Learning Objectives
After studying this chapter you should be able to:
- Distinguish an algebraic identity from an ordinary equation.
- Recognise number patterns that lead to algebraic identities, such as \((n-1)^2+(n+1)^2-2n^2=2\).
- Visualise \((a\pm b)^2\) and \((a+b+c)^2\) using areas of square and rectangles.
- Apply identities to compute squares and products quickly without a calculator.
- Factorise quadratic expressions using identities and by splitting the middle term.
- Understand and state the cubic identities \((a\pm b)^3\), \(a^3\pm b^3\) and the three-variable identity.
- Use algebra tiles as an area model for products such as \((x+3)(x+4)\).
- Simplify rational expressions by cancelling common factors, stating when the denominator is zero.
1. What is an Algebraic Identity?
You already know that an equation is a statement that two expressions are equal, for example \(2x+3=7\). Such an equation is true only for particular values of the variable. Here \(x=2\) makes it true, but \(x=5\) does not.
An algebraic identity is also an equality of two expressions, but it is true for all values of the variables involved. For instance, the statement \((x+y)^2=x^2+2xy+y^2\) stays true whichever numbers you substitute for \(x\) and \(y\).
Definition – Algebraic Identity
An equality that is true for all values of the variables in it is called an algebraic identity. Identities help us to rewrite expressions in shorter or more useful forms.
Compare these two statements:
| Statement | Kind | Why? |
|---|---|---|
| \(x^2-1=24\) | Equation | True only when \(x=5\) or \(x=-5\). |
| \((a+b)^2=a^2+2ab+b^2\) | Identity | True for every pair of values \((a,b)\). |
In this chapter you will learn a set of standard identities, see why each one is true using geometry and algebra, and then use them for two main jobs: computing values quickly and factorising expressions.
2. Patterns That Lead to Identities (Section 4.1)
Identities are often discovered from patterns. Look at the squares of consecutive whole numbers: \(1, 4, 9, 16, 25, 36, 49, \ldots\) Take any three consecutive squares, add the first and the last, and subtract twice the middle one.
Solved Example 1
For the squares \(1, 4, 9\): \(1+9-2\times 4 = 10-8 = 2\).
For the squares \(25, 36, 49\): \(25+49-2\times 36 = 74-72 = 2\).
Answer: In both cases the result is exactly 2.
Is this always true? Write the three consecutive numbers as \(n-1\), \(n\) and \(n+1\). Then (as the book shows) the pattern becomes an identity that is true for every whole number \(n\ge 2\):
Pattern Identity
\[ (n-1)^2+(n+1)^2-2n^2 = 2 \]
Worked check: \( (n-1)^2+(n+1)^2-2n^2 = (n^2-2n+1)+(n^2+2n+1)-2n^2 = 2\).
Think and Reflect
Try the same idea with four consecutive squares. Add the first and the fourth, subtract the second and the third. What do you get? For example, \(1^2+4^2-2^2-3^2 = 1+16-4-9 = 4\).
3. Visualising the Square of a Sum (Section 4.2)
We can see why \((a+b)^2=a^2+2ab+b^2\) is true by drawing a square whose side is the segment \((a+b)\).
Fig. 4.1 – A square of side \((a+b)\) splits into four regions: \(a^2\), \(ab\), \(ab\) and \(b^2\).
The square of side \((a+b)\) has area \((a+b)^2\). It is made of a square of area \(a^2\), a square of area \(b^2\), and two rectangles each of area \(ab\). Therefore,
Identity 1 – Square of a Sum
\[ (a+b)^2 = a^2+2ab+b^2 \]
The same identity can be obtained purely algebraically, using the distributive law:
\[ (a+b)^2 = (a+b)(a+b) = a(a+b)+b(a+b) = a^2+ab+ab+b^2 = a^2+2ab+b^2. \]
The identity works for all numbers — positive, negative, integers, fractions and decimals. Let us verify two such cases.
Solved Example 2
Take \(a=-2\), \(b=-3\). Then \(a+b=-5\), so \((a+b)^2 = (-5)^2 = 25\).
On the other side, \(a^2+2ab+b^2 = 4+12+9 = 25\).
Answer: Both sides give 25, so the identity holds.
Solved Example 3 (rational numbers)
Take \(a=-\dfrac{2}{3}\), \(b=\dfrac{3}{4}\). Then \(a+b=\dfrac{1}{12}\), so \((a+b)^2=\dfrac{1}{144}\).
Now compute each piece: \(a^2=\dfrac{4}{9}=\dfrac{64}{144}\), \(2ab=-\dfrac{12}{12}=-1=-\dfrac{144}{144}\), \(b^2=\dfrac{9}{16}=\dfrac{81}{144}\).
Sum: \(\dfrac{64-144+81}{144}=\dfrac{1}{144}\).
Answer: Both sides equal \(\dfrac{1}{144}\).
An important conclusion from the identity \((a+b)^2=a^2+2ab+b^2\) is that \((a+b)^2\) is generally not equal to \(a^2+b^2\). For \(a=10, b=2\): \((a+b)^2=12^2=144\), while \(a^2+b^2=104\), and \(144 \ne 104\). The two expressions are equal only in special situations.
