Factorising Quadratics 1 - Basic - Numbas at mathcentre.ac.uk

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9 years, 7 months ago

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Katie Lester 9 years, 7 months ago

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Katie Lester 9 years, 7 months ago

Created this as a copy of Factorising Quadratics.

Name	Status	Author	Last Modified
Factorising Quadratics 1 - Basic	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 2 - Intermediate	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 3 - Challenging	Ready to use	Katie Lester	01/06/2016 09:47
Factorising Quadratics 4 - Solving by Factorising or Using Formula	Ready to use	Katie Lester	01/06/2016 09:47
Factoring 2 Quadratics 2 of form: $ax^2+bx+c$	draft	Clare Lundon	05/10/2016 22:45
Factorising Quadratics 1: $x^2+bx+c$ 3 questions READY	draft	Clare Lundon	05/10/2016 22:46
steve's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	steve kilgallon	18/07/2016 16:05
Lucy's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	Lucy Hadfield	03/08/2016 10:50
Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Francis Duah	12/08/2016 17:56
Gareth's copy of Lucy's copy of Factoring Quadratics 2 $ax^2+bx+c$ 4 questions READY	draft	Gareth Woods	26/08/2016 11:21
Kjell's copy of Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Kjell Hansen	05/09/2016 11:01
Kjell's copy of Kjell's copy of Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Kjell Hansen	05/09/2016 11:02
Solving Quadratic Equations using the Quadratic Formula and a Calculator	draft	Francis Duah	14/09/2016 16:55
Francis 's copy of Solving Quadratic Equations using the Quadratic Formula 4 Questions READY	draft	Francis Duah	14/09/2016 16:50
Factoring 1 Quadratic of form: $ax^2+bx+c$	draft	Clare Lundon	13/10/2016 10:06
Solving Quadratic Equations using the Quadratic Formula	draft	Hannah Bartholomew	18/09/2017 11:52
Ugur's copy of Factorising Quadratics 2 - Intermediate	Ready to use	Ugur Efem	01/11/2021 12:31
Ugur's copy of Factorising Quadratics 1 - Basic	Ready to use	Ugur Efem	04/11/2021 15:33

There are 19 other versions that do you not have access to.

Question statement

Simplify the following quadratics into two linear factors of the form $(ax+c)(bx+d)$ .

Remember, the constants in your expression can be positive or negative.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
b	integer	1
c	integer	-3
d	integer	-1
f1	integer	1
f2	integer	-4
f3	integer	3

			Name	Type	Generated Value

a2	integer	1
b2	integer	1
c2	integer	-2
d2	integer	7
f12	integer	1
f22	integer	5
f32	integer	-14

			Name	Type	Generated Value

a3	integer	1
b3	integer	1
c3	integer	-15
d3	integer	12
f13	integer	1
f23	integer	-3
f33	integer	-180

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
1

JME function reference

Description

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:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
1. Step Information only
Part b) Gap-fill
1. Gap Mathematical expression
  Add
Part c) Gap-fill
1. Gap Mathematical expression
  Add

Gap-fill

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

$\simplify{{f1}x^2+{f2}*x+{f3}}$

$=$ Unnamed gap

Advice

A worked example using the equation from Part a:

$\simplify{{f1}x^2+{f2}*x+{f3}}=0$

Firstly, start to find the possible common factors of the coefficient of $x$ (here, $\var{f2}$ ), and the constant ( $\var{f3}$ ).

Here there are some possible factors of $\var{f3}$ , but we need to find two numbers which will sum to make $\var{f2}$ (since the coefficient of $x^2$ is $1$ in this case).

We can use $\var{c}$ and $\var{d}$ , since

$\var{c}*\var{d}=\var{f3}$

and

$\var{c}+\var{d}=\var{f2}$ .

Once you have found the two numbers which will work, put them into two brackets like so:

$\simplify{(x+{c})(x+{d})}$

This is your simplification.

If you wanted to find the solutions for $x$ , you would take each bracket and equate it to zero.

Some students may find this hard to see instantly, so the method is explained below.

Taking the first bracketed term,

( $\simplify{x+{c}}$ )

and equating it to $0$

$\simplify{x+{c}}=0$

then gives us a very simple equation.

From this, we can see $x=\simplify{0-{c}}$ .

The same goes for the second bracket:

$x=\simplify{0-{d}}$ .

A similar technique is used when the coefficient of $x^2$ is not $1$ .

It simply must be taken into account that the factors will be multiplied by this.

For example, taking the equation $2x^2+7x+5$ :

The only factors of $5$ are $5$ and $1$ .

Creating the empty brackets,

$($ $2x+$ $)($ $x+$ $)$

we can see that they must be filled as follows:

$(2x+5)(x+1)$

as multiplying this out gives the original equation.

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?
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Information only
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Factorising Quadratics 1 - Basic

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Katie Lester 9 years, 7 months ago

Katie Lester 9 years, 7 months ago

Error in variable testing condition

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Generated value: `integer`

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Gap-fill

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Factorising Quadratics 1 - Basic

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Katie Lester 9 years, 7 months ago

Katie Lester 9 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`