Title: Default for 1.3 Simplification and Factorisation
Course: HELM Workbooks 2023-2024

1. Addition and subtraction of like terms

Like terms are multiples of the same quantity. For example \(5y\), \(17y\) and \({1\over 2} y\) are all multiples of \(y\) and so are like terms. Similarly, \(3x^2\), \(-5x^2\) and \({1\over 4}x^2\) are all multiples of \(x^2\) and so are like terms.

Further examples of like terms are:

\(kx\) and \(\ell x\) which are both multiples of \(x\),

\(x^2y\), \(6x^2y\), \(-13x^2y\), \(-2yx^2\), which are all multiples of \(x^2y,\)

\(abc^2\), \(-7abc^2\), \(kabc^2\), are all multiples of \(abc^2\)

Like terms can be added or subtracted in order to simplify expressions.

Example

Simplify \(5x-13x+22x\).

Solution

All three terms are multiples of \(x\) and so are like terms. The expression can be simplified to \(14x\).

Example

Simplify \(5z+2x\).

Solution

\(5z\) and \(2x\) are not like terms. They are not multiples of the same quantity. This expression cannot be simplified.

Example

Simplify \(2x^2-7x+11x^2+x\).

Solution

\(2x^2\) and \(11x^2\), both being multiples of \(x^2\), can be collected together and added to give \(13x^2\).

Similarly, \(-7x\) and \(x\) can be added to give \(-6x\).

We get \(2x^2-7x+11x^2+x= 13x^2-6x \) which cannot be simplified further.

Example

Simplify \(3a^2b-7a^2b-2b^2+a^2\).

Solution

Note that \(3a^2b\) and \(7a^2b\) are both multiples of \(a^2b\) and so are like terms. There are no other like terms. Therefore

\(3a^2b-7a^2b-2b^2+a^2 = -4a^2b-2b^2+a^2\)

Exercise

Simplify, if possible. If not possible enter the original expression.

1. \([value]=\)
2. \([value]=\)
3. \([value]=\)
4. \([value]=\)
5. \([value]=\)
Feedback: 1. All the terms are multiples of \(x\) so this expression can be simplified to: \([value] =([value]) \cdot x = [value].\) 2. All the terms are multiples of \(q\) so this expression can be simplified to: \([value] =([value]) \cdot q = [value].\) 3. All the terms are multiples of \(x^2\) so this expression can be simplified to: \([value] = ([value]) \cdot x^2 = [value].\) 4. All the terms are multiples of \(v^2\) so this expression can be simplified to \([value] = ([value]) \cdot v^2 = [value].\) 5. Neither of the terms are similar so this expression cannot be simplified.

Exercise

Simplify, if possible. If not possible enter the original expression.

1. \([value] =\)
2. \([value]=\)
3. \([value]=\)
Feedback: 1. The two terms that are multiples of \(w\) are like terms, and the two terms that are multiples of \(r\) are like terms . The expression can be simplified to: \([value] = ([value]) \cdot w + ([value]) \cdot r = [value].\) 2. The expression [value] cannot be simplified. 3. All three terms are multiples of \(w^2\) and so are like terms. The expression can be simplified to: \([value] = ([value]) \cdot w^2 = [value].\)

Exercise

Simplify, if possible. If not possible enter the original expression.

1. \([value] =\)
2. \([value] =\)
3. \([value]=\)
4. \([value]=\)
5. \([value] =\)
6. \([value] =\)
7. \([value] = \)
Feedback: 1. \([value] = [value].\) 2. \([value] = [value].\) 3. \([value] = [value].\) 4. Cannot be simplified. 5. \([value] = [value].\) (Note that \(ab\) = \(ba\). ) 6. \([value] = [value].\) 7. \([value] = [value].\)

Exercise

Simplify, if possible. If not possible enter the original expression.

1. \([value]x  -[value]x =\)
2. \([value]x([value]x)=\)
3. \([value]x([value]x)=\)
4. \([value]x-[value]x=\)
5. \([value]x([value]x)=\)
Feedback: 1. Both terms are multiples of \(x\) and so are like terms. The expression can be simplified to \([value].\) 2. Here we have \([value]x \cdot [value]x\), so the expression can be simplified to \([value].\) 3. Here we have \([value]x \cdot ([value]x)\), so the expression can be simplified to \([value].\) 4. Both terms are multiples of \(x\) and so are like terms. The expression can be simplified to \([value].\) 5. Here we have \([value]x \cdot [value]x\), so the expression can be simplified to \([value].\)

Exercise

Simplify, if possible. If not possible enter the original expression.

1. \( \displaystyle [value] =\)
2. \( \displaystyle [value] =\)
3. \([value]=\)
4. \( [value] =\)   where \(a\) and \(b\) are constants.
Feedback: 1. Both fractions have a common denominator of \([value]\), so the expression can be simplified to: \(\displaystyle [value] = [value].\) 2. There are two multiples of \( x^2 \) which are like terms. The expression can be simplified to: \( \displaystyle{[value] = [value]}.\) 3. The expression [value] cannot be simplified. 4. Both terms are multiples of \( x^2 \), so they are like terms. The expression can be simplified to: \([value] = [value].\)

2. Removing brackets from expressions \( a(b+c) \) and \( a(b-c) \)

Removing brackets means multiplying out. For example \( 5(2+4)=5\times 2 +5\times 4=10+20=30\). In this simple example we could alternatively get the same result as follows: \(5(2+4)=5\times 6=30\). That is: \[ 5(2+4)=5\times 2+5\times 4. \]

In an expression such as \(5(x+y)\) it is intended that the 5 multiplies both \(x\) and \(y\) to produce \(5x+5y\). Thus the expressions \(5(x+y)\) and \(5x+5y\) are equivalent. In general we have the following rules known as distributive laws:

Key Point

\[ a(b + c) = ab + ac \]

\[ a(b - c) = ab - ac \]

Note that when the brackets are removed both terms in the brackets are multiplied by \(a\).

