Exploring Algebraic Identities

Contents hide

1What Is an Identity?

2Square Identities

3Factoring Quadratics

4Cube Identities

5Simplifying Rational Expressions

Chapter mind map: how it all connects

1 · Square identities(x±y)², (x+y+z)² and x²−y², visualised with area models

2 · Factoring quadraticsSplitting the middle term, matching perfect-square patterns

Exploring Algebraic Identities

3 · Cube identities(x±y)³, x³±y³, and the three-variable identity x³+y³+z³−3xyz

4 · ApplicationsFast squaring/cubing, area word problems, simplifying rational expressions

What you will learn in this chapter

The difference between an equation and an identity
All the standard square and cube identities, and how to visualise them geometrically
How to factor a quadratic by splitting the middle term
How to simplify a rational expression by factoring and cancelling common factors

identityexpansionfactorisationsplitting the middle termperfect squarerational expression

1What Is an Identity?

Learn by heartDefinition 1

An identity is an equation that is true for every value of the variables in it. An equation that is only true for particular values is not an identity. E.g. $x^2-1=24$ is true only for $x=\pm5$ (an equation), but $(x+y)^2=x^2+2xy+y^2$ is true for all $x,y$ (an identity).

Take any three consecutive square numbers, say $1, 4, 9$ . Add the smallest and largest, subtract twice the middle: $(1+9)-2(4)=2$ . Try $25,36,49$ : $(25+49)-2(36)=2$ again. This always gives 2, and algebra explains why: writing the numbers as $(n-1)^2, n^2, (n+1)^2$ , we get $(n-1)^2+(n+1)^2-2n^2=2n^2+2-2n^2=2$ for every $n$ .

2Square Identities

Figure 1: A square of side (a+b) splits into an a×a square, a b×b square, and two ab rectangles, giving (a+b)² = a² + 2ab + b². — **Figure 1**: A square of side (a+b) splits into an a×a square, a b×b square, and two ab rectangles, giving (a+b)² = a² + 2ab + b².

Formula

(x+y)^2 = x^2+2xy+y^2

Also true for negative and rational

x,y

, proved generally using the distributive property:

(x+y)(x+y)=x(x+y)+y(x+y)

Formula

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

Replace

y

with

-y

in the identity above

Solved Example

Q. Find $43^2$ and $29^2$ using identities.

$43^2=(40+3)^2=1600+240+9=1849$ . $29^2=(30-1)^2=900-60+1=\;$ $841$

Formula

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

Every pair of variables contributes a “2×product” cross term

Solved Example

Q. Find $122^2$ using the three-term identity.

$122^2=(100+20+2)^2=10000+400+4+2(2000)+2(200)+2(40)=10404+4000+400+80=\;$ $14884$

Formula: Śhrīdharāchārya’s identity

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

, equivalently

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

Proposed 750 CE as a quick way to square numbers

Solved Example

Q. Use Śhrīdharāchārya’s method to find $55^2$ .

$55^2=(55+5)(55-5)+5^2=60\times50+25=3000+25=\;$ $3025$

3Factoring Quadratics

The identities above run in reverse too: they factor expressions, not just expand them.

Solved Example

Q. Factor $9x^2+30x+25$ .

$9x^2=(3x)^2$ , $25=5^2$ , $30x=2(3x)(5)$ , matching $a^2+2ab+b^2$ with $a=3x,b=5$ . So $9x^2+30x+25=\;$ $(3x+5)^2$

Splitting the middle term

To factor $x^2+(a+b)x+ab$ as $(x+a)(x+b)$ : find two numbers $a,b$ whose sum is the coefficient of $x$ and whose product is the constant term.

Solved Example

Q. Factor $x^2-5x-24$ .

Need two numbers with sum $-5$ and product $-24$ : these are $-8$ and $3$ .

$x^2-5x-24=x^2-8x+3x-24=x(x-8)+3(x-8)=\;$ $(x-8)(x+3)$

Formula (general quadratic)

(ax+b)(cx+d) = acx^2+(ad+bc)x+bd

Common Mistake

When the constant term is negative, the two numbers $a$ and $b$ must have opposite signs; when it is positive, they must have the same sign as the middle-term coefficient’s sign. Getting this backwards is the single most common factoring error.

