factorise a quadratic

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From users who are members of Fundamentals of Mathematics :

Daniel Mansfield

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6 years, 5 months ago

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Daniel Mansfield 6 years, 5 months ago

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Daniel Mansfield 6 years, 7 months ago

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Daniel Mansfield 6 years, 7 months ago

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Daniel Mansfield 6 years, 8 months ago

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Name	Status	Author	Last Modified
factorise a quadratic	Ready to use	Daniel Mansfield	18/12/2019 00:35
Laura's copy of factorise a quadratic	draft	Laura Horton	07/02/2019 13:18
Gizem's copy of factorise a quadratic (order of answers etc)	draft	Gizem Intepe	27/08/2019 00:58
ahmad's copy of factorise a quadratic	draft	ahmad yahya	11/06/2019 10:03
factorize a quadratic	draft	Xiaodan Leng	10/07/2019 22:48
SGHS- factorise a quadratic	Ready to use	john Springall	29/02/2020 23:50
Evaluatie: ontbinden in factoren	draft	Alexander Holvoet	02/06/2024 14:33

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Question statement

Consider the quadratic $\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b	integer	3
c	integer	-1

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Description

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

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Generated value: `integer`

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1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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What are the the two linear factors of this quadratic?

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Advice

You can use the quadratic formula to deduce that $\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ has roots:

$x = \frac{\simplify{-({a}*{c}+{b})}\pm\sqrt{ (\var{a*c+b})^2 - 4\times(\var{a})\times(\var{b*c}) }}{2\times \var{a}} = \var{-1*c} \text{ or } \displaystyle \simplify{-1*{b}/{a}}.$

The roots determine the factors, but only upto a constant. In general, a quadratic with roots $\var{-1*c}$ and $\simplify{-1*{b}/{a}}$ has the form:

$C \times (x + \simplify{{b}/{a}}) \times (x - \var{-1*c})$

for some constant term $C$ . The only thing left to do is determine the value of the constant which makes:

$C \times (x + \simplify{{b}/{a}}) \times (x - \var{-1*c}) = \simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ .

Equating the coefficients of the $x^2$ terms in the left and right hand sides shows that $C=\var{a}$ . So

$(\var{a}x + \var{b}) \times (x-\var{-1*c}) = \simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ .

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Daniel Mansfield 6 years, 5 months ago

Daniel Mansfield 6 years, 7 months ago

Daniel Mansfield 6 years, 7 months ago

Daniel Mansfield 6 years, 8 months ago

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Generated value: `integer`

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factorise a quadratic

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Daniel Mansfield 6 years, 5 months ago

Daniel Mansfield 6 years, 7 months ago

Daniel Mansfield 6 years, 7 months ago

Daniel Mansfield 6 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`