As we have noted above, if you insert numbers instead of letters into these expressions you will see that both left and right hand sides are equivalent. For example

\[ 4(3 + 5) \text{ has the same value as } 4(3) + 4(5); \text {that is } 32, \]

and

\[ 7(8 - 3) \text{ has the same value as } 7(8) - 7(3); \text{ that is } 35. \]

Example

Remove the brackets from

1. \( 9(2 + y) \)
2. \(9(2y)\)

Solution

1. In the expression \( 9(2 + y) \) the \(9\) must multiply both terms in the brackets: \[ 9(2 + y) = 9(2) + 9(y) \] \[ = 18 + 9y. \]
2. Recall that \(9(2y)\) means \(9 \times (2 \times y) \) and that when multiplying numbers together the presence of brackets is irrelevant. Thus \[ 9(2y) = 9 \times 2 \times y = 18y. \]

The crucial distinction between the role of the factor \(9\) in the two expressions \( 9(2 + y) \) and \(9(2y)\) in this example should be noted.

Example

Remove the brackets from \( 9(x + 2y) \).

Solution

In the expression \(9(x + 2y) \) the \(9\) must multiply both the \(x\) and the \(2y\) in the brackets. Thus

\[ 9(x + 2y) = 9x + 9(2y) \]

\[ = 9x + 18y. \]

Example

Remove the brackets from \( -3(5x - z) \).

Solution

The number \(-3\) must multiply both the \(5x\) and the \(z\). \[ -3(5x - z) = (-3)(5x) - (-3)(z) \] \[ = -15x + 3z. \]

Example

Remove the brackets from \( -(3x + 1) \).

Solution

Although the \(1\) is unwritten, the minus sign outside the brackets stands for \(-1\). We must therefore consider the expression \( -1(3x + 1). \) \[ -1(3x + 1) = (-1)(3x) + (-1)(1) \] \[ = -3x + (-1) \] \[ = -3x - 1. \]

Example

Remove the brackets from the expression \( 5x - (3x+1) \) and simplify the result by collecting like terms.

Solution

The brackets in \(-(3x + 1)\) were removed in Example 8. \[ 5x - (3x + 1) = 5x - 1(3x + 1) \] \[ = 5x - 3x - 1 \] \[ = 2x - 1. \]

Example

Show that

\[ \frac{-x-1}{4}, \hspace{0.3cm} \frac{-(x+1)}{4}, \text{ and } -\frac{x+1}{4} \]

are all equivalent expressions.

Solution

Consider \( -(x + 1) \). Removing the brackets we obtain \(-x - 1 \) and so

\[ \frac{-x - 1}{4} \text{ is equivalent to } \frac{-(x + 1)}{4} . \]

A negative quantity divided by a positive quantity will be negative. Hence

\[ \frac{-(x + 1)}{4} \text{ is equivalent to } -\frac{x + 1}{4} . \]

You should study all three expressions carefully to recognise the variety of equivalent ways in which we can write an algebraic expression.

Sometimes the bracketed expression can appear on the left, as in \( (a + b)c \). To remove the brackets here we use the following rules:

Key Point

\[ (a + b)c = ac + bc \]

\[ (a - b)c = ac - bc \]

Note that when the brackets are removed both the terms in the brackets multiply \( c \).

Example

Remove the brackets from \( (2x + 3y)x \).

Solution

Both terms in the brackets multiply the x outside. Thus

\[ (2x + 3y)x = 2x(x) + 3y(x) \]

\[ = 2x^2 + 3yx. \]

3. Removing brackets from expressions of the form \( (a+b)(c+d) \)

Sometimes it is necessary to consider two bracketed terms multiplied together. In the expression \( (a + b)(c + d) \), by regarding the first bracket as a single term we can use the result given previously to write it as \( (a + b)c + (a + b)d \). Removing the brackets from each of these terms produces \( ac + bc + ad + bd \). More concisely:

Key Point

\[ (a + b)(c + d) = (a + b)c + (a + b)d = ac + bc + ad + bd. \].

We see that each term in the first bracketed expression multiplies each term in the second bracketed expression.

Example

Remove the brackets from \( (3 + x)(2 + y) \).

Solution

We find \[ (3 + x)(2 + y) = (3 + x)(2) + (3 + x)y \] \[ = (3)(2) + (x)(2) + (3)(y) + (x)(y) = 6 + 2x + 3y + xy. \]

Example

Remove the brackets from \( (3x + 4)(x + 2) \) and simplify your result.

Solution

\[ (3x + 4)(x + 2) = (3x + 4)(x) + (3x + 4)(2) \] \[ = 3x2 + 4x + 6x + 8 = 3x2 + 10x + 8. \]

Example

Remove the brackets from \( (a + b)^2 \) and simplify your result.

