Learning Goal
Syllabus Requirement
Students need to learn the technique of
completing the square and the solution of
quadratic equations by:

Factorisation.

Use of the formula.

Completing the square.
Study Notes
There are 3 different ways to solve quadratic functions and you should be familiar with them all, as in the exam you may be asked to use a specific method. The methods are the quadratic equation, completing the square and factorisation and they are all explained in detail below.
The 3 Methods for Solving a Quadratic Function
Method 1: Using the formula
Recall that for a quadratic function of the form ??ax^{2} + bx + c = 0?? the roots (solutions) can be found by using the formula:
$$x = \frac{b\pm \sqrt{b^{2}  4ac}}{2a}$$
Method 2: Completing the Square
Completing the square allows you to solve a quadratic function, ??ax^{2} + bx + c = 0??:

First, we need to ensure that ??a = 1?? in our quadratic function, if it does not divide all of the terms in the quadratic function by ??a?? to ensure that ??a = 1??, from now on the terms ??b?? and ??c?? should be taken from the quadratic function where ??a = 1??.

Completing the square relies on knowing that: $$x{^2} + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^{2}$$

To complete the square on the left hand side we need to add ??\left(\frac{b}{2}\right)^{2}??, which we must also add to the right hand side as well to keep the equation valid, we can also move the ??c?? to the right hand side:
$$x^{2} + bx + c = 0$$
$$\implies x^{2} + bx =  c$$
$$\implies x^{2} + bx + \left(\frac{b}{2}\right)^{2} = c + \left(\frac{b}{2}\right)^{2}$$

Next complete the square on the left hand side:
$$x^{2} + bx + \left(\frac{b}{2}\right)^{2} = c + \left(\frac{b}{2}\right)^{2}$$
$$\implies \left(x + \frac{b}{2}\right)^{2} = c + \left(\frac{b}{2}\right)^{2}$$

Now take the square root of both sides:
$$\left(x + \frac{b}{2}\right)^{2} = c + \left(\frac{b}{2}\right)^{2}$$
$$\implies x + \frac{b}{2} = \pm \sqrt{c + \left(\frac{b}{2}\right)^{2}}$$

