Simon's copy of Dividing a polynomial with remainders, using the remainder theorem

Error

There was an error loading the page.

Contributors

History

○

Simon Thomas 6 years, 2 months ago

Created this as a copy of Dividing a polynomial with remainders, using the remainder theorem.

Name	Status	Author	Last Modified
Dividing a polynomial with remainders, using the remainder theorem	Ready to use	Elliott Fletcher	20/11/2019 14:37
Remainder theorem and polynomials	draft	Adrian Jannetta	16/08/2023 14:56
Remainder Theorem	draft	Mark Hodds	24/03/2018 16:00
Encontrar el resto cuando $f(x) = ax^3+bx^2+cx+d$ es dividido por $({{a}x+{k}})$	Ready to use	Luis Hernandez	19/06/2023 13:31
Simon's copy of Dividing a polynomial with remainders, using the remainder theorem	draft	Simon Thomas	28/02/2019 12:14
Dividing a polynomial with remainders, using the remainder theorem	draft	Xiaodan Leng	11/07/2019 01:38

There is one other version that you do not have access to.

Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

coef_x3	integer	3
coef_x2	integer	3
coef_x	integer	1
const	integer	5
k	integer	2
remainder	rational	43/9
a	integer	3

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..4)  

JME function reference

Description

Coefficient of x^3

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

remainder

This variable doesn't seem to be used anywhere.

Advice

The remainder theorem states that if a polynomial $f(x)$ is divided by $(\simplify{a*x-b})$ then the remainder is $f \left( \frac{b}{a} \right)$ .

This means that if we substitute $x = \frac{b}{a}$ into the equation for $f(x)$ , the result will be equal to the remainder when $f(x)$ is divided by $(\simplify{a*x-b})$ .

Therefore, to calculate the remainder when $f(x) = \simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\simplify{{a}*x+{k}})$ , we use this same principle.

As we are dividing $f(x)$ by $(\simplify{{a}*x+{k}})$ , using the remainder theorem tells us that substituting

$\begin{align} x &= \frac{b}{a}\\ &= \simplify{-({k}/{a})} \end{align}$

into our equation for $f(x)$ will give us the remainder when $f(x)$ is divided by $(\simplify{{a}*x+{k}})$ . Substituting this value of $x$ into $f(x)$ gives us

$\begin{align} f(\simplify{-({k}/{a})}) &= \simplify[all,!collectNumbers, fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}\\ &= \simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}. \end{align}$

Therefore, the remainder when $f(x) = \simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\simplify{{a}*x+{k}})$ is $\displaystyle\simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}$ .

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Number entry
Number entry		Hasn't run yet
Information only

This question is used in the following exam:

Factor and remainder theorems by Simon Thomas in Algebra.

Loading...

Error

Simon's copy of Dividing a polynomial with remainders, using the remainder theorem

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Number entry

Advanced settings

Loading...

Error

Simon's copy of Dividing a polynomial with remainders, using the remainder theorem

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Number entry

Advanced settings

Generated value: `integer`