Term 1
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Algebraic Expressions  Exponents Equations and Inequalities  Euclidean Geometry 
ALGEBRA – Algebraic Expressions
Products

 \(\mathbb{N}\): natural numbers are \(\left\{1; 2; 3; \ldots\right\}\)
 \(\mathbb{N}_0\): whole numbers are \(\left\{0; 1; 2; 3; \ldots\right\}\)
 \(\mathbb{Z}\): integers are \(\left\{\ldots; 3; 2; 1; 0; 1; 2; 3; \ldots\right\}\)
 A rational number is any number that can be written as \(\frac{a}{b}\) where \(a\) and \(b\) are integers and
\(b\ne 0\).  The following are rational numbers:
 Fractions with both numerator and denominator as integers
 Integers
 Decimal numbers that terminate
 Decimal numbers that recur (repeat)
 Irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator as
integers.  If the \(n^{\text{th}}\) root of a number cannot be simplified to a rational number, it is called a surd.
 If \(a\) and \(b\) are positive whole numbers, and \(a<b\), then \(\sqrt[n]{a}<\sqrt[n]{b}\).
 A binomial is an expression with two terms.
 The product of two identical binomials is known as the square of the binomial.
 We get the difference of two squares when we multiply \(\left(ax+b\right)\left(axb\right)\)
 Factorising is the opposite process of expanding the brackets.
 The product of a binomial and a trinomial is:\[\left(A+B\right)\left(C+D+E\right)=A\left(C+D+E\right)+B\left(C+D+E\right)\]
 Taking out a common factor is the basic factorisation method.
 We often need to use grouping to factorise polynomials.
 To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.
 The sum of two cubes can be factorised as:
\[{x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}xy+{y}^{2}\right)\]  The difference of two cubes can be factorised as:
\[{x}^{3}{y}^{3}=\left(xy\right)\left({x}^{2}+xy+{y}^{2}\right)\]  We can simplify fractions by incorporating the methods we have learnt to factorise expressions.
 Only factors can be cancelled out in fractions, never terms.
 To add or subtract fractions, the denominators of all the fractions must be the same.
Factorisation
Factorisation is the opposite process of expanding brackets. For example, expanding brackets would require
\(2(x + 1)\) to be written as \(2x + 2\). Factorisation would be to start with \(2x + 2\) and end up with
\(2(x + 1)\).
The two expressions \(2(x + 1)\) and \(2x+2\) are equivalent; they have the same value for all values of \(x\).
In previous grades, we factorised by taking out a common factor and using difference of squares.
Common factors
Factorising based on common factors relies on there being factors common to all the terms.
For example, \(2x – 6{x}^{2}\) can be factorised as follows:
\[2x – 6{x}^{2} = 2x(1 – 3x)\]
And \(2(x – 1) – a(x – 1)\) can be factorised as follows:
\[(x – 1)(2 – a)\]
Difference of two squares
We have seen that \((ax + b)(ax – b)\) can be expanded to \({a}^{2}{x}^{2} – {b}^{2}\).
Therefore \({a}^{2}{x}^{2} – {b}^{2}\) can be factorised as \((ax + b)(ax – b)\).
For example, \({x}^{2} – 16\) can be written as \({x}^{2} – {4}^{2}\) which is a difference of two
squares. Therefore, the factors of \({x}^{2} – 16\) are \((x – 4)\) and \((x + 4)\).
To spot a difference of two squares, look for expressions:
 consisting of two terms;
 with terms that have different signs (one positive, one negative);
 with each term a perfect square.
For example: \({a}^{2} – 1\); \(4{x}^{2} – {y}^{2}\); \(49 + {p}^{4}\).
Factorising by grouping in pairs
The taking out of common factors is the starting point in all factorisation problems. We know that the
factors of \(3x+3\) are \(\text{3}\) and \(\left(x+1\right)\). Similarly, the factors of \(2{x}^{2}+2x\)
are \(2x\) and \(\left(x+1\right)\). Therefore, if we have an expression:
\[2{x}^{2} + 2x + 3x + 3\]
there is no common factor to all four terms, but we can factorise as follows:
\[\left(2{x}^{2} + 2x\right) + \left(3x + 3\right) = 2x\left(x + 1\right) + 3\left(x + 1\right)\]
We can see that there is another common factor \(\left(x+1\right)\). Therefore, we can write:
\[\left(x + 1\right)\left(2x + 3\right)\]
We get this by taking out the \(\left(x+1\right)\) and seeing what is left over. We have \(2x\) from the
first group and \(\text{+3}\) from the second group. This is called factorising by grouping.
Factorising a quadratic trinomial
Factorising is the reverse of calculating the product of factors. In order to factorise a quadratic, we
need to find the factors which, when multiplied together, equal the original quadratic.
Consider a quadratic expression of the form \(a{x}^{2} + bx\). We see here that \(x\) is a common factor
in both terms. Therefore \(a{x}^{2} + bx\) factorises as \(x\left(ax + b\right)\). For example,
\(8{y}^{2} + 4y\) factorises as \(4y\left(2y + 1\right)\).
