1.3.5. Factorising Quadratic Expressions

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Name	Status	Author	Last Modified
1.3.5. Factorising Quadratic Expressions	Ready to use	Merryn Horrocks	28/06/2023 06:02
1.3.5. Task 1	Ready to use	Merryn Horrocks	21/09/2023 04:11
1.3.5. Task 2	Ready to use	Merryn Horrocks	21/09/2023 04:11

Question statement

Factorising Quadratic Expressions

Quadratic expressions commonly occur in many areas of mathematics, physics and engineering. Many quadratic expressions can be written as the product of two linear factors and, in this Section, we examine how these factors can be easily found.

Key Point 18

An expression of the form

$ax^2+bx+c\qquad a\neq 0$

where $a$ , $b$ and $c$ are numbers is called a quadratic expression (in the variable $x$ ).

The numbers $b$ and $c$ may be zero but $a$ must not be zero (for, then, the quadratic reduces to a linear expression or constant). The number $a$ is called the coefficient of $x^2$ , $b$ is the coefficient of $x$ and $c$ is called the constant term.

Case 1

Consider the product $(x+1)(x+2)$ . Removing brackets yields $x^2 +3x+2$ . Conversely, we see that the factors of $x^2+3x+2$ are $(x+1)$ and $(x+2)$ . However, if we were given the quadratic expression first, how would we factorise it? The following examples show how to do this but note that not all quadratic expressions can be easily factorised.

To enable us to factorise a quadratic expression in which the coefficient of $x^2$ equals $1$ , we note the following expansion:

$(x + m)(x + n) = x^2 + mx + nx + mn = x^2 + (m + n)x + mn$

So, given a quadratic expression we can think of the coefficient of $x$ as $m+n$ and the constant term as $mn$ . Once the values of $m$ and $n$ have been found the factors can be easily obtained.

Example

Factorise $x^2+4x-5$

Solution

Writing $x^2+4x-5=(x+m)(x+n)=x^2+(m+n)x+mn$ we seek numbers $m$ and $n$ such that $m+ n = 4$ and $mn = -5$ . By trial and error it is not difficult to find that $m = 5$ and $n =-1$ (or, the other way round, $m = -1$ and $n = 5$ ). So we can write

$x^2 + 4x - 5 = (x + 5)(x - 1)$
The answer can be checked easily by removing brackets.

Give any introductory information the student needs.

			Name	Type	Generated Value

q1expr	expression	x^2 - 4x - 32
q1m	integer	4
q1n	integer	-8

			Name	Type	Generated Value

q2prime	integer	5
q2m	integer	-6
q2n	integer	-6
q2b	integer	-36
q2c	integer	36
q2v	string	k
q2expr	expression	5*k^2 - 36k + 36

			Name	Type	Generated Value

alphabet

list

List of 21 items

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
simplify(expression(
  "x^2 + " + (q1m+q1n) + "x+" + (q1m*q1n)
),["basic","zeroFactor","zeroTerm","unitFactor"])

JME function reference

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

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Generated value: `expression`

x^2 - 4x - 32

← Depends on:

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Advice

Task 1:

$\var{q1expr} =( \var{ simplify(expression("x+"+q1m),"basic") })( \var{ simplify(expression("x+"+q1n),"basic") })$

Task 2:

$\var{q2expr} =( \var{ simplify(expression(q2prime +q2v+ "+"+q2m),"basic") })( \var{ simplify(expression(q2v+"+"+q2n),"basic") })$

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Merryn Horrocks 1 year, 9 months ago

Merryn Horrocks 2 years, 10 months ago

Merryn Horrocks 2 years, 10 months ago

Factorising Quadratic Expressions

Key Point 18

Case 1

Example

Solution

Error in variable testing condition

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Generated value: `expression`

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Mathematical expression

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1.3.5. Factorising Quadratic Expressions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Merryn Horrocks 1 year, 9 months ago

Merryn Horrocks 2 years, 10 months ago

Merryn Horrocks 2 years, 10 months ago

Factorising Quadratic Expressions

Key Point 18

Case 1

Example

Solution

Error in variable testing condition

Error

Warning

Generated value: expression

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `expression`