Finally, rearrange for ??x??:
$$x + \frac{b}{2} = \pm \sqrt{c + \left(\frac{b}{2}\right)^{2}}$$
$$\implies x = \frac{b}{2} \pm \sqrt{c + \left(\frac{b}{2}\right)^{2}} $$
Method 3: Factorisation
Not all quadratic equations can be factorised, but when they can be it is usually the most straightforward and errorfree method for solving an equation of the form ??ax^{2} + bx + c = 0??.
First factorise into the product of two linears:
$$ax^{2} + bx + c =(dx+e)(fx+g)$$
Now since ??(dx+e)(fx+g)=0?? it must be the case that either ??(dx+e)=0?? or ??(fx+g)=0??.
When ??(dx+e)=0?? $$dx=e \implies x=\frac{e}{d}$$
When ??(fx+g)=0?? $$fx=g \implies x=\frac{g}{f}$$
In equations where the coefficient of ??x^2?? is equal to 1, factorisation is simpler: $$x^{2} + bx + c = (x + e)(x + g) = 0$$
Now $$(x+e)=0 \implies x=e$$
and $$(x+g)=0 \implies x=g$$
Worked Examples
We'll now run through some examples that cover each of the methods:
Using the formula
Example 1
Question
Solve the quadratic function ??x^{2}  5x + 6 = 0?? for ??x?? by using the formula.
Show Solution
Worked Solution
Note that ??a = 1??, ??b = 5?? and ??c = 6??, all that remains is to substitute these values into the quadratic equation:
$$x = \frac{b \pm \sqrt{b^{2}4ac}}{2a}$$
$$ = \frac{(5) \pm \sqrt{(5)^{2}  4 \times 1 \times 6}}{2 \times 1}$$
$$ = \frac{5 \pm \sqrt{25  24}}{2}$$
$$ = \frac{5 \pm \sqrt{1}}{2}$$
$$ = \frac{5 \pm 1}{2}$$
$$\therefore x = \frac{5 + 1}{2}\mathrm{\: or \:} x = \frac{5  1}{2}$$
$$\implies x = 3 \mathrm{\: or \:} x = 2$$
Completing The Square
Example 2
Question
Solve the quadratic function ??x^{2}  5x + 6 = 0?? for ??x?? by completing the square.
Show Solution
Worked Solution
Start by moving the ??c?? term to the right hand side of the equation before completing the square on the left hand side and solving:
$$x^{2}  5x + 6 = 0$$
$$\implies x^{2}  5x = 6$$
Recall that ?? \left(x + \frac{b}{2}\right)^{2} = c + \left(\frac{b}{2}\right)^{2}??
$$\implies \left(x  \frac{5}{2}\right)^{2} = 6 + \left(\frac{5}{2}\right)^{2}$$
$$\implies \left(x  \frac{5}{2}\right)^{2} = 6 + 6.25$$
$$\implies \left(x  \frac{5}{2}\right)^{2} = 0.25$$
Next take the square root of both sides and then rearrange for ??x??:
$$\implies \left(x  \frac{5}{2} \right) = \pm \sqrt{0.25} $$
$$\implies x = 2.5 \pm 0.5$$
$$\therefore x = 2.5 + 0.5 \mathrm{\: or \:} x = 2.5  0.5$$
$$\implies x = 3 \mathrm{\: or \:} x = 2$$
Factorisation
Example 3
Question
Solve the quadratic function ??x^{2}  5x + 6 = 0?? for ??x?? by using factorisation.
Show Solution
Worked Solution
First inspect the equation given ??x^{2}  5x + 6 = 0?? and compare to ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c} = 0??
??\implies \color{red}{a} = 1??
??\implies \color{green}{b} = 5??
??\implies \color{blue}{c} = 6??
Let's designate the factorised solution to be ??(dx + e)(fx + g)?? and note the equations linking ??a??, ??b?? and ??c?? to ??d??, ??e??, ??f?? and ??g??:
Recall ??\color{red}{a} = d \times f??
??\implies 1 = d \times f??
We know that ??d?? and ??f?? will be integers and therefore can conclude that ??d = f = 1??.
Recall ??\color{green}{b} = (e \times f) + (g \times d)??
??\implies 5 = e \times 1 + g \times 1??
??\implies 5 = e + g??
We now know that ??e?? and ??g?? must sum to ??5??.
Recall ??\color{blue}{c} = e \times g??
??\implies 6 = e \times g??
We now know that the product of ??e?? and ??g?? must equal ??6??
Consider the factors of ??6?? and test if they meet the criteria above. With more practice you will quickly spot the correct factors.
 ??e =+6?? and ??g =+1??
?? \implies +6 \times +1 = 6?? ??6 + 1 = 7??
 ??e =6?? and ??g =1??
?? \implies 6 \times 1 = 6?? ??6  1 = 7??
 ??e =+2?? and ??g =+3??
?? \implies +2 \times +3 = 6?? ??2 + 3 = 5??
 ??e =2?? and ??g =3??
?? \implies 2 \times 3 = 6?? ??2  3 = 5??
Therefore we can say
??e = 2?? and ??g = 3?? (or the other way around, it doesn't matter).
Substituting into our factorised solution equation gives: $$(x  2)(x  3) = 0$$
$$\therefore x = 2 \mathrm{\: or \:} x = 3$$
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2005 
June 
Core 1 
3 
n/a 
6 



Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
1 
n/a 
3 



Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
9 
(a) 
2 



Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
10 
(a) 
2 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
9 
(a), (b) 
5 



Edexcel 
Maths 
Standard 
2008 
June 
Core 1 
2 
n/a 
3 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
10 
(a) 
3 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
9 
(a) 
3 