Solution

When a quantity is squared it is multiplied by itself. Thus \[ (a + b)^2 = (a + b)(a + b) = (a + b)a + (a + b)b \] \[ = a^2 + ba + ab + b^2 = a^2 + 2ab + b^2. \]

Key Point

\[ (a + b)^2 = a^2 + 2ab + b^2 \] \[ (a - b)^2 = a^2 - 2ab + b^2 \]

Example

Explain the distinction between \( (x + 3)(x + 2)\) and \( x + 3(x + 2). \)

Solution

In the first expression removing the brackets we find \[ (x + 3)(x + 2) = x^2 + 3x + 2x + 6 \] \[ = x^2 + 5x + 6. \] In the second expression we have \[ x + 3(x + 2) = x + 3x + 6 = 4x + 6. \] Note that in the second expression the term \( (x + 2) \) is only multiplied by \(3\) and not by \(x\).

Example

Remove the brackets from \( (s^2 + 2s + 4)(s + 3) \).

Solution

Each term in the first bracket must multiply each term in the second. Working through all combinations systematically we have \[ (s^2 + 2s + 4)(s + 3) = (s^2 + 2s + 4)(s) + (s^2 + 2s + 4)(3) \] \[ = s^3 + 2s^2 + 4s + 3s^2 + 6s + 12 \] \[ = s^3 + 5s^2 + 10s + 12.\]

Exercise

Remove the brackets from each of the following expressions:

1. \( [value](mn)=\)
2. \((am)n=\)
3. \( (a + m)n=\)
4. \((a - m)n =\)
5. \( [value](p+q)=\)
6. \( [value](p-q)=\)
7. \([value]([value]p + q)=\)
8. \( [value]([value]pq)=\)
9. \( [value]([value]p - q)=\)
10. \( [value] (p - [value]q  ) =\)
11. \(  [value] (p + [value]q  ) =\)
12. \( [value] ([value]pq)=\)
Feedback: 1. \( [value](mn)= [value]\) 2. \((am)n= [value]\) 3. \( (a + m)n= [value]\) 4. \((a - m)n = [value]\) 5. \( [value](p+q)= [value]\) 6. \( [value](p-q)= [value]\) 7. \([value]([value]p + q)= [value]\) 8. \( [value]([value]pq)= [value]\) 9. \( [value]([value]p - q)= [value]\) 10. \( [value] (p - [value]q) = [value]\) 11. \([value] (p + [value]q) = [value]\) 12. \( [value] ([value]pq)= [value]\)

Exercise

Remove the brackets from each of the following expressions:

1. \([value](a+b)=\)
2. \([value](m-n)=\)
3. \([value](x-y)=\)
Feedback: 1. In the expression \( [value](a + b) \), the \( [value] \) must multiply both the \(a\) and the \(b\) in the brackets. Thus \[ [value](a + b) = [value]a + [value]b. \] 2. In the expression \( [value](m - n) \), the \( [value] \) must multiply both the \(m\) and the \(-n\) in the brackets. Thus \[ [value](m - n) = [value]m - [value]n. \] 3. In the expression \( [value](x -y) \), the \( [value] \) must multiply both the \(x\) and the \(-y\) in the brackets. Thus \[ [value](x - y) = [value]x - [value]y. \]

Exercise

Remove the brackets from each of the following expressions:

1. \( ([value]+a)([value]+b)=\)
2. \( (x + [value])(x + [value])=\)
3. \((x + [value])(x + [value])=\)
4. \((x + [value])(x - [value])=\)
Feedback: 1. \[ ([value]+a) ([value]+b) =([value]+a) ([value]) + ([value]+a)(b) \] \[ = ([value])([value]) + (a)([value]) + ([value])(b) + (a)(b) \] \[ = [value] + [value]a + [value]b + ab. \] 2. \[ (x + [value]) (x + [value])= (x + [value]) (x) + (x + [value])([value]) \] \[ = (x)(x) + ([value])(x) + ([value])(x) + ([value])([value]) \] \[ = x^2 + [value]x + [value]. \] 3. \[ (x + [value]) (x + [value])= (x + [value]) (x) + (x + [value])([value]) \] \[ = (x)(x) + ([value])(x) + ([value])(x) + ([value])([value]) \] \[ = x^2 + [value]x + [value]. \] 4. \[ (x + [value]) (x - [value])= (x + [value]) (x) - (x + [value])([value]) \] \[ = (x)(x) + ([value])(x) - ([value])(x) - ([value])([value]) \] \[ = x^2 + [value]x - [value]. \]

Exercise

Remove the brackets from each of the following expressions:

1. \( ([value]+x)([value]+x)=\)
2. \( (x + [value])(x + [value])=\)
3. \((x + [value])(x + [value])=\)
4. \((x + [value])(x - [value])=\)
5. \( ([value]+x)x=\)
6. \( ([value]x + 1)x=\)
7. \( ( [value]x + 1)(x + 1)=\)
8. \(( [value]x + 1)( [value]x + 1)=\)
9. \( ([value]x+ [value]) ( [value]x + [value] )=\)
10. \( ([value]x+ [value]) ( [value]x - [value] )=\)
11. \( ( [value] - [value]x)(x + 1)=\)
12. \(( [value] - x)(1 - x)=\)
Feedback: 1. \[ ([value]+x) ([value]+x) = ([value]+x) ([value]) + ([value]+x)(x) \] \[ = ([value])([value]) + (x)([value]) + ([value])(x) + (x)(x) \] \[ = [value] + [value]x + x^2. \] 2. \[ (x + [value]) (x + [value])= (x + [value]) (x) + (x + [value]) ([value]) \] \[ = (x)(x) + ([value])(x) + ([value])(x) + ([value])([value]) \] \[ = x^2 + [value]x + [value]. \] 3. \[ (x + [value]) (x + [value])= (x + [value]) (x) + (x + [value])([value]) \] \[ = (x)(x) + ([value])(x) + ([value])(x) + ([value])([value]) \] \[ = x^2 + [value]x + [value]. \] 4. \[ (x + [value]) (x - [value])= (x + [value]) (x) - (x + [value])([value]) \] \[ = (x)(x) + ([value])(x) - ([value])(x) - ([value])([value]) \] \[ = x^2 + [value]x - [value]. \] 5. In the expression \( ([value]+x)x \), the \( [value] \) outside the bracket must multiply both the \([value]\) and the \(x\) in the brackets. Thus \[ ([value]+x)x = [value]x + x \times x \] \[ = [value]x + x^2 \] 6. In the expression \( ([value]x + 1)x \), the \( [value] \) outside the bracket must multiply both the \([value]x\) and the \(1\) in the brackets. Thus \[ ([value]x + 1)x = [value]x \times x + 1 \times x \] \[ [value] \] 7. \[ ( [value]x + 1)(x + 1) = ( [value]x + 1)(x) + ( [value]x + 1)(1) \] \[ = [value]x(x)+ (1)(x) + [value]x(1) +1(1) \] \[ = [value] \] 8. \[ ( [value]x + 1)( [value]x + 1)= ( [value]x + 1)([value]x) + ( [value]x + 1)(1) \] \[ = [value]x([value]x)+ (1)([value]x) + [value]x(1) +1(1) \] \[ = [value] \] 9. \[ ([value]x+ [value]) ( [value]x + [value] )= ([value]x+ [value])([value]x) + ([value]x+ [value])([value]) \] \[ = [value]x([value]x)+ ([value])([value]x) + [value]x([value]) +[value]([value]) \] \[ = [value] \] 10. \[ ([value]x+ [value]) ( [value]x - [value] )= ([value]x+ [value])([value]x) + ([value]x+ [value])(-[value]) \] \[ = [value]x([value]x)+ ([value])([value]x) - [value]x([value]) - [value]([value]) \] \[ = [value] \] 11. \[ ( [value] - [value]x)(x + 1)= ( [value] - [value]x)(x) + ( [value] - [value]x)(1) \] \[ = [value](x) - [value](x) + [value](1) - [value](1) \] \[ = [value] \] 12. \[ ( [value] - x)(1 - x)= ( [value] - x)(1) + ( [value] - x)(-x) \] \[ = [value](1) - x(1) + [value](-x) - x(-x) \] \[ = [value] \]

Exercise

Remove the brackets from each of the following expression and simplify your answer:

\( (s + [value])(s + [value])(s - [value])=\)
Feedback: \[ (s + [value])(s + [value])(s - [value])=\] \[(s + [value])(s + [value])(s) + (s + [value])(s + [value])(- [value])= \] \[(s + [value])(s(s) + [value](s)) + (s + [value])(s(- [value]) + [value](- [value]))= \] \[(s + [value])(s^2) + (s + [value])([value]s) + (s + [value])(- [value]s) + (s + [value]) ([value])= \] \[ s(s^2) + [value](s^2) + s([value]s) + [value]([value]s) + s(- [value]s) + [value](- [value]s)+ s([value]) + [value] ([value]) \] \[ s^3 + [value]s^2 + [value]s^2 + [value]s - [value]s^2 - [value]s - [value]s - [value] = \] \[ [value] \]

4. Factorisation

A number is said to be factorised when it is written as a product. For example, \(21\) can be factorised into \( 7 \times 3 \). We say that \(7\) and \(3\) are factors of \(21\). Algebraic expressions can also be factorised. Consider the expression \( 7(2x + 1) \). Removing the brackets we can rewrite this as \[ 7(2x + 1) = 7(2x) + (7)(1) = 14x + 7. \] Thus \(14x + 7\) is equivalent to \( 7(2x + 1) \). We see that \( 14x + 7 \) has factors \(7\) and \( (2x + 1) \). The factors \(7\) and \( (2x + 1) \) multiply together to give \( 14x + 7 \). The process of writing an expression as a product of its factors is called factorisation. When asked to factorise \( 14x + 7 \) we write \[ 14x + 7 = 7(2x + 1) \] and so we see that factorisation can be regarded as reversing the process of removing brackets. Always remember that the factors of an algebraic expression are multiplied together.

Example

Factorise the expression 4x + 20.

Solution

Both terms in the expression \(4x + 20\) are examined to see if they have any factors in common. Clearly \(20\) can be factorised as \( (4)(5) \) and so we can write \[ 4x + 20 = 4x + (4)(5) \] The factor \(4\) is common to both terms on the right; it is called a common factor, and is placed at the front and outside brackets to give \[ 4x + 20 = 4(x + 5). \] Note that the factorised form can be checked by removing the brackets again.

Example

Factorise \( z^2 - 5z \).

Solution

\[ (3x + 4)(x + 2) = (3x + 4)(x) + (3x + 4)(2) \] Note that since \( z^2 = z \times z \) we can write \[ z^2 - 5z = z(z) - 5z \] so that there is a common factor of \(z\). Hence \[ z^2 - 5z = z(z) - 5z = z(z - 5). \]

Example

Factorise \( 6x - 9y \).

