Quadratics: Factorisation 2

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Ben McGovern

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3 years, 9 months ago

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Ben McGovern 2 years, 7 months ago

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Ben McGovern 3 years, 9 months ago

Gave some feedback: Ready to use

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Ben McGovern 4 years, 10 months ago

Created this as a copy of Quadratics: Factorisation 1.

Name	Status	Author	Last Modified
Surds: Simplifying Expressions 1a	Ready to use	Ben McGovern	15/08/2023 15:04
Surds: Simplifying Expressions 1b	Ready to use	Ben McGovern	15/08/2023 15:04
Surds: Simplifying Expressions 2b	Ready to use	Ben McGovern	15/08/2023 15:49
Surds: Simplifying Expressions 2a	Ready to use	Ben McGovern	15/08/2023 15:49
Surds: Simplifying Expressions 3a	Ready to use	Ben McGovern	15/08/2023 15:31
Surds: Simplifying Expressions 3b	Ready to use	Ben McGovern	15/08/2023 15:14
Surds: Simplifying Expressions 4a	Ready to use	Ben McGovern	15/08/2023 15:29
Surds: Simplifying Expressions 4b	Ready to use	Ben McGovern	15/08/2023 15:26
Surds: Simplifying Expressions 4c	Ready to use	Ben McGovern	15/08/2023 15:50
Surds: Simplifying Expressions 5a - Rationalising the Denominator	Ready to use	Ben McGovern	15/08/2023 15:55
Surds: Simplifying Expressions 5b - Rationalising the Denominator	Ready to use	Ben McGovern	15/08/2023 16:07
Surds: Simplifying Expressions 6 - Rationalising the Denominator	Ready to use	Ben McGovern	21/09/2022 16:31
Surds: Simplifying Expressions 7 - Rationalising the Denominator	Ready to use	Ben McGovern	21/09/2022 16:31
Quadratics: Using the Quadratic Formula	Ready to use	Ben McGovern	14/09/2022 16:14
Quadratics: Factorisation 1	Ready to use	Ben McGovern	14/08/2023 15:01
Quadratics: Factorisation 2	Ready to use	Ben McGovern	14/08/2023 15:14
Quadratics: Completing the Square 1	Ready to use	Ben McGovern	14/08/2023 14:59
Quadratics: Completing the Square 2	Ready to use	Ben McGovern	14/09/2022 16:03
Quadratics: Interpreting the Discriminant	Ready to use	Ben McGovern	14/09/2022 16:08
Quadratics: Factorisation 3 - Difference of 2 squares	Ready to use	Ben McGovern	14/08/2023 15:33
Indices: Simplification 1a	Ready to use	Ben McGovern	26/06/2023 12:12
Indices: Simplification 1b	Ready to use	Ben McGovern	26/06/2023 14:31
Indices: Simplification 1c	Ready to use	Ben McGovern	26/06/2023 14:30
Indices: Simplification 2a	Ready to use	Ben McGovern	26/06/2023 14:33
Indices: Simplification 2b	Ready to use	Ben McGovern	26/06/2023 14:36
Indices: Simplification 3a	Ready to use	Ben McGovern	26/06/2023 14:37
Indices: Simplification 3b	Ready to use	Ben McGovern	26/06/2023 14:40
Indices: Change of Notation 1	Ready to use	Ben McGovern	15/08/2023 11:43
Indices: Change of Notation 2	Ready to use	Ben McGovern	15/08/2023 11:51
Indices: Change of Notation 3	Ready to use	Ben McGovern	15/08/2023 11:54
Solve Quadratics - Factorisation 1	Ready to use	Ruth Hand	14/08/2023 15:44
Quadratics: Solving a Quadratic Equation	Ready to use	Ben McGovern	09/12/2022 16:45
Quadratics: Solving by Factorisation	Ready to use	Ben McGovern	14/09/2022 16:25
AD2 Quadratics - solve	Ready to use	Megan Oliver	09/04/2024 15:13
AD5 Difference of two squares	Ready to use	Megan Oliver	09/04/2024 15:14
AA2 Indices - divide	Ready to use	Andrew Neate	09/04/2024 15:00
AA4 Indices - Fractional 2	Ready to use	Andrew Neate	09/04/2024 15:01
AA3 Indices - Fractional 1	Ready to use	Andrew Neate	09/04/2024 15:01
AA1 Indices - times	Ready to use	Andrew Neate	28/03/2024 13:17
Calculation with indices - times	Ready to use	Will Morgan	26/06/2024 11:53
Calculation with Indices - divide	Ready to use	Will Morgan	26/06/2024 12:07

There are 209 other versions that do you not have access to.

Question statement

Factorise the following quadratic expression:

$\simplify[unitFactor]{{a}x^2+{b}x+{c}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
b	integer	19
c	integer	-14
m	integer	3
n	integer	1
p	integer	-2
q	integer	7

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Name

Data type

Value

xxxxxxxxxx
 
m*n

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Generated value: `integer`

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Advice

For some quadratic expressions of the form

$ax^2+bx+c,$ where

$a \neq 1$ , we are able to factorise the expression into the form

$(mx+p)(nx+q),$ for integers

$m$ ,

$n$ ,

$p$ , and

$q$ .

This is more difficult than when $a=1$ , but there is a process which allows us to factorise certain quadratics of this type.

The first step is to identify two numbers which multiply to give $ac$ and add to give $b$ .

So, for $\simplify[unitFactor]{{a}x^2+{b}x+{c}}$ , we have

$a \times c = \var{a*c} \quad\text{and}\quad b=\var{b}.$

In this case, the numbers we need are $\var{m*q}$ and $\var{n*p}$ .

Next, we want to rewrite our quadratic expression so that we have 2 linear terms with these coefficients:

$\simplify[unitFactor,!canonicalOrder]{{a}x^2+{m*q}x+{n*p}x+{c}}.$

We are now able to factor the first two terms and the last two terms:

$\simplify[unitFactor,!canonicalOrder]{{m}x({n}x+{q})+{p}({n}x+{q})}.$

We can see that we now have a common factor of $(\simplify{{n}x+{q}})$ , so rewriting our expression in terms of this factor, we get:

$\simplify[unitFactor]{({n}x+{q})({m}x+{p})},$

as required.

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Quadratics: Factorisation 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ben McGovern 2 years, 7 months ago

Ben McGovern 3 years, 9 months ago

Ben McGovern 4 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `integer`