Edexcel 
Maths 
Standard 
2010 
January 
Core 1 
10 
(a) 
3 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
4 
(a) 
2 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
10 
(a) 
1 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
1 
n/a 
3 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
10 
(a) 
3 


Edexcel  Maths  Standard  2005  June  Core 1  Question 3
Question
3.
$$x^{2}  8x  29 \equiv (x + a)^{2} + b$$
Where ??a?? and ??b?? are constants.
(a)
Find the value of ??a?? and the value of ??b??. [3]
(b)
Hence, or otherwise, show that the roots of
$$x^{2}  8x  29 = 0$$
are ??c \pm d \sqrt{5}??, where ??c?? and ??d?? are integers to be found. [3]
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Show Solution
Worked Solution
3. (a)
Start by expanding out the brackets on the right hand side of the equation given:
??x^{2}  8x  29 \equiv (x + a)^{2} + b??
??\equiv x^{2} + ax + ax + a^{2} + b??
??\equiv x^{2} + 2ax + a^{2} + b??
??\equiv x^{2} + 2ax + (a^{2} + b)??
Now comparing coefficients:
$$x^{2} + \color{green}{2a}x + \color{blue}{(a^{2} + b)} \equiv x^{2} \color{green}{ 8}x \color{blue}{ 29}$$
Firstly of the ??x?? term:
??\implies 2a = 8??
??\therefore a = 4??
Secondly the
constant term:
??\implies a^{2} + b = 29??
Substitute in the value of ??a?? from above:
??\implies (4)^{2} + b = 29??
??\implies 16 + b = 29??
??\implies b = 29  16??
??\therefore b = 45??
For completeness we should state
??a = 4?? and
??b = 45??
3. (b)
Start by substituting in the values for ??a?? and ??b?? found in part
3. (a)
??x^{2}  8x  29 = 0 = \left ( x  4 \right )^{2}  45??
??\implies \left ( x  4 \right )^{2}  45 = 0??
??\implies \left ( x  4 \right )^{2} = 45??
??\implies \left ( x  4 \right ) = \pm \sqrt{45}??
??\implies x = 4 \pm \sqrt{45}??
Recall that the expression you are trying to achieve is in the form ??c \pm d \sqrt{5}??.
Therefore, from the
The Use and Manipulation of Surds we will use Law 3, which states:
$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$
To rewrite ??\sqrt{45}?? as a product of two surds, one being ??\sqrt{5}??:
??\sqrt{45}??
??=\sqrt{5 \times 9}??
??=\sqrt{5} \times \sqrt{9}??
??=\sqrt{5} \times 3??
??=3\sqrt{5}??
Therefore the final solution is:
??\implies x = 4 \pm 3\sqrt{5}??
For completeness, we should state that ??c = 4?? and ??d = 3??.
Edexcel  Maths  Standard  2006  January  Core 1  Question 1
Question
1.
Factorise completely ??x^{3}  4x^{2} + 3x??. [3]
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Show Solution
Worked Solution
1.
Start by taking a factor of ??x?? from the equation
??x^{3}  4x^{2} + 3x??
??\implies x (x^{2}  4x + 3)??
Now we need to factorise the following:
??x^{2}  4x + 3??
Recall That where the factor ??\color{red}{a}?? of ??x^{2}?? in the
quadratic is equal to 1 the factorised solution takes the form: ??x^{2} + \color{green}{b}x + \color{blue}{c} = (x + e)(x + g)??
Therefore we are looking for:
??\color{green}{b} = e + g??
??\implies 4 = e + g??
We now know that ??e?? and ??g?? must sum to ??4??.
??\color{blue}{c} = e \times g??
??\implies 3 = e \times g??
We now know that the product of ??e?? and ??g?? must equal ??3??
Consider the the factors of ??3?? and test if they meet the criteria above:
 ??e =+1?? and ??g =+3??
?? \implies +1 \times +3 = 3?? ??1 + 3 = 4??
 ??e =1?? and ??g =3??
?? \implies 1 \times 3 = 3?? ??1  3 = 4??
Therefore we can say
??e = 1?? and
??g = 3?? (or the other way around, it doesn't matter).
Substituting into our factorised solution equation gives: $$(x  1)(x  3)$$
Finally substitute this into the original equation:
$$x(x  1)(x  3)$$
Edexcel  Maths  Standard  2006  January  Core 1  Question 9
Question
9.
Figure 2.
Figure 2 shows part of the curve C with the equation
$$y = (x  1)(x^{2}  4)$$
The curve cuts the ??x??axis at the points ??P??, ??(1, 0)?? and ??Q??, as shown in Figure 2.
(a) Write down the ??x??coordinate of ??P?? and the ??x??coordinate of ??Q??. [2]
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Worked Solution
9. (a)
Start by noting that the question is asking for the ??x??coordinate of ??P?? and ??Q?? and at these points ??y = 0??
$$\therefore (x  1)(x^{2}  4) = 0$$
Factorising the quadratic section of the equation ??(x^{2}  4)??
Note that this quadratic is in the form ??(x^{2}  p^{2})?? can easily be factorised into ??(x + p)(x  p)??.
See exam tips below for more details.
Inspecting ??(x^{2}  4)??, realise that ??4?? can easily be expressed in the form ??p^{2}?? where ??p = 2??
$$\implies (x^{2}  4) = (x^{2}  2^{2})$$
So can easily be factorised in the form ??(x + p)(x  p)?? where ??p = 2?? from above
$$\implies (x^{2}  4) = (x^{2}  2^{2}) = (x + 2)(x  2)$$
Substituting this into the original equation gives:
$$\therefore (x  1)(x^{2}  4) = (x  1)(x + 2)(x  2) = 0$$
For the equation to equal ??0?? either:
??(x  1) = 0??
??\implies x = 1?? which is given in the question
??(x + 2) = 0??
??\implies x = 2??
??(x  2) = 0??
??\implies x = 2??
For completeness we should state that the ??x??coordinate of ??P = 2?? and the ??x??coordinate of ??Q = 2??.
Edexcel  Maths  Standard  2006  January  Core 1  Question 10
Question
10.
$$x^{2} + 2x + 3 \equiv (x + a)^{2} + b$$
(a)
Find the values of the constants ??a?? and ??b??. [2]
Hide Question
Show Solution
Worked Solution
10. (a)
Start expanding the brackets on the right hand side of the equation given:
??x^{2} + 2x + 3 \equiv (x + a)^{2} + b??
??\equiv x^{2} + ax + ax + a^{2} + b??
??\equiv x^{2} + 2ax + a^{2} + b??
??\equiv x^{2} + 2ax + (a^{2} + b)??
Now comparing coefficients:
Firstly of the ??x?? term:
??\implies 2a = 2??
??\therefore a = 1??
Secondly the constant term:
??\implies a^{2} + b = 3??
Substitute in the value of ??a?? from above:
??\implies (1)^{2} + b = 3??
??\implies 1 + b = 3??
??\implies b = 3  1??
??\therefore b = 2??
For completeness we should state ??a = 1?? and ??b = 2??
Edexcel  Maths  Standard  2006  June  Core 1  Question 9
Question
9.
Given that ??f(x) = (x^{2}  6x)(x  2) + 3x??
(a)
Express ??f(x)?? in the form ??x(ax^{2} + bx + c)??, where ??a??, ??b?? and ??c?? are constants. [3]
(b)
Hence factorise ??f(x)?? completely. [2]
Hide Question
Show Solution
Worked Solution
9. (a)
Lets start by expanding the brackets on the right hand side of the equation given:
??f(x) = (x^{2}  6x)(x  2) + 3x??
??\implies x^{3}  2x^{2}  6x^{2} + 12x + 3x??
Now collect the like terms together:
??\implies x^{3} + (2  6)x^{2} + (12 + 3)x??
??\implies x^{3}  8x^{2} + 15x??
Remember The question asked for ??f(x)?? in the form ??x(ax^{2} + bx + c)??, therefore we need to take a factor of ??x?? out:
??\implies x(x^{2}  8x + 15)??
$$f(x) = x(x^{2}  8x + 15)$$
For completeness we should state
??a = 1??,
??b = 8?? and
??c = +15??
9. (b)
Now we need to factorise the quadratic section of the equation:
$$f(x) = x(\color{red}{x^{2}  8x + 15})$$
So now lets factorise ??x^{2}  8x + 15??
Recall That where the factor ??\color{red}{a}?? of ??x^{2}?? in the
quadratic is equal to 1 the factorised solution takes the form: ??x^{2} + \color{green}{b}x + \color{blue}{c} = (x + e)(x + g)??
Therefore we are looking for:
??\color{green}{b} = e + g??
