Algebraic identities are useful for expanding and factorising algebraic expressions, simplifying calculations, recognising patterns and simplifying rational expressions. This chapter develops identities through algebra, geometrical models and algebra tiles.

4.1 Introduction

In earlier chapters, we studied linear polynomials and linear equations. We now move to algebraic identities—mathematical rules that remain true for every permissible value of the variables involved.

A pattern involving consecutive squares

Take three consecutive square numbers, such as $1,4,9$. Add the smallest and largest squares and subtract twice the middle square:

$$(1+9)-2\left(4\right)=10-8=2.$$

The same happens with $25,36,49$:

$$(25+49)-2\left(36\right)=74-72=2.$$

To explain why, write three consecutive numbers as $n-1,n,n+1$. Their squares are ${(n-1)}^2,n^2,{(n+1)}^2$. Then

$$\begin{array}{cc}\hfill \hfill & \hfill {(n-1)}^2+{(n+1)}^2-2n^2=(n^2-2n+1)+(n^2+2n+1)-2n^2\hfill \\ \hfill \hfill & \hfill =2.\hfill \end{array}$$

Thus, for any three consecutive square numbers:

$$\mathrm{\text{smallest square}}+\mathrm{\text{largest square}}-2\left(\mathrm{\text{middle square}}\right)=2.$$

This example shows how algebra can turn an observed numerical pattern into a proof.

4.2 Visualising Identities

Meaning of an algebraic identity

An algebraic identity is an equation that is true for all permissible values of the variables occurring in it.

Equation and identity

Geometrical model for ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$

Construct a square whose side is $a+b$. Its area is ${(a+b)}^2$. Divide it into:

• one square of area $a^2$;
• one square of area $b^2$; and
• two rectangles, each of area $ab$.

The sum of the areas of these parts equals the area of the larger square. Therefore,

$${(a+b)}^2=a^2+2ab+b^2.$$

The geometrical model verifies the identity when $a$ and $b$ represent lengths. Algebra proves that it also works for negative numbers, rational numbers and other real numbers.

Algebraic derivation

Using the distributive property:

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b)}^2=(a+b)(a+b)\hfill \\ \hfill \hfill & \hfill =a(a+b)+b(a+b)\hfill \\ \hfill \hfill & \hfill =a^2+ab+ba+b^2\hfill \\ \hfill \hfill & \hfill =a^2+2ab+b^2.\hfill \end{array}$$

Checking negative values

Let $a=-2$ and $b=-3$. Then

$${(a+b)}^2={(-5)}^2=25,$$

while

$$a^2+2ab+b^2=4+12+9=25.$$

Thus, the identity remains true even when both values are negative.

Comparing ${(𝐚\mathbf{+}𝐛)}^{\mathrm{𝟐}}$ and ${𝐚}^{\mathrm{𝟐}}\mathbf{+}{𝐛}^{\mathrm{𝟐}}$

Since

$${(a+b)}^2=a^2+b^2+2ab,$$

the term $2ab$ determines which expression is greater:

• If $ab>0$, then ${(a+b)}^2>a^2+b^2$. Thus, $a$ and $b$ have the same non-zero sign.
• If $ab<0$, then ${(a+b)}^2<a^2+b^2$. Thus, $a$ and $b$ have opposite signs.
• If $ab=0$, then ${(a+b)}^2=a^2+b^2$. Thus, $a=0$, $b=0$, or both are zero.

Caution: In general, ${(a+b)}^2\ne a^2+b^2$. The middle term $2ab$ must not be omitted.

Using the identity to expand an expression

For example,

$$\begin{array}{cc}\hfill \hfill & \hfill {(5x+2y)}^2={\left(5x\right)}^2+2\left(5x\right)\left(2y\right)+{\left(2y\right)}^2\hfill \\ \hfill \hfill & \hfill =25x^2+20xy+4y^2.\hfill \end{array}$$

Using the identity for numerical calculation

$$\begin{array}{cc}\hfill \hfill & \hfill {43}^2={(40+3)}^2\hfill \\ \hfill \hfill & \hfill ={40}^2+2\left(40\right)\left(3\right)+3^2\hfill \\ \hfill \hfill & \hfill =1600+240+9\hfill \\ \hfill \hfill & \hfill =1849.\hfill \end{array}$$

4.3 Factorisation of Algebraic Expressions Using Identities

Expansion expresses a product as a sum of terms. Factorisation is the reverse process: it expresses an algebraic expression as a product of its factors.

The identity

$$a^2+2ab+b^2={(a+b)}^2$$

helps us recognise a perfect-square trinomial.

