\documentclass[10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[version=4]{mhchem}
\usepackage{stmaryrd}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\graphicspath{ {./images/} }

\title{Factorisation }

\author{}
\date{}


\begin{document}
\maketitle



\section*{Factorisation of simple quadratics}
A simple quadratic expression is an expression of the form \(x^{2}+b x+c\), where \(b\) and \(c\) are given numbers. When we expand \((x+3)(x+4)\), we obtain a simple quadratic.

\[
\begin{gathered}
x(x+4)+3(x+4)=x^{2}+4 x+3 x+12 \\
=x^{2}+7 x+12
\end{gathered}
\]

We want to develop a method for reversing this process.

In the expansion \((x+3)(x+4)=x^{2}+7 x+12\), notice that the coefficient of \(x\) is \(3+4=7\). The term that is independent of \(x\), the constant term, is \(3 \times 4=12\). This suggests a method of factorising.

In general, when we expand \((x+p)(x+q)\), we obtain

\[
x^{2}+p x+q x+p q=x^{2}+(p+q) x+p q
\]

The coefficient of \(x\) is the sum of \(p\) and \(q\), and the constant term is the product of \(p\) and \(q\).

\section*{Factorisation of simple quadratics}
To factorise a simple quadratic, look for two numbers that add to give the coefficient of \(x\), and that multiply together to give the constant term.

For example, to factorise \(x^{2}+8 x+15\), we look for two numbers that multiply to give 15 and add to give 8 . Of the pairs that multiply to give \(15(15 \times 1,5 \times 3,-5 \times(-3)\) and \(-15 \times(-1))\), only 5 and 3 add to give 8 .

Therefore, \(x^{2}+8 x+15=(x+3)(x+5)\)

The result can be checked by expanding \((x+3)(x+5)\).

2 Factorise these quadratric expressions.

3 Factorise these quadratric expressions.

4 Factorise these quadratic expressions.

Example 85 Factorise these quadratic expressions.

\section*{1 Factorisation using perfect squares}
In Chapter 4, Example 8, we used the methods discussed previously to show that \(x^{2}+8 x+16=(x+4)^{2}\). This form of quadratic factorisation is called a perfect square.

We recall from Chapter 1 that a perfect square is an expression such as \((x+3)^{2},(x-5)^{2}\) or \((2 x+7)^{2}\).


Note that the constant term 9 is the square of half the coefficient of \(x\).

The quadratic \(x^{2}+10 x+25\) is a perfect square since the "constant term is equal to the square of half of the coefficient of \(x\) '. So:

\[
x^{2}+10 x+25=(x+5)^{2}
\]

We recognise a perfect square such as this in the following way: the constant term is the square of half of the coefficient of \(x\). For example:

\[
\begin{array}{ll}
x^{2}+12 x+36=(x+6)^{2} & x^{2}+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^{2} \\
x^{2}-14 x+49=(x-7)^{2} & x^{2}-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^{2}
\end{array}
\]

\section*{Factorising using the perfect square identities}
\begin{itemize}
  \item In general, \(a^{2}+2 a b+b^{2}=(a+b)^{2}\)
  \item Similarly, \(a^{2}-2 a b+b^{2}=(a-b)^{2}\)
\end{itemize}

When a quadratic expression has the form of a perfect square, factorisation can occur immediately by application of the relevant identity.


Example 9 1 Factorise using the appropriate perfect square identity.


\section*{\(4 E\) Quadratics with common factors}
Sometimes a common factor can be taken out of a quadratic expression so that the expression inside the brackets becomes a simple quadratic that can be factorised.


2 Factorise:

3 Factorise:



\section*{Review exercise}
1 Factorise:
a \(4 x+16\)
b \(7 x-21\)
c \(6 a-9\)
d \(4 a b+7 a\)
e \(6 p q-11 p\)
f \(5 m n-10 n\)
g \(4 u v-8 v\)
h \(a^{2}+9 a\)
i \(4 m^{2} n-12 m n\)
j \(a^{2} b-4 a b^{2}\)
k \(3 p q-6 p^{2}\)
l \(6 p^{3} q-18 p q\)

2 Factorise:
a \(x^{2}-9\)
b \(x^{2}-16\)
c \(9 a^{2}-25\)
d \(16 m^{2}-1\)
e \(9-4 b^{2}\)
f \(100-81 b^{2}\)
g \(16 x^{2}-y^{2}\)
h \(2 m^{2}-50\)
i \(3 a^{2}-27\)
j \(1-36 b^{2}\)
k \(4 y^{2}-\frac{1}{4}\)
l \(p^{2} q^{2}-1\)

3 Factorise:

a \(x^{2}+8 x+12\)

b \(x^{2}+9 x+18\)

c \(x^{2}+11 x+30\)

d \(x^{2}-11 x+24\)

e \(x^{2}-10 x+24\)

f \(x^{2}-14 x+24\)

g \(x^{2}-25 x+24\)

h \(x^{2}+x-20\)

i \(x^{2}-2 x-48\)

j \(x^{2}-4 x-12\)

k \(x^{2}+3 x-40\)

l \(x^{2}-7 x-8\)

m \(x^{2}-x-132\)

n \(a^{2}+19 a+60\)

o \(x^{2}-50 x+96\)

4 Factorise:
a \(a^{2}-22 a+121\)
b \(m^{2}-14 m+49\)
c \(s^{2}+8 s+16\)
d \(a^{2}+24 a+144\)
e \(a^{2}-12 a+36\)
f \(z^{2}-40 z+400\)
g \(x^{2}+5 x+\frac{25}{4}\)
h \(y^{2}-\frac{2 y}{3}+\frac{1}{9}\)
i \(a^{2}+\frac{3 a}{2}+\frac{9}{16}\)