??\implies 8 = e + g??
We now know that ??e?? and ??g?? must sum to ??8??.
??\color{blue}{c} = e \times g??
??\implies 15 = e \times g??
We now know that the product of ??e?? and ??g?? must equal ??15??
Consider the the factors of ??15?? and test if they meet the criteria above:
 ??e =+1?? and ??g =+15??
?? \implies +1 \times +15 = 15?? ??1 + 15 = 16??
 ??e =1?? and ??g =15??
?? \implies 1 \times 15 = 15?? ??1  15 = 16??
 ??e =+5?? and ??g =+3??
?? \implies +5 \times +3 = 15?? ??5 + 3 = 8??
 ??e =5?? and ??g =3??
?? \implies 5 \times 3 = 15?? ??5  3 = 8??
Therefore we can say
??e = 5?? and
??g = 3?? (or the other way around, it doesn't matter).
Substituting into our factorised equation gives: $$(x + e)(x + g) = (x  5)(x  3)$$
Finally substituting this into the original equation we can say:
$$f(x) = x(x  5)(x  3)$$
Edexcel  Maths  Standard  2008  June  Core 1  Question 2
Question
2.
Factorise completely. [3]
$$(x^{3}  9x)$$
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Worked Solution
2.
Lets start by removing a factor of ??x?? from the equation given
$$(x^{3}  9x) = x(x^{2}  9)$$
Now we need to factorise the quadratic section of the equation:
$$x(\color{red}{x^{2}  9})$$
Note that this quadratic is in the form ??(x^{2}  p^{2})?? and this can easily be factorised into ??(x + p)(x  p)??.
See exam tips below for more details.
Inspecting ??(x^{2}  9)??, realise that ??9?? can easily be expressed in the form ??p^{2}?? where ??p = 3??
$$\implies (x^{2}  9) = (x^{2}  3^{2})$$
So can easily be factorised in the form ??(x + p)(x  p)?? where ??p = 3?? from above
$$\implies (x^{2}  9) = (x^{2}  3^{2}) = (x + 3)(x  3)$$
Substituting this into the original equation gives us the fully factorised solution:
$$\therefore (x^{3}  9x) = x(x + 3)(x  3)$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 10
Question
10. (a)
Factorise completely ??x^{3}  6x^{2} + 9x??. [3]
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Show Solution
Worked Solution
10. (a)
Lets start by removing a factor of ??x?? from the equation given
$$x^{3}  6x^{2} + 9x = x(x^{2}  6x + 9)$$
Now we need to factorise the quadratic section of the equation:
$$x(\color{red}{x^{2}  6x + 9})$$
Recall That where the factor ??\color{red}{a}?? of ??x^{2}?? in the
quadratic is equal to 1 the factorised solution takes the form: ??x^{2} + \color{green}{b}x + \color{blue}{c} = (x + e)(x + g)??
Therefore we are looking for:
??\color{green}{b} = e + g??
??\implies 6 = e + g??
We now know that ??e?? and ??g?? must sum to ??6??.
??\color{blue}{c} = e \times g??
??\implies 9 = e \times g??
We now know that the product of ??e?? and ??g?? must equal ??9??
Consider the the factors of ??9?? and test if they meet the criteria above:
 ??e =+1?? and ??g =+9??
?? \implies +1 \times +9 = 9?? ??1 + 9 = 10??
 ??e =1?? and ??g =9??
?? \implies 1 \times 9 = 9?? ??1  9 = 10??
 ??e =+3?? and ??g =+3??
?? \implies +3 \times +3 = 9?? ??3 + 3 = 6??
 ??e =3?? and ??g =3??
?? \implies 3 \times 3 = 9?? ??3  3 = 6??
Therefore we can say
??e = 3?? and
??g = 3??.
Substituting into our factorised equation gives: $$(x + e)(x + g) = (x  3)(x  3)$$
Finally substituting this into the original equation we can say:
$$x^{3}  6x^{2} + 9x = x(x  3)(x  3)$$
Edexcel  Maths  Standard  2010  January  Core 1  Question 9
Question
9. (a)
Factorise completely ??x^{3}  4x??. [3]
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Worked Solution
9. (a)
Lets start by removing a factor of ??x?? from the equation given
$$(x^{3}  4x) = x(x^{2}  4)$$
Now we need to factorise the quadratic section of the equation:
$$x(\color{red}{x^{2}  4})$$
Note that this quadratic is in the form ??(x^{2}  p^{2})?? and this can easily be factorised into ??(x + p)(x  p)??.
See exam tips below for more details.
Inspecting ??(x^{2}  4)??, realise that ??4?? can easily be expressed in the form ??p^{2}?? where ??p = 2??
$$\implies (x^{2}  4) = (x^{2}  2^{2})$$
So can easily be factorised in the form ??(x + p)(x  p)?? where ??p = 2?? from above
$$\implies (x^{2}  4) = (x^{2}  2^{2}) = (x + 2)(x  2)$$
Substituting this into the original equation gives us the fully factorised solution:
$$\therefore (x^{3}  4x) = x(x + 2)(x  2)$$
Edexcel  Maths  Standard  2010  January  Core 1  Question 10
Question
10.
??f(x) = x^{2} + 4kx + (3 + 11k)??, where ??k?? is a constant.
(a)
Express ??f(x)?? in the form ??(x + p)^{2} + q??, where ??p?? and ??q?? are constants to be found in terms of k. [3]
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Worked Solution
10. (a)
Start by expanding out the brackets of the equation of the required form given in the question. This will allow us to compare coefficients and hence find ??p?? and ??q??:
$$(x + p)^{2} + q$$
??\implies x^{2} + px + px + p^{2} + q??
Now collect the like terms together:
??\implies x^{2} + 2px + (p^{2} + q)??
Now comparing coefficients in this equation to the function given in the question, ??f(x) = x^{2} + 4kx + (3 + 11k)??:
$$f(x) = x^{2} + \color{green}{2p}x + \color{blue}{(p^{2} + q)} = x^{2} + \color{green}{4k}x + \color{blue}{(3 + 11k)}$$
Firstly of the ??x?? term:
??\implies \color{green}{2p} = \color{green}{4k}??
??\therefore p = 2k??
Secondly the constant term:
??\implies \color{blue}{p^{2} + q} = \color{blue}{3 + 11k}??
Substitute in the value of ??p?? from above:
??\implies (2k)^{2} + q = 3 + 11k??
??\implies 4k^{2} + q = 3 + 11k??
??\implies q = 4k^{2} + 3 + 11k??
Remember The question asked us to express ??f(x)?? in the form ??(x + p)^{2} + q??.
Substitute in the values of ??p?? and ??q?? found above to give the final solution:
$$\implies f(x) = (x + p)^{2} + q = (x + 2k)^{2}  4k^{2} + 11k + 3$$
Edexcel  Maths  Standard  2010  June  Core 1  Question 4
Question
4. (a)
Show that ??x^{2} + 6x + 11?? can be written as
$$(x + p)^{2} + q$$
where ??p?? and ??q?? are integers to be found. [2]
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Worked Solution
4. (a)
Start by expanding out the brackets of the equation of the required form given in the question. This will allow us to compare coefficients and hence find ??p?? and ??q??:
$$(x + p)^{2} + q$$
??\implies x^{2} + px + px + p^{2} + q??
Now collect the like terms together:
??\implies x^{2} + 2px + (p^{2} + q)??
Now comparing coefficients to the equation given in the question, ??x^{2} + 6x + 11??:
$$f(x) = x^{2} + \color{green}{2p}x + \color{blue}{(p^{2} + q)} = x^{2} + \color{green}{6}x + \color{blue}{11}$$
Firstly of the ??x?? term:
??\implies \color{green}{2p} = \color{green}{6}??
??\therefore p = 3??
Secondly the constant term:
??\implies \color{blue}{p^{2} + q} = \color{blue}{11}??
Substitute in the value of ??p?? from above:
??\implies (3)^{2} + q = 11??
??\implies 9 + q = 11??
??\implies q = 11  9??
??\implies q = 2??
For completeness we should state ??p = 3?? and ??q = 2??
Edexcel  Maths  Standard  2012  June  Core 1  Question 10
Question
10.
Figure 1.
Figure 1 shows a sketch of the curve C with equation ??y =f(x)?? where
$$f(x) = x^{2}(9  2x)$$
There is a minimum at the origin, a maximum at the point ??(3, 27)?? and ??C?? cuts the ??x??axis at the point ??A??.