Recognising a perfect-square trinomial

Check whether:

1. the first term is a perfect square;
2. the last term is a perfect square; and
3. the middle term is twice the product of their square roots.

Example 1: ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟒}𝐱\mathbf{+}\mathrm{𝟒}$

Here,

$$x^2={\left(x\right)}^2,4=2^2,4x=2\left(x\right)\left(2\right).$$

Therefore,

$$x^2+4x+4={(x+2)}^2.$$

Example 2: $\mathrm{𝟑𝟔}{𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟐}𝐱\mathbf{+}\mathrm{𝟏}$

$$\begin{array}{cc}\hfill \hfill & \hfill 36x^2+12x+1={\left(6x\right)}^2+2\left(6x\right)\left(1\right)+1^2\hfill \\ \hfill \hfill & \hfill ={(6x+1)}^2.\hfill \end{array}$$

Taking out a common factor first

Consider $50p^2+60pq+18q^2$. All three terms have the common factor $2$:

$$\begin{array}{cc}\hfill \hfill & \hfill 50p^2+60pq+18q^2=2(25p^2+30pq+9q^2)\hfill \\ \hfill \hfill & \hfill =2[{\left(5p\right)}^2+2\left(5p\right)\left(3q\right)+{\left(3q\right)}^2]\hfill \\ \hfill \hfill & \hfill =2{(5p+3q)}^2.\hfill \end{array}$$

Caution: Before applying an identity, check whether the terms have a common factor. Taking it out first often makes the perfect-square pattern visible.

Identity for the square of a difference

Replace $b$ by $-b$ in ${(a+b)}^2=a^2+2ab+b^2$:

$$\begin{array}{cc}\hfill \hfill & \hfill {[a+(-b)]}^2=a^2+2a(-b)+{(-b)}^2\hfill \\ \hfill \hfill & \hfill {(a-b)}^2=a^2-2ab+b^2.\hfill \end{array}$$

Thus,

$${(a-b)}^2=a^2-2ab+b^2.$$

The last term remains positive because ${(-b)}^2=b^2$.

Numerical example

$$\begin{array}{cc}\hfill \hfill & \hfill {29}^2={(30-1)}^2\hfill \\ \hfill \hfill & \hfill ={30}^2-2\left(30\right)\left(1\right)+1^2\hfill \\ \hfill \hfill & \hfill =900-60+1\hfill \\ \hfill \hfill & \hfill =841.\hfill \end{array}$$

Geometrical interpretation of ${(𝐚\mathbf{-}𝐛)}^{\mathrm{𝟐}}$

Begin with a square of side $a$, having area $a^2$. The smaller square of side $a-b$ is obtained after subtracting rectangles of areas $ab$ and $b(a-b)$. Therefore,

$$\begin{array}{cc}\hfill \hfill & \hfill {(a-b)}^2=a^2-ab-b(a-b)\hfill \\ \hfill \hfill & \hfill =a^2-ab-ab+b^2\hfill \\ \hfill \hfill & \hfill =a^2-2ab+b^2.\hfill \end{array}$$

4.4 More Identities

Square of the sum of three terms

To find ${(a+b+c)}^2$, let $d=b+c$. Then

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b+c)}^2={(a+d)}^2\hfill \\ \hfill \hfill & \hfill =a^2+2ad+d^2\hfill \\ \hfill \hfill & \hfill =a^2+2a(b+c)+{(b+c)}^2\hfill \\ \hfill \hfill & \hfill =a^2+2ab+2ac+b^2+2bc+c^2.\hfill \end{array}$$

Hence,

$${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca.$$

Geometrically, a square of side $a+b+c$ can be divided into three squares of areas $a^2,b^2,c^2$ and six rectangles representing the pairwise products $ab,bc,ca$, with each product occurring twice.

Example: Finding ${\mathrm{𝟏𝟏𝟗}}^{\mathrm{𝟐}}$

$$\begin{array}{cc}\hfill \hfill & \hfill {119}^2={(100+10+9)}^2\hfill \\ \hfill \hfill & \hfill ={100}^2+{10}^2+9^2+2\left(100\right)\left(10\right)+2\left(100\right)\left(9\right)+2\left(10\right)\left(9\right)\hfill \\ \hfill \hfill & \hfill =10000+100+81+2000+1800+180\hfill \\ \hfill \hfill & \hfill =14161.\hfill \end{array}$$

Difference of two squares

The identity introduced in Grade 8 is

$$a^2-b^2=(a+b)(a-b).$$

It can be rearranged as

$$a^2=(a+b)(a-b)+b^2.$$

In 750 CE, Śhrīdharāchārya proposed this form as a method for calculating squares quickly.