Solution

By observation, we see that there is a common factor of 3. Thus \[ 6x - 9y = 3(2x - 3y). \]

Exercise

Factorise:

1. \( [value]x + [value]y =\)
2. \( [value]x - [value]y =\)
3. \( [value]x + [value]y =\)
4. \( [value]x + [value]z + [value]y =\)
Feedback: 1. We observe that there is a common factor of \([value]\). Thus \[ [value]x + [value]y = [value].\] 2. We observe that there is a common factor of \([value]\). Thus \[ [value]x - [value]y = [value].\] 3. We observe that there is a common factor of \([value]\). Thus \[ [value]x + [value]y = [value].\] 4. We observe that there is a common factor of \([value]\). Thus \[ [value]x + [value]z + [value]y= [value].\]

Exercise

Factorise:

1. \( [value]ab + a^2 =\)
2. \( xy + [value]xyz =\)
3. \( [value]x^2 - [value]x =\)
Feedback: 1. We observe that there is a common factor of \(a\). Thus \[ [value]ab + a^2= [value].\] 2. We observe that there is a common factor of \( xy \). Thus \[ xy + [value]xyz = [value].\] 3. We observe that there is a common factor of \([value]x\). Thus \[ [value]x^2 - [value]x = [value].\]

Exercise

Factorise:

1. \( [value]ab + a =\)
Feedback: We observe that there is a common factor of \(a\). Thus \[ [value]ab + a= [value].\]

Exercise

Factorise:

1. \( [value]x^2 + [value]x^3y + [value]yx^4=\)
Feedback: We observe that there is a common factor of \(x^2\). Thus \[ [value] \cdot x^2 + [value] \cdot x^3 \cdot y + [value] \cdot y \cdot x^4= [value].\]

5. Factorising quadratic expressions

Quadratic expressions commonly occur in many areas of mathematics, physics and engineering. Many quadratic expressions can be written as the product of two linear factors and, in this Section, we examine how these factors can be easily found:

Key Point

An expression of the form \[ ax^2 + bx + c, \hspace{1cm} a \neq 0, \] where \(a\), \(b\) and \(c\) are numbers is called a quadratic expression (in the variable x).

The numbers \(b\) and \(c\) may be zero but a must not be zero (for, then, the quadratic reduces to a linear expression or constant). The number a is called the coefficient of \(x^2\), \(b\) is the  coefficient of \(x\) and \(c\) is called the constant term.

Case 1

Consider the product \( (x+1)(x+2)\). Removing brackets yields \( x^2 +3x+2 \). Conversely, we see that the factors of \( x^2+3x+2\) are \( (x+1)\) and \( (x+2)\). However, if we were given the quadratic expression first, how would we factorise it? The following examples show how to do this but note that not all quadratic expressions can be easily factorised. To enable us to factorise a quadratic expression in which the coefficient of \(x^2\^ equals 1, we note the following expansion: \[ (x + m)(x + n) = x^2 + mx + nx + mn = x^2 + (m + n)x + mn \] So, given a quadratic expression we can think of the coefficient of \(x\) as \(m+n\) and the constant term as \(mn\). Once the values of \(m\) and \(n\) have been found the factors can be easily obtained.

Example

Factorise \(x^2 + 4x - 5\).

Solution

Writing \(x^2 + 4x - 5 = (x + m)(x + n) = x2 + (m + n)x + mn \), we seek numbers \(m\) and \(n\) such that \(m+ n = 4\) and \(mn = -5\). By trial and error it is not difficult to find that \(m = 5\) and \(n = -1 \) (or, the other way round, \( m = -1 \) and \(n = 5\) ). So we can write

\[ x2 + 4x - 5 = (x + 5)(x - 1) \] The answer can be checked easily by removing brackets.

Case 2

When the coefficient of \(x^2\) is not equal to \(1\) it may be possible to extract a numerical factor. For example, note that \(3x^2 + 18x + 24\) can be written as \(3(x^2 + 6x + 8)\) and then factorised as in the previous Example 56. Sometimes no numerical factor can be found and a slightly different approach may be taken. We will demonstrate a technique which can always be used to transform the given expression into one in which the coefficient of the squared variable equals 1.

Example

Factorise \(2x^2 + 5x + 3\).

Solution

First note the coefficient of \(x^2\); in this case is \(2\). Multiply the whole expression by this number and rearrange as follows: \[ 2(2x^2 + 5x + 3) = 2(2x^2) + 2(5x) + 2(3) = (2x^)2 + 5(2x) + 6. \] We now introduce a new variable \(z\) such that \(z = 2x\) Thus we can write \[ (2x)^2 + 5(2x) + 6 \text{ as } z^2 + 5z + 6. \] This can be factorised to give \( (z + 3)(z + 2)\). Returning to the original variable by replacing \(z\) by \(2x\) we find \[ 2(2x^2 + 5x + 3) = (2x + 3)(2x + 2) \] A factor of \(2\) can be extracted from the second bracket on the right so that \[ 2(2x^2 + 5x + 3) = 2(2x + 3)(x + 1), \] so that \[ 2x^2 + 5x + 3 = (2x + 3)(x + 1). \]

As an alternative to the technique of Example 56, experience and practice can often help us to identify factors. For example suppose we wish to factorise \(3x^2 + 7x + 2\). We write \[ 3x^2 + 7x + 2 = ( \hspace{1cm} )( \hspace{1cm} ) \] In order to obtain the term \(3x^2\) we can place terms \(3x\) and \(x\) in the brackets to give \[ 3x^2 + 7x + 2 = (3x + ? )(x + ? ) \] In order to obtain the constant \(2\), we consider the factors of \(2\). These are \(1\),\(2\) or \(-1\),\(-2\). By placing these factors in the brackets we can factorise the quadratic expression. Various possibilities exist: we could write \( (3x+2)(x+1)\) or \((3x+1)(x+2)\) or \((3x - 2)(x - 1)\) or \( (3x - 1)(x -2)\), only one of which is correct. By removing brackets from each in turn we look for the factorisation which produces the correct middle term, \(7x\). The correct factorisation is found to be \[ 3x^2 + 7x + 2 = (3x + 1)(x + 2). \] With practice you will be able to carry out this process quite easily.