Think and Reflect
Since \((a+b)^2=a^2+b^2+2ab\), compare \((a+b)^2\) and \(a^2+b^2\) by looking at the sign of \(2ab\).
- When \(ab>0\) (same signs), \(2ab>0\), so \((a+b)^2 > a^2+b^2\).
- When \(ab<0\) (opposite signs), \(2ab<0\), so \((a+b)^2 < a^2+b^2\).
- When \(ab=0\) (at least one is zero), the two are equal.
Solved Example 4
Expand \((5x+2y)^2\).
Take \(a=5x\) and \(b=2y\). Then \(2ab=2(5x)(2y)=20xy\), so
\[ (5x+2y)^2 = (5x)^2+2(5x)(2y)+(2y)^2 = 25x^2+20xy+4y^2. \]
Answer: \(25x^2+20xy+4y^2\).
Because the identity is about numbers arranged in the form \((a+b)^2\), it is extremely useful for squaring large numbers mentally.
Solved Example 5
Compute \(43^2\) using an identity.
Write \(43=40+3\). Then \(a=40\), \(b=3\) and
\[ 43^2=(40+3)^2=40^2+2(40)(3)+3^2=1600+240+9=1849. \]
Answer: \(43^2=1849\).
Solved Example 6 (original)
Expand \((x-7y)^2\).
Here \(a=x\), \(b=7y\), so \(2ab=2(x)(7y)=14xy\):
\[ (x-7y)^2 = x^2-2(x)(7y)+(7y)^2 = x^2-14xy+49y^2. \]
Answer: \(x^2-14xy+49y^2\).
Solved Example 7 (original)
Expand \((-3a+2b)^2\).
\[ (-3a+2b)^2 = (-3a)^2+2(-3a)(2b)+(2b)^2 = 9a^2-12ab+4b^2. \]
Answer: \(9a^2-12ab+4b^2\).
Solved Example 8 (original)
Expand \(\left(\dfrac{x}{2}+y\right)^2\).
\[ \left(\frac{x}{2}+y\right)^2 = \left(\frac{x}{2}\right)^2+2\left(\frac{x}{2}\right)(y)+y^2 = \frac{x^2}{4}+xy+y^2. \]
Answer: \(\dfrac{x^2}{4}+xy+y^2\).
3.1 The Square of a Difference
Replace \(b\) by \(-b\) in the identity \((a+b)^2=a^2+2ab+b^2\). Since \((-b)^2=b^2\), we get:
Identity 2 – Square of a Difference
\[ (a-b)^2 = a^2-2ab+b^2 \]
This identity can also be seen geometrically. A square of side \(a\), whose area is \(a^2\), is reduced by removing a strip of area \(ab\) and another strip of area \(b(a-b)\), leaving the inner square of area \((a-b)^2\):
Fig. 4.2 – The square \((a-b)^2\) "fits inside" the square \(a^2\). Removing two strips of areas \(ab\) and \(b(a-b)\) shows \(a^2-(a-b)^2=2ab-b^2\).
Solved Example 9
Compute \(29^2\) using an identity.
Write \(29=30-1\), so \(a=30\) and \(b=1\):
\[ 29^2=(30-1)^2=30^2-2(30)(1)+1^2=900-60+1=841. \]
Answer: \(29^2=841\).
Solved Example 10 (original)
Compute \(79^2\) and \(193^2\) using identities.
\(79^2=(80-1)^2=80^2-2(80)(1)+1^2=6400-160+1=6241\).
\(193^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249\).
Answer: \(79^2=6241\) and \(193^2=37249\).
Solved Example 11 (original)
Compute \(299^2\).
\[ 299^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401. \]
Answer: \(299^2=89401\).
4. Important Algebraic Identities at a Glance
The list below collects the core identities of this chapter (exactly as summarised in the textbook). Keep them on a revision card — you will use them constantly.
1. Square of sum
\((x+y)^2=x^2+2xy+y^2\)
2. Square of 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 squares
\((x+y)(x-y)=x^2-y^2\)
5. Product with common term
\((x+a)(x+b)=x^2+(a+b)x+ab\)
6. General linear product
\((ax+b)(cx+d)=acx^2+(ad+bc)x+bd\)
7. Difference of cubes
\(x^3-y^3=(x-y)(x^2+xy+y^2)\)
8. Sum of cubes
\(x^3+y^3=(x+y)(x^2-xy+y^2)\)
9. Cube of a sum
\((x+y)^3=x^3+3x^2y+3xy^2+y^3\)
10. Cube of a difference
\((x-y)^3=x^3-3x^2y+3xy^2-y^3\)
11. Three-variable identity
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\)
12. Shrīdhārāchārya square trick
\(a^2=(a+b)(a-b)+b^2\)
Note: identity 2 is a special case of identity 1 (put \(-y\) in place of \(y\)), and identity 12 is discussed in Section 6.3.
5. Factorisation Using Identities (Section 4.3)
Reading an identity "backwards" turns a sum (or difference) of terms into a product. That process — writing an expression as a product of its factors — is called factorisation.
Solved Example 12
Factorise \(x^2+4x+4\).
Notice \(x^2+4x+4 = x^2+2(x)(2)+2^2\), which matches \(a^2+2ab+b^2\) with \(a=x\), \(b=2\).
Answer: \(x^2+4x+4=(x+2)^2\).