Example: Finding ${\mathrm{𝟓𝟓}}^{\mathrm{𝟐}}$

$$\begin{array}{cc}\hfill \hfill & \hfill {55}^2=(55+5)(55-5)+5^2\hfill \\ \hfill \hfill & \hfill =60\times 50+25\hfill \\ \hfill \hfill & \hfill =3025.\hfill \end{array}$$

Pattern for numbers ending in $\mathrm{𝟓}$

If a number ending in $5$ is written as $10n+5$, then

$${(10n+5)}^2=100n(n+1)+25.$$

Therefore, multiply the part before $5$ by its successor and place $25$ at the end. For example,

$${65}^2=6\times 7\mathrm{\text{followed by}}25=4225.$$

An additional identity represented in the chapter

The arrangement in Fig. 4.6 represents:

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b+c)}^2+{(a+b-c)}^2\hfill \\ \hfill \hfill & \hfill +{(a-b+c)}^2+{(a-b-c)}^2\hfill \\ \hfill \hfill & \hfill =4(a^2+b^2+c^2).\hfill \end{array}$$

4.5 Factorisation Using Algebra Tiles

Algebra tiles give a geometrical meaning to algebraic terms:

• an $x^2$-tile is a square of dimensions $x\times x$;
• an $x$-tile is a rectangle of dimensions $x\times 1$; and
• a unit tile is a square of dimensions $1\times 1$.

Example: Factorising ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟏𝟐}$

Consider a rectangle whose sides are $x+3$ and $x+4$. Its area is

$$\begin{array}{cc}\hfill \hfill & \hfill (x+3)(x+4)=x^2+3x+4x+12\hfill \\ \hfill \hfill & \hfill =x^2+7x+12.\hfill \end{array}$$

Using algebra tiles:

• one $x^2$-tile represents $x^2$;
• seven $x$-tiles represent $7x$; and
• twelve unit tiles represent $12$.

The seven $x$-tiles are split as $3x+4x$, and the twelve unit tiles form a $3\times 4$ array. The complete arrangement forms a rectangle with sides $x+3$ and $x+4$. Therefore,

$$x^2+7x+12=(x+3)(x+4).$$

If $7x$ were split as $2x+5x$, the unit tiles would have to form a $2\times 5$ array containing only $10$ tiles, not $12$. Therefore, that split would not produce the required rectangle.

Further tile patterns

$$(x+2)(x+3)=x^2+5x+6.$$

Also, since $5+6=11$ and $5\times 6=30$,

$$x^2+11x+30=(x+5)(x+6).$$

General product identities

From the same pattern,

$$(x+a)(x+b)=x^2+(a+b)x+ab.$$

More generally,

$$(px+a)(qx+b)=pq{x}^2+(pb+aq)x+ab.$$

For example,

$$\begin{array}{cc}\hfill \hfill & \hfill (2x+3)(3x+1)=6x^2+2x+9x+3\hfill \\ \hfill \hfill & \hfill =6x^2+11x+3.\hfill \end{array}$$

4.6 Factorisation Without Using Algebra Tiles

The tile method can be converted into an algebraic method by splitting the middle term.

Factorising ${𝐱}^{\mathrm{𝟐}}\mathbf{+}𝐁𝐱\mathbf{+}𝐂$

Find two numbers $a$ and $b$ such that

$$a+b=B\mathrm{\text{and}}ab=C.$$

Then

$$x^2+Bx+C=(x+a)(x+b).$$

Example 1: ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟕}𝐱\mathbf{+}\mathrm{𝟏𝟐}$

We need two numbers whose sum is $7$ and product is $12$. They are $3$ and $4$. Therefore,

$$x^2+7x+12=(x+3)(x+4).$$

Example 2: ${𝐱}^{\mathrm{𝟐}}\mathbf{+}\mathrm{𝟏𝟏}𝐱\mathbf{+}\mathrm{𝟑𝟎}$

The required numbers are $5$ and $6$, because

$$5+6=11\mathrm{\text{and}}5\times 6=30.$$

Hence,

$$x^2+11x+30=(x+5)(x+6).$$

Although $2\times 15=30$ and $3\times 10=30$, these pairs do not work because their sums are not $11$.

Example 3: ${𝐱}^{\mathrm{𝟐}}\mathbf{-}\mathrm{𝟓}𝐱\mathbf{+}\mathrm{𝟔}$

We need two numbers whose sum is $-5$ and product is $6$. They are $-2$ and $-3$. Therefore,

$$x^2-5x+6=(x-2)(x-3).$$

Sign guide

• If $C>0$, the two numbers have the same sign. Their sum determines whether both are positive or both are negative.
• If $C<0$, the two numbers have opposite signs.
• Always check both the sum and the product before choosing the pair.

