Factorise a quadratic

Error

There was an error loading the page.

Contributors

Feedback

From users who are members of Content created by Newcastle University :

Christian Lawson-Perfect

said

Has some problems

1 year, 11 months ago

From users who are not members of Content created by Newcastle University :

Tess Lynn

said

Ready to use

3 years ago

History

○

Christian Lawson-Perfect commented 1 year, 11 months ago

The pattern restriction doesn't cope with repeated factors

○

Christian Lawson-Perfect 1 year, 11 months ago

Gave some feedback: Has some problems

○

Tess Lynn 3 years ago

Gave some feedback: Ready to use

○

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Created this.

There is only one version of this question that you have access to.

There are 13 other versions that do you not have access to.

Question statement

Factorise the following quadratic expression $q(x)$ into linear factors i.e. input $q(x)$ in the form $(ax+b)(cx+d)$ or of the form $a(x+b)(cx+d)$ for suitable integers $a$ , $b$ , $c$ and $d$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
c	integer	3
b	integer	2
d	integer	1
f	integer	2
s3	integer	1
s2	integer	1
s1	integer	-1
n4	number	5
n2	integer	5
disc	number	25
rdis	string	The roots exist and are distin
n1	integer	7
rep	string
n3	integer	4
c1	integer	3
n5	integer	2
d1	integer	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(1..2)

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`

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Mathematical expression
  Add
1. Step Information only

Gap-fill

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

$q(x)=\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}$

$q(x)=\;$ Unnamed gap

You can get more information on factorising a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Advice

Direct Factorisation.

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

$\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}$

Factorisation by finding the roots.

For if $x=r$ and $x=s$ and are the roots then $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ .

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

$x = \frac{-b +\sqrt{b^2-4ac}}{2a}\mbox{ and } x = \frac{-b -\sqrt{b^2-4ac}}{2a}$
there are three main types of solutions depending upon the value of the discriminant

$\Delta=b^2-4ac$

1. $\Delta \gt 0$ . The roots are real and distinct

2. $\Delta=0$ . The roots are real and equal. Their common value is $-\frac{b}{2a}$

3. $\Delta \lt 0$ . There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\Delta = \simplify{{-(b*c+a*d)}^2-4*{a*b}*{c*d}={disc}}$

{rdis}.

So the {rep} roots are:

$\begin{eqnarray} x = \frac{\var{n1} + \sqrt{\var{disc}}}{\var{n3}} &=& \frac{\var{n1} + \var{n4} }{\var{n3}} &=& \simplify{{n1 + n4}/ {n3}}\\ x = \frac{\var{n1} - \sqrt{\var{disc}}}{\var{n3}} &=& \frac{(\var{n1} - \var{n4}) }{\var{n3}} &=& \simplify{{n1 - n4}/ {n3}} \end{eqnarray}$
So we see that:

$q(x)=\simplify{{a*b}}\left(\simplify{x-{n1 + n4}/ {n3}}\right)\left(\simplify{x-{n1 - n4}/ {n3}}\right)=\simplify{({b} * x + { -d}) * ({a} * x + { -c})}$

Completing the square.

First we complete the square for the quadratic expression $\simplify{{a*b}x^2+{-n1}x+{c*d}}$

$\begin{eqnarray} \simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\var{n5}\left(\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2+ \simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\right)\\ &=&\var{n5}\left(\left(\simplify{x+({-n1}/{2*a*b})}\right)^2 -\simplify{ {n2^2}/{4*(a*b)^2}}\right) \end{eqnarray}$
So to solve

$\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:

$\begin{eqnarray} \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}& -\simplify{ {n2^2}/{4*(a*b)^2}}=0\Rightarrow\\ \left(\simplify{x+({-n1}/{2*a*b})}\right)^2&\phantom{{}}&=\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \end{eqnarray}$
So we get the two {rep} solutions:

$\begin{eqnarray} \simplify{x+({-n1}/{2*a*b})}&=&\simplify{{abs(n2)}/{2*a*b}} \Rightarrow &x& = \simplify{({abs(n2)+n1}/{2*a*b})}\\ \simplify{x+({-n1}/{2*a*b})}&=&\simplify{-({abs(n2)}/{2*a*b})} \Rightarrow &x& = \simplify{({n1-abs(n2)}/{2*a*b})} \end{eqnarray}$
Finding these roots then gives the factorisation as before.

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?
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet
Information only

This question is used in the following exams:

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect commented 1 year, 11 months ago

Christian Lawson-Perfect 1 year, 11 months ago

Tess Lynn 3 years ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Loading...

Error

Factorise a quadratic

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect commented 1 year, 11 months ago

Christian Lawson-Perfect 1 year, 11 months ago

Tess Lynn 3 years ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`