Solved Example 13
Factorise \(36x^2+12x+1\).
Here \((6x)^2+2(6x)(1)+1^2\) matches \((a+b)^2\) with \(a=6x\), \(b=1\).
Answer: \(36x^2+12x+1=(6x+1)^2\).
Solved Example 14
Factorise \(50p^2+60pq+18q^2\).
First take out the common factor 2:
\[ 50p^2+60pq+18q^2 = 2(25p^2+30pq+9q^2). \]
Now \((5p)^2+2(5p)(3q)+(3q)^2=(5p+3q)^2\), so
Answer: \(2(5p+3q)^2\). Always look for a common factor first.
Solved Example 15 (original)
Factorise \(9x^2+24xy+16y^2\).
\[ 9x^2+24xy+16y^2 = (3x)^2+2(3x)(4y)+(4y)^2 = (3x+4y)^2. \]
Answer: \((3x+4y)^2\).
If the middle term is negative, use the \((a-b)^2\) form. For example \(16s^2-40st+25t^2 = (4s)^2-2(4s)(5t)+(5t)^2 = (4s-5t)^2\), and \(49x^2+28xy+4y^2=(7x+2y)^2\).
Think and Reflect
Verify that replacing \(b\) by \(-b\) in \((a+b)^2\) gives \((a-b)^2\): \((a+(-b))^2=a^2+2a(-b)+(-b)^2=a^2-2ab+b^2\). The same trick will give the cubic identities in Section 9.
6. Squares and Products Using Identities (Sections 4.2–4.4)
Identities turn tricky-looking calculations into easy ones. Here are three families of quick mental computations.
6.1 Difference of two squares for products
Identity 4 – Difference of Squares
\[ a^2-b^2 = (a+b)(a-b) \]
Reading from right to left: to multiply two numbers whose average is nice, write them as the product \((a+b)(a-b)\).
Solved Example 16 (original)
Compute (i) \(104\times 96\), (ii) \(23\times 17\).
(i) \(104\times 96 = (100+4)(100-4)=100^2-4^2=10000-16=9984\).
(ii) \(23\times 17 = (20+3)(20-3)=400-9=391\).
Answer: 9984 and 391.
6.2 The square of a three-term sum
Identity 3 – Square of Three Terms
\[ (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]
It is obtained from \((a+b)^2\) by treating \(b+c\) as a single quantity, or directly by expanding \((a+b+c)(a+b+c)\). The textbook confirms it by splitting a square of side \((a+b+c)\).
Solved Example 17
Compute \(119^2\) using the three-term identity.
Write \(119=100+10+9\). Then
\[ 119^2=(100+10+9)^2 = 100^2+10^2+9^2+2(100)(10)+2(10)(9)+2(100)(9) \]
\[ = 10000+100+81+2000+180+1800 = 14161. \]
Answer: \(119^2=14161\).
Solved Example 18 (original)
Compute \(214^2\).
\[ 214^2=(200+14)^2=40000+2(200)(14)+196 = 40000+5600+196 = 45796. \]
Answer: \(214^2=45796\).
6.3 Shrīdhārāchārya’s method for squares
The textbook tells us that around 750 CE the Indian mathematician Shrīdhārāchārya proposed a quick way to square a number. Rearranged, it is the identity
Identity 12 – Shrīdhārāchārya’s Square Rule
\[ a^2 = (a+b)(a-b)+b^2 \]
Because \((a+b)(a-b)=a^2-b^2\), adding \(b^2\) recovers \(a^2\).
Solved Example 19
Compute \(55^2\) using Shrīdhārāchārya’s rule.
Take \(a=55\), \(b=5\). Then \(55+5=60\), \(55-5=50\) and
\[ 55^2=(55+5)(55-5)+5^2=60\times 50+25=3000+25=3025. \]
Answer: \(55^2=3025\).
Special Trick – Numbers Ending in 5
Using the same idea, \((10n+5)^2 = 100n(n+1)+25\). Multiply the part before the "5" by the next whole number, then write 25 after it.
\(35^2 \to 3\times 4=12 \Rightarrow 1225\), \(65^2 \to 6\times 7=42 \Rightarrow 4225\), \(85^2 \to 8\times 9=72 \Rightarrow 7225\), \(105^2 \to 10\times 11=110 \Rightarrow 11025\).
7. Factorisation Using Algebra Tiles (Section 4.5)
Algebra tiles are area models made of three tiles. Think of a strip of length \(1\) and a square of side \(x\):
- the \(x^2\) tile is a square of side \(x\),
- the \(x\) tile is a rectangle \(x\times 1\),
- the unit tile is a square \(1\times 1\).
Placing these tiles to form rectangles shows products visually. Here is the tile picture for \((x+3)(x+4)\):
Fig. 4.3 – Tile layout for \((x+3)(x+4)\): one \(x^2\)-tile, seven \(x\)-tiles (three on the right column, four along the top row) and twelve unit squares in a \(3\times 4\) grid.
Think and Reflect
Could the seven \(x\)-tiles be arranged as \(2x+5x\) instead of \(3x+4x\)? Both give seven \(x\)-tiles, but the unit squares change: \(2\times 5=10\) rather than \(3\times 4=12\). So \((x+2)(x+5)=x^2+7x+10\) is a different product. The split of the middle term must always match the constant term.