When the coefficient of ${𝐱}^{\mathrm{𝟐}}$ is not $\mathrm{𝟏}$

For $A{x}^2+Bx+C$, look for two numbers whose product is $AC$ and whose sum is $B$. Split $Bx$ using these numbers and factor by grouping.

For example,

$$\begin{array}{cc}\hfill \hfill & \hfill 6x^2+7x+2=6x^2+3x+4x+2\hfill \\ \hfill \hfill & \hfill =3x(2x+1)+2(2x+1)\hfill \\ \hfill \hfill & \hfill =(3x+2)(2x+1).\hfill \end{array}$$

Caution: A pair is correct only when it satisfies both conditions. Matching the product alone is not enough.

4.7 Finding New Identities

Cube of a sum

Using ${(a+b)}^2=a^2+2ab+b^2$ and the distributive property:

$$\begin{array}{cc}\hfill \hfill & \hfill {(a+b)}^3=(a+b){(a+b)}^2\hfill \\ \hfill \hfill & \hfill =(a+b)(a^2+2ab+b^2)\hfill \\ \hfill \hfill & \hfill =a^3+3a^{2}b+3a{b}^2+b^3.\hfill \end{array}$$

Thus,

$${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3.$$

Geometrical interpretation

A cube of edge $a+b$ can be divided into:

• one cube of volume $a^3$;
• one cube of volume $b^3$;
• three cuboids, each of volume $a^{2}b$; and
• three cuboids, each of volume $a{b}^2$.

Their total volume is $a^3+3a^{2}b+3a{b}^2+b^3$.

Cube of a difference

Replace $b$ by $-b$ in the cube-of-a-sum identity:

$${(a-b)}^3=a^3-3a^{2}b+3a{b}^2-b^3.$$

The signs of the four terms alternate.

Recognising a perfect cube

$$\begin{array}{cc}\hfill \hfill & \hfill p^3+6p^{2}q+12p{q}^2+8q^3=p^3+3p^2\left(2q\right)+3p{\left(2q\right)}^2+{\left(2q\right)}^3\hfill \\ \hfill \hfill & \hfill ={(p+2q)}^3.\hfill \end{array}$$

Therefore, a cube having this volume has side $p+2q$ units.

Similarly,

$$8n^3-60n^{2}m+150n{m}^2-125m^3={(2n-5m)}^3.$$

Difference of two cubes

$$\begin{array}{cc}\hfill \hfill & \hfill (x-y)(x^2+xy+y^2)=x^3+x^{2}y+x{y}^2-x^{2}y-x{y}^2-y^3\hfill \\ \hfill \hfill & \hfill =x^3-y^3.\hfill \end{array}$$

Hence,

$$x^3-y^3=(x-y)(x^2+xy+y^2).$$

Sum of two cubes

Similarly,

$$x^3+y^3=(x+y)(x^2-xy+y^2).$$

Sign reminder: In the factorisation of a sum or difference of cubes, the sign in the first factor is the same as the original sign, while the middle term in the quadratic factor has the opposite sign.

The factor $𝐱\mathbf{-}𝐲$ in higher powers

We know that

$$x^2-y^2=(x-y)(x+y)$$

and

$$x^3-y^3=(x-y)(x^2+xy+y^2).$$

Also,

$$x^4-y^4=(x-y)(x+y)(x^2+y^2),$$

and

$$x^5-y^5=(x-y)(x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4).$$

Thus, $x-y$ is a factor of each of these differences.

Identity involving three cubes

Expanding the product and combining like terms gives:

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$

Application

Suppose

$$x+y+z=10,xyz=25,x^2+y^2+z^2=38.$$

First use

$${(x+y+z)}^2=x^2+y^2+z^2+2(xy+xz+yz).$$

Therefore,

$$100=38+2(xy+xz+yz),$$

so

$$xy+xz+yz=31.$$

Now apply the three-cube identity:

$$\begin{array}{cc}\hfill \hfill & \hfill x^3+y^3+z^3-3xyz=10(38-31)\hfill \\ \hfill \hfill & \hfill =70.\hfill \end{array}$$

Since $3xyz=75$,

$$x^3+y^3+z^3=145.$$

4.8 Simplifying Rational Expressions

A rational algebraic expression can often be simplified by factorising its numerator and denominator and cancelling their common non-zero factors.

Steps for simplification

1. Factorise the numerator completely.
2. Factorise the denominator completely.
3. Identify common multiplicative factors.
4. Cancel only those common factors that are non-zero.
5. Retain all restrictions obtained from the original denominator.