4Cube Identities

Identity	Expression
Sum cubed	$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$
Difference cubed	$(x-y)^3 = x^3-3x^2y+3xy^2-y^3$
Difference of cubes	$x^3-y^3 = (x-y)(x^2+xy+y^2)$
Sum of cubes	$x^3+y^3 = (x+y)(x^2-xy+y^2)$
Three-variable cube identity	$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Where the cube identities come from

$(x+y)^3=(x+y)(x+y)^2=(x+y)(x^2+2xy+y^2)$ , expanded by distributivity, gives $x^3+3x^2y+3xy^2+y^3$ . Replacing $y$ with $-y$ gives the difference-cubed identity, with signs alternating $+,-,+,-$ .

Solved Example

Q. Find $103^3$ using an identity.

$103^3=(100+3)^3=100^3+3(100)^2(3)+3(100)(3)^2+3^3=1000000+90000+2700+27=\;$ $1092727$

Solved Example

Q. Factor $8n^3-60n^2m+150nm^2-125m^3$ .

Rewrite as $(2n)^3-3(2n)^2(5m)+3(2n)(5m)^2-(5m)^3$ , matching $x^3-3x^2y+3xy^2-y^3$ with $x=2n,y=5m$ . So it equals $\;$ $(2n-5m)^3$

Special case worth remembering

If $x+y+z=0$ , then $x^3+y^3+z^3=3xyz$

Because the first factor $(x+y+z)$ in the identity becomes zero, leaving $x^3+y^3+z^3-3xyz=0$ .

Given

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

Find

$x^3+y^3+z^3$

Solution

From $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ : $100=38+2(xy+yz+zx) \Rightarrow xy+yz+zx=31$

From the cube identity: $x^3+y^3+z^3-3(25)=10\big(38-31\big) \Rightarrow x^3+y^3+z^3=70+75=\;$ $145$

5Simplifying Rational Expressions

Factor the numerator and denominator fully, then cancel any factor common to both (assuming the denominator is never zero).

Given

$\dfrac{x^2-7x+12}{5x^2+5x-100}$ , denominator $\ne 0$

Simplify

Solution

Numerator: need sum $-7$ , product $12$ : these are $-3,-4$ . So $x^2-7x+12=(x-3)(x-4)$

Denominator: $5x^2+5x-100=5(x^2+x-20)$ ; need sum $1$ , product $-20$ : these are $5,-4$ . So $=5(x-4)(x+5)$

$\dfrac{(x-3)(x-4)}{5(x-4)(x+5)} = \;$ $\dfrac{x-3}{5(x+5)}$ (cancelling $x-4$ , valid since $x\ne4$ )

Solved Example: area word problem

Q. A rectangular pool’s breadth is 4 m less than its length, and its area is 96 sq. m. Find its dimensions.

Let length $=x$ m, breadth $=x-4$ m. Then $x(x-4)=96 \Rightarrow x^2-4x-96=0$

Need sum $-4$ , product $-96$ : these are $-12,8$ . So $(x-12)(x+8)=0 \Rightarrow x=12$ (rejecting $x=-8$ , a length can’t be negative)

Length $=\;$ $12$ m, breadth $=8$ m

Every identity in this chapter comes from expanding a product with the distributive property, just applied to squares, cubes, and sums of several variables.

the single idea to carry out of this chapter

All identities in one place

$(x+y)^2$

x^2+2xy+y^2

$(x-y)^2$

x^2-2xy+y^2

$(x+y+z)^2$

x^2+y^2+z^2+2xy+2yz+2zx

$x^2-y^2$

(x+y)(x-y)

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

(x+a)(x+b)

$(ax+b)(cx+d)$

acx^2+(ad+bc)x+bd

$x^3-y^3$

(x-y)(x^2+xy+y^2)

$x^3+y^3$

(x+y)(x^2-xy+y^2)

$(x+y)^3$

x^3+3x^2y+3xy^2+y^3

$(x-y)^3$

x^3-3x^2y+3xy^2-y^3

$x^3+y^3+z^3-3xyz$

(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

Check yourself before the exam

Can I state and apply all 11 identities without looking them up?
Can I factor a quadratic by splitting the middle term, choosing the correct signs?
Can I use Śhrīdharāchārya’s method or the (x+y+z)² identity to square a number mentally?
Can I simplify a rational expression and state the value that must be excluded?