Example

Factorise

1. \( x^2 - 36z^2,\)
2. \(25x^2 - 9z^2,\)
3. \(\alpha^2 - 1\)

Solution

In each case we are required to find the difference of two squared terms.

1. Note that \(x^2 - 36z^2 = x^2 - (6^z)2.\) This factorises as \( (x + 6z)(x - 6z).\)
2. Here \(25x^2 - 9z^2 = (5x)^2 - (3z)^2.\) This factorises as \((5x + 3z)(5x - 3z).\)
3. \( \alpha^2 - 1 = (\alpha + 1)(\alpha - 1). \)

Exercise

Factorise:

1. \( x^2 + [value]x + [value] =\)
2. \( x^2 + [value]x - [value] =\)
3. \( x^2 + [value]x + [value] =\)
4. \( x^2 - [value]x + [value]=\)
Feedback: 1. We need to write \( x^2 + [value]x + [value] \) in the form \((x + m)(x + n)\). Thus we seek numbers \(m\) and \(n\) such that \(m+ n = [value]\) and \(mn = [value]\). By trial and error it is not difficult to find that \(m = 1\) and \(n = [value]\) (or, the other way round, \(m = [value]\) and \(n = 1\)). So we can write \[ x^2 + [value]x + [value] = (x + 1)(x + [value]) . \] The answer can be checked easily by removing brackets. 2. We need to write \(x^2 + [value]x - [value] \) in the form \((x + m)(x + n)\). Thus we seek numbers \(m\) and \(n\) such that \(m+ n = [value]\) and \(mn = -[value]\). By trial and error it is not difficult to find that \(m = -1\) and \(n = [value]\) (or, the other way round, \(m = [value]\) and \(n = -1\)). So we can write \[x^2 + [value]x - [value] = (x-1)(x + [value]) . \] The answer can be checked easily by removing brackets. 3. We need to write \( x^2 + [value]x + [value] \) in the form \((x + m)(x + n)\). Thus we seek numbers \(m\) and \(n\) such that \(m+ n = [value]\) and \(mn = [value]\). By trial and error it is not difficult to find that \(m = [value]\) and \(n = [value]\) (or, the other way round, \(m = [value]\) and \(n = [value] \)). So we can write \[ x^2 + [value]x + [value] = (x+[value])(x+[value]) . \] The answer can be checked easily by removing brackets. 4. We need to write \( x^2 - [value]x + [value] \) in the form \((x + m)(x + n)\). Thus we seek numbers \(m\) and \(n\) such that \(m+ n = -[value]\) and \(mn = [value]\). By trial and error it is not difficult to find that \(m = -[value]\) and \(n = -[value]\). So we can write \[ x^2 - [value]x + [value] = (x-[value])(x-[value]) . \] The answer can be checked easily by removing brackets.

Exercise

Factorise:

1. \( [value]x^2 + [value]x + 1 =\)
2. \( [value]x^2 + [value]x +[value] =\)
3. \( [value]x^2 + [value]x - [value] =\)
4. \( [value]x^2 - [value]x - 1=\)
5. \( [value]x^2  - 1=\)
6. \( [value]=\)
7. \( [value]x^2 + [value]x + [value]=\)
Feedback: 1. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + 1) = [value]([value]x^2) + [value]([value])x + [value](1) = ([value]x)^2 + [value]([value]x) + [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value]([value]x) + [value] \text{ as } z^2 + [value]z + [value]. \] This can be factorised to give \( (z + [value])(z + 1)\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + 1) = ([value]x + [value])([value]x + 1) \] A factor of \([value]\) can be extracted from the first bracket on the right so that \[ [value]([value]x^2 + [value]x + 1) = [value](x + 1)([value]x + 1), \] so that \[ [value]x^2 + [value]x + 1 = ([value]x + 1)(x + 1). \] 2. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + [value]) = [value]([value]x^2) + [value]([value])x + [value]([value]) = ([value]x)^2 + [value]([value]x) + [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value]([value]x) + [value]. \text{ as } z^2 + [value]z + [value]. \] This can be factorised to give \( (z + [value])(z + [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + [value]) = ([value]x + [value])([value]x + [value]) \] A factor of \([value]\) can be extracted from both of the brackets on the right so that \[ [value]([value]x^2 + [value]x + [value]) = [value]^2(x + 1)(x + 1), \] so that \[ [value]x^2 + [value]x + [value] = [value](x + 1)(x + 1). \] 3. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value](3x^2 + [value]x - [value]) = [value]([value]x^2) + [value]([value])x - [value]([value]) = ([value]x)^2 + [value] ([value]x) - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value] ([value]x) - [value] \text{ as } z^2 +[value] z - [value]. \] This can be factorised to give \( (z + [value])(z - [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value](3x^2 + [value]x - [value]) = ([value]x + [value])([value]x - [value]) \] A factor of \([value]\) can be extracted from both brackets on the right so that \[ [value](3x^2 + [value]x - [value]) = [value] (x + [value])(x - [value]), \] so that \[ 3x^2 + [value]x - [value] = [value] (x + [value])(x - [value]). \] 4. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 - [value]x -1) = [value]([value]x^2) - [value]([value])x - [value](1) = ([value]x)^2 - [value] ([value]x) - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 - [value] ([value]x) - [value]. \text{ as } z^2 -[value] z - [value]. \] This can be factorised to give \( (z - [value])(z + 1)\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 - [value]x -1) = ([value]x - [value])([value]x + 1) \] A factor of \([value]\) can be extracted from the first bracket on the right so that \[ [value]([value]x^2 - [value]x -1) = [value](x - 1)([value]x + 1), \] so that \[ [value]x^2 - [value]x -1 = (x - 1)([value]x + 1). \] 5. First note the coefficient of \(x^2\); in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 -1) = [value]([value]x^2) - [value](1) = ([value]x)^2 - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\) Thus we can write \[ ([value]x)^2 - [value] \text{ as } z^2 - [value]. \] This can be factorised to give \( (z + [value])(z - [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 -1) = ([value]x + [value])([value]x - [value]) \] A factor of \([value]\) can be extracted from both of the brackets on the right so that \[ [value]([value]x^2 -1) = [value]([value]x + 1)([value]x - 1), \] so that \[ [value]x^2 -1 = ([value]x + 1)([value]x - 1). \] 6. First note the coefficient of \(x^2\); in this case is \(-1\). Multiply the whole expression by this number and rearrange as follows: \[ -1(-x^2 + 1) = -1(-x^2) -1(1) = x^2 - 1. \] This can be factorised to give \( (x + 1)(x - 1)\). Thus, \[ -(-x^2 + 1) = (x + 1)(x - 1) \] A factor of \(-1\) can be extracted from the second bracket on the right so that \[ -(-x^2 + 1) = -(x + 1)(1 - x), \] so that \[ -x^2 + 1 = (x + 1)(1 - x). \] 7. First note the coefficient of \(x^2\) in this case is \(-[value]\). Multiply the whole expression by this number and rearrange as follows: \[ -[value](-[value]x^2 + [value]x + [value]) = -[value](-[value]x^2) - [value]([value])x - [value]([value]) = ([value]x)^2 - [value]([value]x) - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 - [value]([value]x) - [value] \text{ as } z^2 - [value]z - [value]. \] This can be factorised to give \( (z - [value])(z + [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ -[value](-[value]x^2 + [value]x + [value]) = ([value]x - [value])([value]x + [value]) \] A factor of \([value]\) can be extracted from the second bracket, and a factor of \(-1\) from the first bracket on the right so that \[ -[value](-[value]x^2 + [value]x + [value]) = -[value] (-[value]x + [value])(x + 1), \] so that \[ (-[value]x^2 + [value]x + [value]) =([value] - [value]x)(x + 1). \]

Exercise

Factorise:

1. \( [value]x^2 + [value] =\)
2. \( [value]x^2 + [value]x + 1 =\)
3. \( [value]x^2 + [value]x + 1 =\)
4. \( [value]x^2 + [value]x + [value]=\)
5. \( [value]x^2 + [value]x  - [value]=\)
6. \( [value]x^2 + [value]x + [value]=\)
Feedback: 1. Note that \(x\) is a common factor of both terms. Thus \[ [value]x^2 + x = x([value]x + 1). \] 2. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + 1) = [value]([value]x^2) + [value]([value])x + [value](1) = ([value]x)^2 + [value]([value]x) + [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value]([value]x) + [value] \text{ as } z^2 + [value]z + [value]. \] This can be factorised to give \( (z + [value])(z + 1\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + 1) = ([value]x + [value])([value]x + 1) \] A factor of \([value]\) can be extracted from the first bracket on the right so that \[ [value]([value]x^2 + [value]x + 1) = [value](x + 1)([value]x + 1), \] so that \[ [value]x^2 + [value]x + 1 = (x + 1)([value]x + 1). \] 3. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + 1) = [value]([value]x^2) + [value]([value])x - [value](1) = ([value]x)^2 + [value] ([value]x) - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value] ([value]x) - [value] \text{ as } z^2 +[value] z - [value]. \] This can be factorised to give \( (z + [value])(z + [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + 1) = ([value]x + [value])([value]x + [value]). \] A factor of \([value]\) can be extracted from the first bracket on the right, and a factor of \([value]\) can be extracted from the second bracket on the right, so that \[ [value]([value]x^2 + [value]x + 1) = [value]([value]x + 1)([value]x + 1), \] so that \[ [value]x^2 + [value]x + 1 = ([value]x + 1)([value]x + 1). \] 4. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + [value]) = [value]([value]x^2) + [value]([value])x + [value]([value]) = ([value]x)^2 + [value] ([value]x) + [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value] ([value]x) + [value] \text{ as } z^2 +[value] z + [value]. \] This can be factorised to give \( (z + [value])(z + [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + [value]) = ([value]x + [value])([value]x + [value]) \] A factor of \([value]\) can be extracted from the first bracket on the right, and a factor of \([value]\) can be extracted from the second bracket on the right, so that/p> \[ [value]([value]x^2 + [value]x + [value]) =([value]([value]x + [value])([value]x + [value]) \] so that \[ [value]x^2 + [value]x + [value] = ([value]x + [value])([value]x + [value]). \] 5. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x - [value]) = [value]([value]x^2) + [value]([value])x - [value]([value]) = ([value]x)^2 + [value] ([value]x)- [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 + [value] ([value]x) - [value] \text{ as } z^2 +[value] z - [value]. \] This can be factorised to give \( (z - [value])(z + [value])\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x - [value]) = ([value]x - [value])([value]x + [value]) \] A factor of \([value]\) can be extracted from the first bracket on the right, and a factor of \([value]\) can be extracted from the second bracket on the right, so that \[ [value]([value]x^2 + [value]x - [value]) =([value]([value]x - [value])([value]x + [value]) \] so that \[ [value]x^2 + [value]x - [value] = ([value]x - [value])([value]x + [value]). \] 6. First note the coefficient of \(x^2\) in this case is \([value]\). Multiply the whole expression by this number and rearrange as follows: \[ [value]([value]x^2 + [value]x + [value]) = [value]([value]x^2) [value]([value])x + [value]([value] = ([value]x)^2 -[value]([value]x) - [value]. \] We now introduce a new variable \(z\) such that \(z = [value]x\). Thus we can write \[ ([value]x)^2 -[value]([value]x); - [value] \text{ as } z^2 -[value]z - [value]. \] This can be factorised to give \( (z + [value])(z - [value]\). Returning to the original variable by replacing \(z\) by \([value]x\) we find \[ [value]([value]x^2 + [value]x + [value]) = ([value]x + [value])([value]x - [value]) \] A factor of \([value]\) can be extracted from the first bracket on the right, and a factor of /(-1/) can be extracted from the second bracket on the right so that \[ [value]([value]x^2 + [value]x + [value]) = -[value](x + 1)(-[value]x + [value]), \] so that \[ [value]x^2 + [value]x + [value] = (x + 1)(-[value]x + [value]). \]

