the remainder:

2x^3+x^2-22x+20 divided by 2x-3

Thanks!

#### topsquark

Math Team
Polynomial Division Problem

Calculate the following and state the remainder:

2x^3+x^2-22x+20 divided by 2x-3

Thanks!
How are you to do this? You could use long division or sythnetic division.

What have you been able to work out?

-Dan I calculated using long division and got the following:

x^2+2x-8 remainder -4

This question is from textbook and solution used synthetic division, but answer is different than mine...

Textbook solution gives: x^2+2x-8 remainder -2

#### skeeter

Math Team
$\dfrac{2x^3 +x^2-22x+20}{2x-3} = \dfrac{x^3 + \frac{x^2}{2} - 11x + 10}{x-\frac{3}{2}}$

synthetic division for the right side ...

Code:
[3/2]    1     1/2     -11    10
3/2        3    -12
------------------------------
1      2       -8   | -2
$\dfrac{x^3 + \frac{x^2}{2} - 11x + 10}{x-\frac{3}{2}} = x^2+2x-8 - \dfrac{2}{x-\frac{3}{2}} = x^2 + 2x - 8 - \dfrac{4}{2x-3}$

• 2 people

@ skeeter

Thanks, I see that (-2)/(x-3/2) is equivalent to (-4)/(2x-3)

So is the textbook incorrect? Should the remainder be (-2)/(x-3/2) instead of just -2?

Also, how to solve using long division then?

Here's txt solution using synthetic division:

https://imgur.com/BF0ZgTH

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#### skeeter

Math Team
Code:
              x^2 + 2x  - 8
--------------------------
2x - 3 |  2x^3 + x^2 - 22x + 20
2x^3 - 3x^2
--------------------
4x^2 - 22x + 20
4x^2 -  6x
---------------
-16x + 20
-16x + 24
----------
-4
according to the remainder theorem, the remainder is -4, because ...

$f(x) = 2x^3 + x^2 - 22x + 20 \implies f\left(\frac{3}{2}\right) = -4$

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• 2 people

@ Skeeter

Okay, so textbook is incorrect. Thanks for clarifying!

:-D

But I still don't understand how synthetic division gave incorrect remainder? Doesn't the -2 at the end of your synthetic division represent the remainder?

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#### skeeter

Math Team
-2 is the remainder for $g(x) = x^2 + \dfrac{x^2}{2} - 11x + 10 \, \text{ since }\, g\left(\frac{3}{2}\right) = -2$

$2 \cdot g(x) = f(x) \implies$ the remainder for g(x) is half that for f(x).

It's all a matter of which polynomial is used to determine the remainder ... I support -4 as the remainder since f(x) divided by (2x-3) was the original problem.

• bearnutella