Identity 5 – Product \((x+a)(x+b)\)
\[ (x+a)(x+b)=x^2+(a+b)x+ab \]
Solved Example 20 (original)
Expand \((x+6)(x+7)\).
With \(a=6\), \(b=7\): \(a+b=13\), \(ab=42\).
Answer: \((x+6)(x+7)=x^2+13x+42\).
Solved Example 21 (original)
Expand \((2x+3)(3x+1)\).
\[ (2x+3)(3x+1)=2x\cdot 3x+2x\cdot 1+3\cdot 3x+3\cdot 1=6x^2+2x+9x+3=6x^2+11x+3. \]
Answer: \(6x^2+11x+3\).
Generalising the tile layout to two different linear factors gives the last product identity:
Identity 6 – General Product
\[ (px+a)(qx+b)=pqx^2+(pb+qa)x+ab \]
8. Factorisation by Splitting the Middle Term (Section 4.6)
When you cannot immediately see a perfect square or a difference of squares, you can still factorise a quadratic \(x^2+(a+b)x+ab\) if you can find two numbers \(a\) and \(b\) whose sum is the middle coefficient and whose product is the last term. This is exactly identity 5 used in reverse.
Solved Example 22
Factorise \(x^2+7x+12\).
We need two numbers with sum 7 and product 12: \(3\) and \(4\). Split the middle term:
\[ x^2+7x+12 = x^2+3x+4x+12 = x(x+3)+4(x+3) = (x+3)(x+4). \]
Answer: \((x+3)(x+4)\).
Solved Example 23
Factorise \(x^2+11x+30\).
Sum 11, product 30: \(5\) and \(6\). So \(x^2+11x+30=x^2+5x+6x+30=(x+5)(x+6)\).
Answer: \((x+5)(x+6)\).
Solved Example 24
Factorise \(x^2-5x+6\).
We need sum \(-5\) and product \(6\). Choose \(-2\) and \(-3\), because \((-2)+(-3)=-5\) and \((-2)(-3)=6\).
\[ x^2-5x+6=x^2-2x-3x+6=(x-2)(x-3). \]
Answer: \((x-2)(x-3)\).
Solved Example 25 (original)
Factorise \(x^2-x-42\).
Sum \(-1\), product \(-42\). The pair \(6\) and \(-7\) works: \(6+(-7)=-1\), \(6(-7)=-42\).
\[ x^2-x-42=x^2+6x-7x-42=(x+6)(x-7). \]
Answer: \((x+6)(x-7)\).
When the coefficient of \(x^2\) is not 1, use identity 6 backwards. Find numbers to split the middle term so that grouping works out.
Solved Example 26 (original)
Factorise \(2x^2+7x+3\).
Split \(7x\) as \(x+6x\):
\[ 2x^2+7x+3=2x^2+x+6x+3=x(2x+1)+3(2x+1)=(2x+1)(x+3). \]
Check: \((2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3\).
Answer: \((2x+1)(x+3)\).
Solved Example 27 (original)
Factorise \(3x^2+18x+27\).
First take out the common factor 3:
\[ 3x^2+18x+27=3(x^2+6x+9)=3(x+3)^2. \]
Answer: \(3(x+3)^2\).
Watch Out
Never start splitting the middle term before checking for a common factor. In \(50p^2+60pq+18q^2\) and \(3x^2+18x+27\), the common factor makes the rest a perfect square.
9. Cubic and Three-Variable Identities (Section 4.7)
9.1 The cube of a sum
Identity 9 – Cube of a Sum
\[ (a+b)^3=a^3+3a^2b+3ab^2+b^3 \]
Derivation: \((a+b)^3=(a+b)(a+b)^2=(a+b)(a^2+2ab+b^2)\), which expands to \(a^3+3a^2b+3ab^2+b^3\). The textbook also draws a cube of edge \((a+b)\), built from one cube of volume \(a^3\), one of volume \(b^3\), three boxes of volume \(a^2b\) and three boxes of volume \(ab^2\).
9.2 The cube of a difference
Replace \(b\) by \(-b\) (the same substitution as in Section 5):
Identity 10 – Cube of a Difference
\[ (a-b)^3=a^3-3a^2b+3ab^2-b^3 \]
Solved Example 28
A cubical box has volume \(p^3+6p^2q+12pq^2+8q^3\). Find the length of its edge.
Recognise the cube pattern with \(a=p\), \(b=2q\): \(p^3+3p^2(2q)+3p(2q)^2+(2q)^3=(p+2q)^3\). Check: \(3p^2(2q)=6p^2q\), \(3p(4q^2)=12pq^2\), \((2q)^3=8q^3\).
Answer: Edge = \(p+2q\).
Solved Example 29
Factorise \(8n^3-60n^2m+150nm^2-125m^3\).
This matches \((2n)^3-3(2n)^2(5m)+3(2n)(5m)^2-(5m)^3\) with \(a=2n\), \(b=5m\). Check: \(3(2n)^2(5m)=3(4n^2)(5m)=60n^2m\), and \(3(2n)(25m^2)=150nm^2\).
Answer: \((2n-5m)^3\).