(a) Write down the coordinates of ??A??. [1]
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Worked Solution
10. (a)
Start by noting that the question is asking for the coordinates of ??A?? where the curve C crosses the ??x??axis, at which point ??y = 0??
$$\therefore x^{2}(9  2x) = 0$$
For the equation to equal ??0?? either:
$$x^{2} = 0$$
??\implies x = 0?? which is given in the question. (Minimum at the origin)
Or
$$9  2x = 0$$
??\implies 2x = 9??
??\implies 2x = 9??
??\implies x = \frac{9}{2} = 4.5??
For completeness we should state that the coordinates of ??A = (4.5, 0)??.
Edexcel  Maths  Standard  2013  January  Core 1  Question 1
Question
1.
Factorise completely ??x  4x^{3}??. [3]
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1.
Lets start by removing a factor of ??x?? from the cubic equation given
$$x  4x^{3} = x(1  4x^{2})$$
Now we need to factorise the quadratic section of the equation:
$$x(\color{red}{1  4x^{2}})$$
Taking this quadratic section ??1  4x^{2}?? and rewriting in the more standard form ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c}?? makes it easier to inspect.
Note that this particular quadratic has a zero ??x?? term therefore ??\color{green}{b} = 0??
$$\color{red}{4}x^{2} + \color{green}{0}x + \color{blue}{1}$$
Recall we are trying to rewrite the
quadratic as the product of two linear expressions; ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c} = (dx + e)(fx + g)?? and since the ??x^{2}?? term is non zero we are looking for:
 ??\color{red}{a} = d \times f??
??\implies 4 = d \times f??
 ??\color{green}{b} = (d \times g) + (e \times f)??
??\implies 0 = (d \times g) + (e \times f)??
 ??\color{blue}{c} = e \times g??
??\implies 1 = e \times g??
??\implies 1 = e = g??
We know from equation (3) that
??e = g = 1??, which we can therefore use to use to simplify equation (2):
??0 = (d \times g) + (e \times f)??
??\implies 0 = (d \times 1) + (1 \times f)??
??\implies 0 = d + f??
??\implies d =  f??
Now considering this with equation (1), ??4 = d \times f?? and the factors of 4 it is clear that
??d = 2?? and
??f = 2??.
Substituting the values of ??d = 2??, ??e = 1??, ??f = 2?? and ??g = 1?? into our factorised equation gives: $$(dx + e)(fx + g) = (2x + 1)(2x + 1) = (1 + 2x)(1  2x)$$
Finally substituting this into the original equation our fully factorised solution is:
$$x  4x^{3} = x(1 + 2x)(1  2x)$$
Edexcel  Maths  Standard  2013  January  Core 1  Question 10
Question
10.
$$4x^{2} + 8x + 3 \equiv a(x + b)^{2} + c$$
(a)
Find the values of ??a??, ??b?? and ??c??. [3]
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10. (a)
Start by expanding out the brackets on the right hand side of the equation given:
??4x^{2} + 8x + 3 \equiv a(x + b)^{2} + c??
??\equiv a(x^{2} + bx + bx + b^{2}) + c??
??\equiv a(x^{2} + 2bx + b^{2}) + c??
??\equiv ax^{2} + a2bx + ab^{2} + c??
Now comparing coefficients:
$$\color{red}{a}x^{2} + \color{green}{a2b}x + \color{blue}{(ab^{2} + c)} \equiv \color{red}{4}x^{2} + \color{green}{8}x + \color{blue}{3}$$
Firstly of the ??x^{2}?? term:
??\implies \color{red}{a} = 4??
Secondly the ??x?? term:
??\implies \color{green}{a2b} = 8??
Substitute in the value of ??a?? from above:
??\implies (4) \times 2b = 8??
??\implies 8b = 8??
??\therefore b = 1??
Finally the constant term:
??\implies \color{blue}{(ab^{2} + c)} = 3??
Substitute in the values of ??a?? and ??b?? from above:
??\implies (4) \times (1)^{2} + c = 3??
??\implies 4 + c = 3??
??\implies c = 3  4??
??\therefore c = 1??
For completeness we should state ??a = 4??, ??b = 1?? and ??c = 1??
Exam Tips