Engineering Example 1: Reliability in a communication network

Introduction

The reliability of a communication network depends on the reliability of its component parts. The reliability of a component can be represented by a number between \(0\) and \(1\) which represents the probability that it will function over a given period of time. A very simple system with only two components \(C_1\) and \(C_2\) can be configured in series or in parallel. If the components are in series then the system will fail if one component fails (see the Figure below).

                [Image]

If the components are in parallel then only one component need function properly and we have built-in redundancy (see the Figure below).

                      [Image]

The reliability of a system with two units in parallel is given by \( 1 - (1 - R_1)(1 - R_2) \) which is the same as \( R_1 + R_2 - R_1R_2 \), where \( R_i \) is the reliability of component \( C_i \). The reliability of a system with \(3\) units in parallel, (see the Figure below), is given by

\[ 1 - (1 - R_1)(1 - R_2)(1 - R_3). \]

[Image]

Problem in words

(a) Show that the expression for the system reliability for three components in parallel is equal to

\[ R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3. \]

(b) Find an expression for the reliability of the system when the reliability of each of the components is the same i.e. \( R_1 = R_2 = R_3 = R. \)

(c) Find the system reliability when \( R = 0:75.\)

(d) Find the system reliability when there are two parallel components each with reliability \( R = 0:75.\)

Mathematical statement of the problem.

(a) Show that \(1-(1-R_1)(1-R_2)(1-R_3) - R_1+R_2+R_3-R_1R_2-R_1R_3-R_2R_3+R_1R_2R_3.\)

(b) Find \(1 - (1 - R_1)(1 - R_2)(1 - R_3)\) in terms of \(R\) when \(R_1 = R_2 = R_3 = R.\)

(c) Find the value of (b) when \( R = 0.75.\)

(d) Find \(1 - (1 - R_1)(1 - R_2)\) when \(R_1 = R_2 = 0.75.\)

Mathematical analysis

(a) \[ 1 - (1 - R_1)(1 - R_2)(1 - R_3) - 1 - (1 - R_1 - R_2 + R_1R_2)(1 - R_3) \]

\[ = 1 - ((1 - R_1 - R_2 + R_1R_2) - 1 - (1 - R_1 - R_2 + R_1R_2) - R_3) \]

\[= 1 - (1 - R_1 - R_2 + R_1R_2 - (R_3 - R_1R_3 - R_2R_3 + R_1R_2R_3))\]

\[= 1 - (1 - R_1 - R_2 + R_1R_2 - R_3 + R_1R_3 + R_2R_3 - R_1R_2R_3) \]

\[= 1 - 1 + R_1 + R_2 - R_1R_2 + R_3 - R_1R_3 - R_2R_3 + R_1R_2R_3 \]

\[= R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3.\]

(b) When \( R_1 = R_2 = R_3 = R\) the reliability is

\[ 1 - (1 - R)^3 \text{   which is equivalent to   } 3R - 3R^2 + R^3. \]

(c) When \( R_1 = R_2 = R_3 = 0.75\) we get

\[ 1 - (1 - 0.75)^3 = 1 - 0.25^3 = 1 - 0.015625 = 0.984375.\]

(d) \(1 - (0.25)^2 = 0.9375.\)

Interpretation

The mathematical analysis confirms the expectation that the more components there are in parallel then the more reliable the system becomes (1 component: 0.75; 2 components: 0.9375; 3 components: 0.984375). With three components in parallel, as in part (c), although each individual component is relatively unreliable (\(R = 0.75\) implies a one in four chance of failure of an individual component) the system as a whole has an over 98% probability of functioning (under 1 in 50 chance of failure).