Factorise a quadratic

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Christian Lawson-Perfect 4 years, 7 months ago

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Bill Foster 55 years, 2 months ago

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Question statement

Factorise the following quadratic expression $q(x)$ into linear factors i.e. input $q(x)$ in the form

$(ax+b)(cx+d)$ for suitable integers

$a$ ,

$b$ ,

$c$ and

$d$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
c	integer	5
b	integer	4
d	integer	1
f	integer	8
s3	integer	1
s2	integer	-1
s1	integer	-1
n5	integer	8
disc	number	324
c1	integer	5
rep	string
n1	integer	22
n2	integer	18
n3	integer	16
n4	number	18
rdis	string	The roots exist and are distin
d1	integer	1

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

$q(x)=\;$ 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.

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This question is used in the following exams:

mathcentre: Factorising quadratics by Bill Foster in Bill's workspace.
Test by Letitia Carr in Letitia's workspace.
Tony's test by Tony Croft in Tony's workspace.
Andy Pomfret's Test by Andy Pomfret in Andy's workspace.
Practice exam aesf500 by Adriana Ford-Thompson in Adriana's workspace.
First Exam by Hayley Bishop in Hayley's workspace.
Peter's test exam by Peter Samuels in Peter's workspace.
Maths Support: Factorising quadratics by Christian Lawson-Perfect in Maths Support Wiki.
MLC Support: Factorising quadratics by Paraic Treacy in Paraic's workspace.
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mathcentre: Diagnostic test - Basic maths by Bill Foster in Bill's workspace.
ACC1012 CBAs 1-3 by Man Isha Kaur in Man Isha's workspace.
C1 Revision by Kwame D Dwamena in Kwame D's workspace.
Jack's copy of ACC1012 CBAs 1-3 by Jack Aiston in Jack's workspace.
Bradford: Diagnostic test - Basic maths by Bill Foster in Bill's workspace.
Basic Maths Revision by Gareth Woods in Gareth's workspace.

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Christian Lawson-Perfect 4 years, 7 months ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Factorise a quadratic

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Christian Lawson-Perfect 4 years, 7 months ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`