Solved Example 30 (original)
Compute \(201^3\).
\[ 201^3=(200+1)^3=200^3+3(200)^2(1)+3(200)(1)^2+1^3 \]
\[ = 8000000+120000+600+1 = 8120601. \]
Answer: \(201^3=8120601\).
9.3 Sum and difference of cubes
Identities 7 and 8
\[ x^3-y^3=(x-y)(x^2+xy+y^2) \]
\[ x^3+y^3=(x+y)(x^2-xy+y^2) \]
The first comes from multiplying out: \((x-y)(x^2+xy+y^2)=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3\). Replacing \(y\) by \(-y\) gives the second.
Think and Reflect
Since \(x-y\) is a common factor of \(x^2-y^2\) and \(x^3-y^3\), guess the pattern for \(x^n-y^n\). For example \(x^4-y^4=(x^2-y^2)(x^2+y^2)\), so \(x-y\) continues to be a factor.
9.4 The three-variable identity
Identity 11 – Three-Variable Identity
\[ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) \]
When \(x+y+z=0\), the right side is zero, giving the famous special case \(x^3+y^3+z^3=3xyz\).
Solved Example 31 (original)
Verify the three-variable identity for \(x=1\), \(y=2\), \(z=3\).
Left: \(1+8+27-3(1)(2)(3)=36-18=18\).
Right: \((1+2+3)(1+4+9-2-2-3)=6(3)=18\).
Answer: Both sides equal 18, so the identity holds.
Solved Example 32 (original)
If \(x+y+z=12\), \(xy+xz+yz=44\) and \(xyz=48\), find \(x^3+y^3+z^3\).
Using identity 11, \(x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-xz-yz)\).
Here \(x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)=144-88=56\), so the bracket is \(56-44=12\).
\[ x^3+y^3+z^3=3(48)+12(12)=144+144=288. \]
Answer: \(x^3+y^3+z^3=288\).
10. Simplifying Rational Expressions (Section 4.8)
A rational expression is a division of one algebraic expression by another. To simplify it, factorise numerator and denominator fully, cancel the common factors, and always state that the original denominator must not be zero.
Solved Example 33
Simplify \(\dfrac{x^2-7x+12}{5x^2+5x-100}\).
Denominator: \(5x^2+5x-100=5(x^2+x-20)=5(x+5)(x-4)\). So \(x\ne -5\) and \(x\ne 4\).
Numerator: \(x^2-7x+12=(x-3)(x-4)\). Cancelling the common factor \(x-4\):
\[ \frac{x^2-7x+12}{5x^2+5x-100}=\frac{(x-3)(x-4)}{5(x+5)(x-4)}=\frac{x-3}{5(x+5)}. \]
Answer: \(\dfrac{x-3}{5(x+5)}\), where \(x\ne -5\) and \(x\ne 4\).
Solved Example 34 (original)
Simplify \(\dfrac{x^2+6x+8}{x^2+4x}\).
Denominator \(x^2+4x=x(x+4)\), so \(x\ne 0\) and \(x\ne -4\). Numerator \(x^2+6x+8=(x+2)(x+4)\).
Answer: \(\dfrac{(x+2)(x+4)}{x(x+4)}=\dfrac{x+2}{x}\), where \(x\ne 0\) and \(x\ne -4\).
Solved Example 35 (original)
Simplify \(\dfrac{x^2-9}{x+3}\).
\[ \frac{x^2-9}{x+3}=\frac{(x-3)(x+3)}{x+3}=x-3, \]
Answer: \(x-3\), where \(x\ne -3\).
Watch Out!
Only factors can be cancelled — never terms. \(\dfrac{x+3}{x+5}\) cannot be "simplified" to \(\dfrac{3}{5}\) because \(x\) is a term, not a factor. And cancellation of \(x-4\) above is valid only while \(x\ne 4\).
The methods of this section apply directly to word problems. For example (both from the textbook):
Solved Example 36
A square floor of side \(x\), eight strips of size \(x\times 1\) and fifteen unit tiles together make a rectangle. Express the area and find the length and breadth.
Area \(=x^2+8x+15\). Since \(8=3+5\) and \(15=3\times 5\), \(x^2+8x+15=(x+3)(x+5)\).
Answer: Area \(=(x+3)(x+5)\); length \(x+5\), breadth \(x+3\).
Solved Example 37
A rectangular pool’s breadth is 4 m less than its length, and its area is \(96\ \text{m}^2\). Find the dimensions.
Let length be \(x\). Then breadth \(x-4\) and \(x(x-4)=96\), i.e. \(x^2-4x-96=0\).
Split: two numbers with sum \(-4\) and product \(-96\) are \(-12\) and \(8\). So \(x^2-12x+8x-96=(x-12)(x+8)=0\), giving \(x=12\) (the negative answer is not possible for a length).
Answer: Length 12 m, breadth 8 m. Check: \(12\times 8=96\).
11. Applications of Algebraic Identities
Application 1 – Path around a square park
A square park has side 40 m. Around it, a path of width \(s\) m is built. Write the area of the path.
Concept: Outer square has side \(40+2s\).
Steps: \((40+2s)^2-40^2=(1600+160s+4s^2)-1600=160s+4s^2\).
Path area \(=4s^2+160s\) m². If \(s=2\), the path area is \(16+320=336\) m².
Application 2 – Edge of a cubical box
A cubical box has volume \(8a^3+36a^2b+54ab^2+27b^3\). Find its edge, and its volume when \(a=1\) and \(b=2\).