5 Factorise:
a \(2 x^{2}+18 x+40\)
b \(3 x^{2}-30 x+63\)
c \(5 x^{2}-50 x+120\)
d \(2 x^{2}+8 x-90\)
e \(3 x^{2}-6 x-105\)
f \(2 x^{2}-6 x-260\)

6 Factorise:
a \(25 a^{2}-16 b^{2}\)
b \(a^{2}+14 a+49\)
c \(a^{2}-a-20\)
d \(1-36 m^{2}\)
e \(4-9 x^{2} y^{2}\)
f \(\frac{1}{9}-\frac{a^{2}}{25}\)
g \(m^{2}-\frac{1}{4}\)
h \(x^{3}-49 x y^{2}\)
i \(3 a^{2}-75\)
j \(b^{2}-20 b+96\)
k \(n^{2}-31 n+150\)
l \(m^{2}+20 m+91\)
\(\mathbf{m} a^{2}-7 a b-98 b^{2}\)
n \(m^{2}-4 m-165\)
o \(x^{2}+3 x y-4 y^{2}\)
p \(20 m^{2} n-5 n^{3}\)
q \(a^{2}-2 a-63\)
r \(5 q^{2}-5 p q-30 p^{2}\)
s \(x^{2}-3 x-130\)
t \(42-x-x^{2}\)
u \(x^{2}+7 x-18\)

\section*{Challenge exercise}
1 Factorise:
a \(x^{4}-1\)
b \(x^{4}-16\)
c \(x^{2}-3\)
d \(x^{2}-5\)
e \(x^{2}-2 \sqrt{2} x+2\)
f \(x^{2}+2 \sqrt{2} x+2\)
g \(x^{2}+x+\frac{1}{4}\)
h \(x^{2}+3 x+\frac{9}{4}\)
i \(x^{2}-x+\frac{1}{4}\)
j \(x^{2}-3 x+\frac{9}{4}\)
k \(x^{2}-x y-2 y^{2}\)
l \(x^{2}+x y-2 y^{2}\)

2 Simplify:

a \(\frac{x^{2}+x-2}{x^{2}-x-20} \times \frac{x^{2}+5 x+4}{x^{2}-x} \div\left(\frac{x^{2}+3 x+2}{x^{2}-2 x-15} \times \frac{x+3}{x^{2}}\right)\)

b \(\frac{x^{2}-64}{x^{2}+24 x+128} \times \frac{x^{2}+12 x-64}{x^{2}-16} \div \frac{x^{2}-16 x+64}{x^{2}-10 x+16}\)

c \(\frac{x^{2}-18 x+80}{x^{2}-5 x-50} \times \frac{x^{2}-6 x-7}{x^{2}-15 x+56} \div \frac{x-1}{x+5}\)

3 Factorise:
a \((x+2)^{2}-8 x\)
\(\mathbf{b}(x-3)^{2}+12 x\)
c \((x+a)^{2}-4 a x\)
d \((x-a)^{2}+4 a x\)

4 Simplify \(\left(\frac{a}{b}+\frac{c}{d}\right) \div\left(\frac{a}{b}-\frac{c}{d}\right)\).

5 Factorise:
a \(\left(x^{2}+1\right)^{2}-4\)
b \(\left(x^{2}-2\right)^{2}-4\)
c \(x^{2}+4 x+4-y^{2}\)
d \(x^{2}+8 x+16-a^{2}\)
e \(m^{2}-2 m+1-n^{2}\)
f \(p^{2}-5 p+\frac{25}{4}-q^{2}\)

6 a By adding and subtracting \(4 x^{2}\), factorise \(x^{4}+4\).

b Factorise \(x^{4}+4 a^{4}\).

7 A swimming pool is designed in an L-shape with dimensions in metres as shown.

The pool is enlarged or reduced depending on the value of \(x\).

a Find, in terms of \(x\), the length of:
i \(A F\)
ii \(C D\)

\begin{center}
\includegraphics[max width=\textwidth]{2023_11_30_c996b59107099c56822bg-14}
\end{center}

b Show that the perimeter is equal to \((10 x+50) \mathrm{m}\).

c What is the perimeter if \(x=3\) ?

d Find the area of the swimming pool in terms of \(x\). Expand and simplify your answer.

e It is decided that a square swimming pool would be a better use of space.

i By factorising your answer to part \(\mathbf{d}\), find the dimensions, in terms of \(x\), of a square swimming pool with the same area as the L-shaped swimming pool.

ii What is the perimeter, in terms of \(x\), of this square swimming pool?

8 For positive whole numbers \(a\) and \(b\), prove that:
a if \(\frac{a}{b}<1\) then \(\frac{a+1}{b+1}>\frac{a}{b}\)
b if \(\frac{a}{b}>1\) then \(\frac{a+1}{b+1}<\frac{a}{b}\)

9 A right-angled triangle has a hypotenuse of length \(b \mathrm{~cm}\) and one other side of length \(a \mathrm{~cm}\). If \(b-a=1\), find the length of the third side in terms of \(a\) and \(b\).

10 Factorise \(\left(1+\frac{y^{2}+z^{2}-x^{2}}{2 y z}\right) \div\left(1-\frac{x^{2}+y^{2}-z^{2}}{2 x y}\right)\).


\end{document}