When attempting to find coefficients of an expression:
Start by multiplying out any brackets. Take care when doing this with sign convention and correct terms.
Rewriting into a standard form ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c}?? makes comparisons easier.

Remember, when factorising we are we are trying to rewrite the quadratic as the product of two linear expressions; ??\color{red}{a}x^{2} + \color{green}{b}x + \color{blue}{c} = (dx + e)(fx + g)??:
Where the ??x^{2}?? term, ??\color{red}{a} = d \times f??
Where the ??x?? term, ??\color{green}{b} = (d \times g) + (e \times f)??
Where the constant term, ??\color{blue}{c} = e \times g??

However in typical factorisation questions the factor ??\color{red}{a}?? of ??x^{2}?? is equal to 1 therefore factorised solution simplifies to: ??x^{2} + \color{green}{b}x + \color{blue}{c} = (x + e)(x + g)??
Where the ??x?? term, ??\color{green}{b} = e + g??
Where the constant term, ??\color{blue}{c} = e \times g??
So in this case we are looking for two numbers that sum to give ??\color{green}{b}?? and multiply to give ??\color{blue}{c}??

Factorisation takes practice to become quick at spotting the solution but look out for special cases like the 'difference of two squares'.
When the equation takes the form ??(x^{2}  p^{2})?? this factorises into ??(x + p)(x  p)??.
For example ??(x^{2}  9) = (x^{2}  3^{2}) = (x + 3)(x  3)??
When the equation takes the form ??(1  p^{2}x^{2})?? this factorises into ??(1 + px)(1  px)??.
For example ??(1  9x^{2}) = (1  3^{2}x^{2}) = (1 + 3x)(1  3x)??

You are less likely to make arithmetic errors using factorisation, but unfortunately some expressions can't be factorised. You can always use the formula, but be very careful substituting in your values for a, b and c, particularly with negative values.