Another type of quadratic is made up of the difference of squares. We know that:
\[\left(a + b\right)\left(a – b\right) = {a}^{2} – {b}^{2}\]
So \({a}^{2} – {b}^{2}\) can be written in factorised form as \(\left(a + b\right)\left(a – b\right)\).
This means that if we ever come across a quadratic that is made up of a difference of squares, we can
immediately write down the factors. These types of quadratics are very simple to factorise. However,
many quadratics do not fall into these categories and we need a more general method to factorise
quadratics.
We can learn about factorising quadratics by looking at the opposite process, where two binomials are
multiplied to get a quadratic. For example:
\begin{align*}
\left(x + 2\right)\left(x + 3\right) & = {x}^{2} + 3x + 2x + 6 \\
& = {x}^{2} + 5x + 6
\end{align*}
We see that the \({x}^{2}\) term in the quadratic is the product of the \(x\)terms in each bracket.
Similarly, the \(\text{6}\) in the quadratic is the product of the \(\text{2}\) and \(\text{3}\) in the
brackets. Finally, the middle term is the sum of two terms.
So, how do we use this information to factorise the quadratic?
Let us start with factorising \({x}^{2} + 5x + 6\) and see if we can decide upon some general rules.
Firstly, write down the two brackets with an \(x\) in each bracket and space for the remaining terms.
\[\left(x \qquad \right)\left(x \qquad\right)\]
Next, decide upon the factors of \(\text{6}\). Since the \(\text{6}\) is positive, possible combinations
are: 1 and 6, 2 and 3, \(\text{1}\) and \(\text{6}\) or \(\text{2}\) and \(\text{3}\).
Therefore, we have four possibilities:
Option 1  Option 2  Option 3  Option 4 
\(\left(x+1\right)\left(x+6\right)\)  \(\left(x1\right)\left(x6\right)\)  \(\left(x+2\right)\left(x+3\right)\)  \(\left(x2\right)\left(x3\right)\) 
Next, we expand each set of brackets to see which option gives us the correct middle term.
Option 1  Option 2  Option 3  Option 4 
\(\left(x+1\right)\left(x+6\right)\)  \(\left(x1\right)\left(x6\right)\)  \(\left(x+2\right)\left(x+3\right)\)  \(\left(x2\right)\left(x3\right)\) 
\({x}^{2}+7x+6\)  \({x}^{2}7x+6\)  \({x}^{2}+5x+6\)  \({x}^{2}5x+6\) 
We see that Option 3, \(\left(x+2\right)\left(x+3\right)\), is the correct solution.
The process of factorising a quadratic is mostly trial and error but there are some strategies that can
be used to ease the process.
General procedure for factorising a trinomial
 Take out any common factor in the coefficients of the terms of the expression to obtain an
expression of the form \(a{x}^{2} + bx + c\) where \(a\), \(b\) and \(c\) have no common factors
and \(a\) is positive.  Write down two brackets with an \(x\) in each bracket and space for the remaining terms:\[\left(x \qquad \right)\left(x \qquad\right)\]
 Write down a set of factors for \(a\) and \(c\).
 Write down a set of options for the possible factors for the quadratic using the factors of \(a\)
and \(c\).  Expand all options to see which one gives you the correct middle term \(bx\).
If \(c\) is positive, then the factors of \(c\) must be either both positive or both negative. If
\(c\) is negative, it means only one of the factors of \(c\) is negative, the other one being
positive. Once you get an answer, always multiply out your brackets again just to make sure it
really works.
Sum and difference of two cubes
We now look at two special results obtained from multiplying a binomial and a trinomial:
Sum of two cubes:
\begin{align*}
\left(x + y\right)\left({x}^{2} – xy + {y}^{2}\right) & = x\left({x}^{2} – xy + {y}^{2}\right) +
y\left({x}^{2} – xy + {y}^{2}\right) \\
& = \left[x\left({x}^{2}\right) + x\left(xy\right) + x\left({y}^{2}\right)\right] +
\left[y\left({x}^{2}\right) + y\left(xy\right) + y\left({y}^{2}\right)\right]\\
& = {x}^{3} – {x}^{2}y + x{y}^{2} + {x}^{2}y x{y}^{2} + {y}^{3}\\
& = {x}^{3} + {y}^{3}
\end{align*}
Difference of two cubes:
\begin{align*}
\left(x – y\right)\left({x}^{2} + xy + {y}^{2}\right) & = x\left({x}^{2} + xy + {y}^{2}\right) –
y\left({x}^{2} + xy + {y}^{2}\right)\\
& = \left[x\left({x}^{2}\right) + x\left(xy\right) + x\left({y}^{2}\right)\right] –
\left[y\left({x}^{2}\right) + y\left(xy\right) + y\left({y}^{2}\right)\right]\\
& = {x}^{3} + {x}^{2}y + x{y}^{2} – {x}^{2}y – x{y}^{2} – {y}^{3}\\
& = {x}^{3} – {y}^{3}
\end{align*}
So we have seen that:
\begin{align*}
{x}^{3} + {y}^{3} & = \left(x + y\right)\left({x}^{2} – xy + {y}^{2}\right) \\
{x}^{3} – {y}^{3} & = \left(x – y\right)\left({x}^{2} + xy + {y}^{2}\right)
\end{align*}
We use these two basic identities to factorise more complex examples.