Example

Simplify

$$\frac{x^2-7x+12}{5x^2+5x-100},$$

assuming that the denominator is not zero.

Factorise the numerator:

$$x^2-7x+12=(x-3)(x-4).$$

Factorise the denominator:

$$\begin{array}{cc}\hfill \hfill & \hfill 5x^2+5x-100=5(x^2+x-20)\hfill \\ \hfill \hfill & \hfill =5(x-4)(x+5).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill \hfill & \hfill \frac{x^2-7x+12}{5x^2+5x-100}=\frac{(x-3)(x-4)}{5(x-4)(x+5)}\hfill \\ \hfill \hfill & \hfill =\frac{x-3}{5(x+5)}.\hfill \end{array}$$

The original denominator is zero for $x=4$ and $x=-5$. Hence the restrictions are

$$x\ne 4,-5.$$

Even though the factor $x-4$ is cancelled, $x=4$ remains excluded because it was not allowed in the original expression.

Another example

$$\frac{36s^2-12st+t^2}{t^2+2ts-48s^2}=\frac{{(t-6s)}^2}{(t+8s)(t-6s)}=\frac{t-6s}{t+8s},$$

where

$$t\ne 6s,-8s.$$

Caution: Only complete factors can be cancelled. Terms connected by addition or subtraction cannot be cancelled separately.

Application 1: Finding the dimensions of a rectangle

Saira uses a square of side $x$, eight $x\times 1$ strips and fifteen unit squares. Their total area is

$$x^2+8x+15.$$

Factorising,

$$x^2+8x+15=(x+3)(x+5).$$

Therefore, the possible dimensions of the rectangle are

$$x+3\mathrm{\text{units and}}x+5\mathrm{\text{units}}.$$

Application 2: Rectangular pool

Let the length of the pool be $x$ metres. Its breadth is $x-4$ metres and its area is $96{\mathrm{\text{m}}}^2$. Therefore,

$$x(x-4)=96.$$

$$x^2-4x-96=0.$$

$$(x-12)(x+8)=0.$$

Thus, $x=12$ or $x=-8$. A physical length cannot be negative, so $x=-8$ is rejected. Hence,

$$\mathrm{\text{Length}}=12\mathrm{\text{m}},\mathrm{\text{breadth}}=8\mathrm{\text{m}}.$$

Correction to the printed working: The factors used for splitting the middle term satisfy $(-12)\times 8=-96$, not $96$.

Chapter Summary

• An algebraic identity is an equation that is true for every permissible value of its variables.
• Identities can be visualised using geometrical models and algebra tiles.
• Identities help us expand and factorise algebraic expressions.
• A perfect-square trinomial can be recognised from its first term, last term and middle term.
• A common factor should be taken out before applying an identity when necessary.
• Quadratic expressions can be factorised by splitting the middle term.
• Identities make calculations such as squares, cubes and products easier.
• Rational expressions can be simplified by factorising and cancelling common non-zero factors.
• Restrictions from the original denominator must be retained after simplification.
• A negative solution must be rejected when a variable represents a physical length.

The 11 identities studied in this chapter

1. Square of a sum

$${(x+y)}^2=x^2+2xy+y^2$$

2. Square of a difference

$${(x-y)}^2=x^2-2xy+y^2$$

3. Square of three terms

$${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

4. Difference of two squares

$$(x+y)(x-y)=x^2-y^2$$

5. Product of two monic linear expressions

$$(x+a)(x+b)=x^2+(a+b)x+ab$$

6. Product of two general linear expressions

$$(ax+b)(cx+d)=ac{x}^2+(ad+bc)x+bd$$

7. Difference of two cubes

$$x^3-y^3=(x-y)(x^2+xy+y^2)$$

8. Sum of two cubes

$$x^3+y^3=(x+y)(x^2-xy+y^2)$$

9. Cube of a sum

$${(x+y)}^3=x^3+3x^{2}y+3x{y}^2+y^3$$

10. Cube of a difference

$${(x-y)}^3=x^3-3x^{2}y+3x{y}^2-y^3$$

11. Identity involving three cubes

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$

Quick revision checklist

• Identify whether the expression matches a known identity.
• While expanding, write every middle or cross-product term.
• While factorising, first remove any common factor.
• For $x^2+Bx+C$, choose two numbers whose sum is $B$ and product is $C$.
• For $A{x}^2+Bx+C$, use the product $AC$ to split the middle term.
• Cancel only common factors, never separate terms.
• State excluded values whenever a denominator is present.