Quick Revision: read this the night before the exam

Identity = true for all values; equation = true only for specific values
$(x\pm y)^2=x^2\pm2xy+y^2$ ; $(x+y+z)^2$ adds all three squares plus three cross terms
$x^2-y^2=(x+y)(x-y)$ ; use it for fast squaring: $a^2=(a+b)(a-b)+b^2$
Splitting the middle term: find two numbers with the given sum and product
$(x\pm y)^3=x^3\pm3x^2y+3xy^2\pm y^3$ (signs alternate for the minus case)
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ ; equals 0 when $x+y+z=0$
Simplify rational expressions by factoring both parts and cancelling, noting what can’t equal zero

Practice Questions

1 markExpand $(3x-5)^2$ using an identity.
1 markEvaluate $998^2$ using a suitable identity.
2 marksFactor $25x^2+70xy+49y^2$ .
2 marksFactor $x^2+2x-63$ by splitting the middle term.
2 marksEvaluate $52\times48$ using $x^2-y^2=(x+y)(x-y)$ .
3 marksExpand $(2p-3q+r)^2$ using the three-term identity.
3 marksFind $a$ if $x^3+12x^2q+48xq^2+64q^3=(x+a)^3$ .
2 marksSimplify $\dfrac{9s^2-25t^2}{3s^2+2st-5t^2}$ , assuming the denominator is never 0.
3 marksIf $x+y+z=0$ , show that $x^3+y^3+z^3=3xyz$ , and use it to find $2^3+(-5)^3+3^3$ .
4 marksA square garden of side 30 m has a path of uniform width $s$ m built around it. Find an expression for the area of the path in terms of $s$ .
3 marksA number plus its reciprocal equals $\dfrac{13}{6}$ . Find the number(s).

Multiple Choice

1 mark $(x+7)(x-7)$ equals:
(a) $x^2-49$ (b) $x^2+49$ (c) $x^2-14x+49$ (d) $x^2+14x-49$
1 markWhich is an identity, not just an equation?
(a) $x^2=16$ (b) $2x+3=11$ (c) $(x+1)^2=x^2+2x+1$ (d) $x^2-x=6$
1 markTo factor $x^2-9x+20$ , the two numbers needed are:
(a) 4 and 5(b) −4 and −5(c) 2 and 10(d) −2 and −10

Assertion (A):

(a+b)^2

is always greater than or equal to

a^2+b^2

when

a,b>0

.
Reason (R):

(a+b)^2-(a^2+b^2)=2ab

, which is non-negative when

a,b>0

Case Study

Saira arranges one square tile of side x units, 8 rectangular strips of size x by 1 units, and 15 unit squares to build one large rectangle, with total area x² + 8x + 15 square units.

1 markWhich two numbers have sum 8 and product 15?
1 markFactor x² + 8x + 15 into two linear factors.
2 marksState the length and breadth of Saira’s rectangle in terms of x.

Answer Key

9x^2-30x+25

998^2=(1000-2)^2=1000000-4000+4=996004

(5x+7y)^2

Q4Sum 2, product −63: numbers 9, −7.

(x+9)(x-7)

52\times48=(50+2)(50-2)=2500-4=2496

4p^2+9q^2+r^2-12pq+4pr-6qr

a=4q

(since

64q^3=(4q)^3

and the pattern matches

x^3+3x^2(4q)+3x(4q)^2+(4q)^3

)

\dfrac{(3s-5t)(3s+5t)}{(3s+5t)(s-t)}=\dfrac{3s-5t}{s-t}

Q9Proof follows from the factored identity;

2+(-5)+3=0

, so sum of cubes

=3(2)(-5)(3)=-90

Q10Outer square

(30+2s)^2

minus garden

30^2

: path area

=120s+4s^2

sq. m

Q11

x+\frac1x=\frac{13}{6} \Rightarrow 6x^2-13x+6=0 \Rightarrow (3x-2)(2x-3)=0 \Rightarrow x=\frac23

\frac32

MCQ 1–3(a) · (c) · (b)

A/RBoth A and R are true, and R is the correct explanation of A

Case Study3 and 5 ·

(x+3)(x+5)

· length

=x+5

, breadth

=x+3