Concept: \(8a^3+36a^2b+54ab^2+27b^3=(2a+3b)^3\).
Steps: Edge \(=2a+3b\). For \(a=1,b=2\), edge \(=8\); volume \(=8^3=512\).
Edge \(=2a+3b\); volume \(=512\) cubic units.
Application 3 – Rectangular field
A field’s area is \((2x^2+7x+3)\) m² and its width is \((2x+1)\) m. Find its length.
Concept: Factorise the area using identity 6.
Steps: \(2x^2+7x+3=(2x+1)(x+3)\).
Length \(=(x+3)\) m.
Application 4 – Handy estimate for a square sheet
A square sheet has side 4.9 m. Find its area using an identity.
Concept: \(4.9^2=(5-0.1)^2\).
Steps: \((5-0.1)^2=25-1+0.01=24.01\).
Area \(=24.01\ \text{m}^2\).
Application 5 – Prove \(n^3-n\) is divisible by 6
Show that \(n^3-n\) is divisible by 6 for every whole number \(n\).
Concept: \(n^3-n=n(n^2-1)=n(n-1)(n+1)\).
Steps: Three consecutive numbers contain a multiple of 3 and at least one even number, so their product is divisible by 6.
Hence \(n^3-n\) is always divisible by 6.
Application 6 – Carpet area without multiplying
A rectangular carpet measures 14 m by 16 m.
Concept: \(14\times 16=(15-1)(15+1)\).
Steps: \(15^2-1^2=225-1=224\).
Area \(=224\ \text{m}^2\).
Application 7 – Tiling a hall floor
A hall floor of area \(x^2+11x+30\) is split into a length and breadth that sum nicely. Write the dimensions that multiply back to the area.
Concept: \(x^2+11x+30=(x+5)(x+6)\).
Dimensions \((x+6)\times(x+5)\).
Application 8 – Consecutive squares, reversed
Use the pattern \((n-1)^2+(n+1)^2-2n^2=2\) to check that \(99^2+101^2-2\cdot 100^2=2\).
Concept: \(99=100-1\) and \(101=100+1\).
\(9801+10201-20000=2\). The identity works.
12. Competency-Based Questions
Competency 1
Which of the following statements is an identity? (a) \(x^2+6x+9=0\) (b) \((a+b)^2=a^2+2ab+b^2\) (c) \(2x+3=7\) (d) \(y^2-4=0\).
Solution: (b) is true for all values; the others hold only for particular values. Answer: (b).
Competency 2
A tile picture for \((x+2)(x+3)\) uses one \(x^2\)-tile and six unit squares. How many \(x\)-tiles does it need, and what product does it show?
Solution: Number of \(x\)-tiles \(=2+3=5\). Answer: 5 tiles; \((x+2)(x+3)=x^2+5x+6\).
Competency 3
Which is the most complete factorisation of \(36x^2-4\)? (a) \((6x-2)(6x+2)\) (b) \(4(3x-1)(3x+1)\) (c) \((6x-2)^2\) (d) \(36x^2-4\).
Solution: \(36x^2-4=4(9x^2-1)=4(3x-1)(3x+1)\). (a) is valid but not completely factored since both factors share 2. Answer: (b).
Competency 4
Verify \((a-b)^2=a^2-2ab+b^2\) for \(a=5\) and \(b=-3\).
Solution: Left: \((5+3)^2=64\). Right: \(25-2(5)(-3)+9=25+30+9=64\). Both 64, so the identity holds.
Competency 5 – Spot the error
A student wrote \((x-5)^2=x^2-25\). Is this correct?
Solution: No. The correct expansion is \((x-5)^2=x^2-10x+25\). The mistake is dropping the middle term \(2ab\). Correct form: \(x^2-10x+25\).
Competency 6 – Choose a strategy
Without a calculator, which identity should you use for \(104\times 96\)?
Solution: \(104=100+4\), \(96=100-4\), so use \((a+b)(a-b)=a^2-b^2\). \(104\times 96=10000-16=9984\).
13. Case-Based Questions
Case 1 – Field and path (3 questions)
A square field has side 40 m. A path of width \(s\) m is added around it.
(1) Write the area of the path. \((40+2s)^2-40^2=4s^2+160s\).
(2) Find the path area when \(s=2\) m. \(16+320=336\ \text{m}^2\).
(3) If the path area is 336 m², how wide is the path? \(4s^2+160s=336 \Rightarrow s^2+40s-84=0 \Rightarrow (s-2)(s+42)=0 \Rightarrow s=2\) m (negative rejected).
Case 2 – Algebra tiles (3 questions)
Ravi lays algebra tiles to build a rectangle of length \((x+4)\) and breadth \((x+3)\).
(1) How many \(x^2\)-tiles, \(x\)-tiles and unit tiles does he need? 1, \(4+3=7\), and \(4\times 3=12\).
(2) Write the area of the rectangle. \(x^2+7x+12\).
(3) Use this to factorise \(x^2+7x+12\). \((x+4)(x+3)\).
Case 3 – Rectangular areas (4 questions)
A hall has area \(x^2+11x+30\); a pool has breadth 4 m less than its length and area 96 m².
