Simon's copy of Finding unknown coefficients of a polynomial, using the remainder theorem

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Simon Thomas 6 years, 1 month ago

Created this as a copy of Finding unknown coefficients of a polynomial, using the remainder theorem.

Name	Status	Author	Last Modified
Finding unknown coefficients of a polynomial, using the remainder theorem	Ready to use	Elliott Fletcher	20/11/2019 14:39
Considere el polinomio $p(x)$ con parametros $t$ y $s$ . $p(x) = ax^3+sx^2+t$ , halle los valores de $s$ y $t$ .	Ready to use	Luis Hernandez	19/06/2023 13:40
Simon's copy of Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Simon Thomas	28/02/2019 12:15
Christian's copy of Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Christian Lawson-Perfect	20/05/2019 13:55
Finding unknown coefficients of a polynomial, using the remainder theorem	draft	Xiaodan Leng	10/07/2019 21:44
Finding unknown coefficients of a polynomial by the remainder theorem	Ready to use	Wan Mekwi	11/10/2020 21:25

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Question statement

Consider the polynomial

$p(x) = \simplify{{coef2_x3}x^3+s*x^2+{coef2_x}x+t}\text{.}$

The polynomial:

has remainder $\var{rem1}$ when divided by $(\simplify{x+{c}})$
has remainder $\var{rem2}$ when divided by $(\simplify{x+{d}})$

Give any introductory information the student needs.

			Name	Type	Generated Value

c	integer	2
d	integer	-1
coef2_x3	integer	1
coef2_x	integer	-2
rem1	integer	3
rem2	integer	2
numerator	number	4
denominator	number	3
x	number	4
y	number	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(-3..3 except 0 except 1)

JME function reference

Description

Dividing term 1.

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:

coef2_x
d
denominator
numerator
x

Statement
Advice
"Part a)" - prompt
"Part a)" → "Gap 0." - Minimum accepted value
"Part a)" → "Gap 0." - Maximum accepted value
"Part a)" → "Gap 1." - Minimum accepted value
"Part a)" → "Gap 1." - Maximum accepted value
"Part c)" → "Unnamed gap" - Minimum accepted value
"Part c)" → "Unnamed gap" - Maximum accepted value
"Part d)" → "Unnamed gap" - Minimum accepted value
"Part d)" → "Unnamed gap" - Maximum accepted value

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
1. Step Information only
Part b) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
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Part c) Gap-fill
1. Gap Number entry
  Add
1. Step Information only
Part d) Gap-fill
1. Gap Number entry
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1. Step Information only

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Using the remainder theorem for the remainder when $p(x)$ is divided by $(\simplify{x+{c}})$ , create an equation involving $s$ and $t$ .

Gap 0. $s + t$ = Gap 1..

Advice

We are told that the polynomial:

has a remainder $\var{rem1}$ when divided by $(\simplify{x+{c}})$
has a remainder $\var{rem2}$ when divided by $(\simplify{x+{d}})$

a)

Firstly, substituting $x = \simplify{-{c}}$ into $p(x)$ gives us

$\begin{align} p(\simplify{-{c}}) &= \simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t},\\ &= \simplify[all,fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t}. \end{align}$

But, by the remainder theorem $p(\simplify{-{c}}) = \var{rem1}$ (using the first bullet point), so this becomes

$\begin{align} \simplify[all,fractionnumbers]{{coef2_x3*(-{c})^3}+s*{(-{c})^2}+{coef2_x*(-{c})}+t} &= \var{rem1},\\ \simplify[all,fractionnumbers]{s*{x}+t} &= \simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}}. \end{align}$

b)

Similarly, substituting $x = \simplify{-{d}}$ into $p(x)$ , gives us

$\begin{align} p(\simplify{-{d}}) &= \simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t},\\ &= \simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t}. \end{align}$

But, by the remainder theorem $p(\simplify{-{d}}) = \var{rem2}$ (using the second bullet point), so this becomes

$\begin{align} \simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+s*{(-{d})^2}+{coef2_x*(-{d})}+t} &= \var{rem2},\\ \simplify[all,fractionnumbers]{s*{y}+t} &= \simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}. \end{align}$

c)

We now have two simultaneous equations for $s$ and $t$ :

$\begin{align} \simplify[all,fractionnumbers]{s*{x}+t} = \simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}} \\ \simplify[all,fractionnumbers]{s*{y}+t} = \simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}} \end{align}$

Next, we subtract the second equation from the first equation.

This allows us to cancel out the terms involving $t$ and gives us an equation only in terms of $s$ , which we can then rearrange to find the value of $s$ .

Subtracting the two equations gives

$\simplify{s*{(-{c})^2-(-{d})^2}} = \simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}}.$

Then, we can rearrange this equation so that

$s = \simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}/{{(-c)^2-(-d)^2}}}.$

d)

We can calculate $t$ by substituting our value of $s$ into one of our original simultaneous equations. For example, let's use the equation

$\simplify[all,fractionnumbers]{s*{(-{d})^2}+t} = \simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.$

Substituting our value of $s$ into this equation gives us

$\begin{align} \simplify[all,fractionnumbers,!noleadingMinus]{{numerator/denominator}+t} &= \simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d)}},\\ t &= \simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d) - numerator/denominator}}. \end{align}$

This same answer would've also been obtained if we had substituted our value of $s$ into the other equation instead.

Give a worked solution to the whole question.

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This question is used in the following exam:

Factor and remainder theorems by Simon Thomas in Algebra.

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Simon's copy of Finding unknown coefficients of a polynomial, using the remainder theorem

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Simon Thomas 6 years, 1 month ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

d)

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Simon's copy of Finding unknown coefficients of a polynomial, using the remainder theorem

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

d)

Generated value: `integer`