(1) Factorise the hall’s area. \((x+5)(x+6)\).
(2) What are the hall’s length and breadth? \(x+6\) and \(x+5\).
(3) Set up the pool condition and factorise it. \(x(x-4)=96\), i.e. \(x^2-4x-96=0=(x-12)(x+8)\).
(4) What are the pool’s dimensions? Length 12 m, breadth 8 m.
Case 4 – Cuboid volumes (3 questions)
A cuboid’s volume is \(8a^3+36a^2b+54ab^2+27b^3\). A second prism has volume \(6a^2-24b^2\).
(1) Show the cuboid is actually a cube and find its edge. \(8a^3+36a^2b+54ab^2+27b^3=(2a+3b)^3\), so edge \(=2a+3b\).
(2) Factorise \(6a^2-24b^2\) completely. \(6(a^2-4b^2)=6(a-2b)(a+2b)\).
(3) Find the edge and volume of the first prism when \(a=1\) and \(b=1\). Edge \(=2(1)+3(1)=5\); volume \(=5^3=125\).
Case 5 – Simplifying expressions safely (3 questions)
Consider \(\dfrac{x^2-7x+12}{5x^2+5x-100}\).
(1) Which values of \(x\) are not allowed? \(x\ne -5\) and \(x\ne 4\).
(2) Simplify the expression. \(\dfrac{x-3}{5(x+5)}\).
(3) After simplifying, is it fine to plug in \(x=4\)? No — cancellation was valid only because \(x\ne 4\); the original denominator is still zero at \(x=4\).
14. Common Mistakes to Avoid
| Mistake | Correct version / reason |
|---|---|
| \((a+b)^2=a^2+b^2\) | \((a+b)^2=a^2+2ab+b^2\) — never forget the middle term. |
| \((a-b)^2=a^2-2ab+b^2\), sign error in \(2ab\) | Use \((a+(-b))^2\); the middle term is always negative here. |
| \((a-b)^2=a^2-b^2\) | These are different identities. \((a-b)^2\ne a^2-b^2\) in general. |
| Dropping \(2xy,2yz,2zx\) in \((x+y+z)^2\) | \((x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\). |
| \((a+b)^3=a^3+b^3\) | \((a+b)^3=a^3+3a^2b+3ab^2+b^3\). |
| Wrong signs in \(x^3-y^3\) factors | \(x^3-y^3=(x-y)(x^2+xy+y^2)\); the middle sign is \(+xy\), not \(-xy\). |
| Guessing the split of the middle term | The two numbers must satisfy both sum and product conditions. |
| Forgetting the common factor first | Always factor out common factors before applying identities. |
| Cancelling terms instead of factors | \((x+3)/(x+5)\) cannot be reduced; only common factors cancel. |
| Cancelling without stating \(x\ne \) restrictions | Write the domain: original denominator must not be zero. |
| Ignoring negative lengths or breadths | Length and breadth are always positive; reject negative solutions. |
15. Exam Focus
- Memorise the identities – the twelve cards in Section 4 are the toolkit for this chapter.
- Practise mental squares using \((a+b)^2\), \((a-b)^2\) and Shrīdhārāchārya’s rule — a favourite topic for quick computation questions.
- Recognise perfect squares: check that the first and last terms are squares and the middle term is twice their product.
- Factorise systematically: common factor first, then identity, then split the middle term.
- Rational expressions: factorise, cancel factors, and always give the excluded values.
- Check your answers by expanding the factorised form back — it is quick and catches sign errors.
16. Quick Revision
Identity vs equation
An identity is true for all values of the variables; an equation only for some.
Starting pattern
\((n-1)^2+(n+1)^2-2n^2=2\) for any whole number \(n\ge 2\).
Squares of sums/differences
\((a\pm b)^2=a^2\pm 2ab+b^2\).
Three-term square
\((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\).
Difference of squares
\((a+b)(a-b)=a^2-b^2\); great for mental products.
Linear products
\((x+a)(x+b)=x^2+(a+b)x+ab\).
Shrīdhārāchārya’s rule
\(a^2=(a+b)(a-b)+b^2\); numbers ending in 5 are easy this way.
Algebra tiles
Area model: \(x^2\)-tile, \(x\)-tiles, unit squares → products like \((x+3)(x+4)\).
Split the middle term
Find two numbers with the right sum and product; group; factor.
Cubic identities
\((a\pm b)^3\), \(a^3\pm b^3\), and \(a^3+b^3+c^3-3abc\) identity.
Rational expressions
Factorise, cancel factors, state excluded values where the denominator is zero.
17. Self-Practice Questions
SP 1
Expand \((7x+4y)^2\).
SP 2
Expand \(\left(\dfrac{3}{4}s+8t\right)^2\).
SP 3
Expand \((2.5p+1.5q)^2\).
SP 4
Compute \(64^2\) using \((60+4)^2\).
SP 5
Compute \(105^2\).
SP 6
Compute \(205^2\).
SP 7
Compute \(79^2\).
SP 8
Compute \(193^2\).
SP 9
Compute \(299^2\).
SP 10
Factorise \(49x^2+28xy+4y^2\).
SP 11
Factorise \(16s^2-40st+25t^2\).
SP 12
Factorise \(9x^2+24xy+16y^2\).
SP 13
Factorise \(25a^2+20ab+4b^2\).
SP 14
Compute \(41^2\) and \(27^2\) using identities.
SP 15
Compute \(97^2\).
SP 16
Expand \((p+3q+7r)^2\).
SP 17
Expand \((3x-2y+4z)^2\).
SP 18
Compute \(117^2\).
SP 19
Compute \(78^2\).
SP 20
Compute \(214^2\).
SP 21
Factorise \(9a^2+b^2+4c^2-6ab+12ac-4bc\).
SP 22
Factorise \(r^2-r-42\).
SP 23
Factorise \(6a^2-24b^2\) completely.
SP 24
Factorise \(36x^2-4\) completely.
SP 25
Compute \(23\times 17\) and \(104\times 96\).
SP 26
Expand \((2x+3)(3x+1)\).
SP 27
Factorise \(2x^2+7x+3\).
SP 28
Compute \(35^2\) and \(65^2\) using the "ends in 5" trick.
SP 29
Compute \(101^3\).
SP 30
Factorise \(8a^3+36a^2b+54ab^2+27b^3\).
SP 31
Simplify \(\dfrac{x^2-7x+12}{5x^2+5x-100}\).
SP 32
Simplify \(\dfrac{x^2+6x+8}{x^2+4x}\).
SP 33
Simplify \(\dfrac{x^2-9}{x+3}\).
SP 34
A garden’s length is 9 m more than its breadth and its area is 220 m². Find the dimensions.
SP 35
A number and its reciprocal add to \(\dfrac{17}{4}\). Find the number.
SP 36
Compute \((a+b)^2\) for \(a=7\) and \(b=-4\), using the identity.
SP 37
Expand \((x-5y)^3\).
SP 38
Compute \(98^3\).
SP 39
Match the factor pair: \(x^3-y^3\).
SP 40
Simplify \(\dfrac{x^2-16}{x-4}\).
18. Self-Practice Answers
| No. | Answer |
|---|---|
| SP 1 | \(49x^2+56xy+16y^2\) |
| SP 2 | \(\dfrac{9}{16}s^2+12st+64t^2\) |
| SP 3 | \(6.25p^2+7.5pq+2.25q^2\) |
| SP 4 | 4096 |
| SP 5 | 11025 |
| SP 6 | 42025 |
| SP 7 | 6241 |
| SP 8 | 37249 |
| SP 9 | 89401 |
| SP 10 | \((7x+2y)^2\) |
| SP 11 | \((4s-5t)^2\) |
| SP 12 | \((3x+4y)^2\) |
| SP 13 | \((5a+2b)^2\) |
| SP 14 | 1681, 729 |
| SP 15 | 9409 |
| SP 16 | \(p^2+9q^2+49r^2+6pq+42qr+14pr\) |
| SP 17 | \(9x^2+4y^2+16z^2-12xy-16yz+24xz\) |
| SP 18 | 13689 |
| SP 19 | 6084 |
| SP 20 | 45796 |
| SP 21 | \((3a-b+2c)^2\) |
| SP 22 | \((r+6)(r-7)\) |
| SP 23 | \(6(a-2b)(a+2b)\) |
| SP 24 | \(4(3x-1)(3x+1)\) |
| SP 25 | 391, 9984 |
| SP 26 | \(6x^2+11x+3\) |
| SP 27 | \((2x+1)(x+3)\) |
| SP 28 | 1225, 4225 |
| SP 29 | 1030301 |
| SP 30 | \((2a+3b)^3\) |
| SP 31 | \(\dfrac{x-3}{5(x+5)}\), \(x\ne -5,4\) |
| SP 32 | \(\dfrac{x+2}{x}\), \(x\ne 0,-4\) |
| SP 33 | \(x-3\), \(x\ne -3\) |
| SP 34 | Length 20 m, breadth 11 m |
| SP 35 | \(4\) or \(\dfrac{1}{4}\) |
| SP 36 | \(3^2=9\); identity gives \(49+2(7)(-4)+16=49-56+16=9\) |
| SP 37 | \(x^3-15x^2y+75xy^2-125y^3\) |
| SP 38 | 941192 |
| SP 39 | \((x-y)(x^2+xy+y^2)\) |
| SP 40 | \(x+4\), \(x\ne 4\) |
Chapter Checklist
- I can explain what an algebraic identity is and how it differs from an equation.
- I know the pattern \((n-1)^2+(n+1)^2-2n^2=2\).
- I can draw the square picture for \((a+b)^2=a^2+2ab+b^2\).
- I can compute squares such as 29², 43², 79², 119² using identities.
- I know and can apply all twelve identities in Section 4.
- I can factorise perfect-square trinomials like \(x^2+4x+4\) and \(36x^2+12x+1\).
- I can factorise using \((a-b)^2\) when the middle term is negative.
- I can use Shrīdhārāchārya’s rule \(a^2=(a+b)(a-b)+b^2\).
- I can split the middle term to factorise quadratics like \(x^2-5x+6\).
- I can read and draw algebra tiles for \((x+a)(x+b)\).
- I can expand and factorise \((a\pm b)^3\) and \(a^3\pm b^3\).
- I can use the three-variable identity \(x^3+y^3+z^3-3xyz\).
- I can simplify rational expressions and state excluded values.
- I always check my factorisation by expansion.