1. What Are Algebraic Identities?
An algebraic identity is a special rule that is true for every value you put into it. This is different from a normal equation, which may only work for one or two numbers.
Here is a fun pattern to start with. Pick any three consecutive square numbers, for example 1, 4, and 9.
- Add the smallest and largest: 1 + 9 = 10
- Subtract twice the middle square: 10 − 2(4) = 2
The answer is always 2. Using algebra, if the middle number is n, the three squares are (n−1)², n², and (n+1)².
2. Identity: (a + b)² = a² + 2ab + b²
This is one of the most useful identities. It helps you expand the square of a sum quickly.
Look at the big square above. Its side length is (a + b), so its total area is (a + b)². Inside it you can see:
- One yellow square with area a²
- One pink square with area b²
- Two blue rectangles, each with area ab
(5x)² + 2(5x)(2y) + (2y)² = 25x² + 20xy + 4y²
3. Identity: (a − b)² = a² − 2ab + b²
If you replace b with −b in the first identity, you get this second rule. It is perfect for finding squares of differences.
30² − 2(30)(1) + 1² = 900 − 60 + 1 = 841
4. Identity: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
When three terms are squared together, every pair multiplies twice.
5. Identity: a² − b² = (a + b)(a − b)
This is called the difference of squares. It is very helpful for factorising and quick multiplication.
6. Identity: (x + a)(x + b) = x² + (a + b)x + ab
When you multiply two brackets that both start with x, the result follows a simple pattern.
A more general form is:
7. Identity: (a + b)³ = a³ + 3a²b + 3ab² + b³
Just like a square can be split into flat pieces, a cube can be split into smaller cubes and blocks.
8. Identity: (a − b)³ = a³ − 3a²b + 3ab² − b³
Replace b with −b in the cube identity above. Notice how the plus and minus signs alternate.
9. Identity: x³ − y³ = (x − y)(x² + xy + y²)
This helps you factorise the difference of two cubes. You can check it by multiplying the right-hand side.
10. Identity: x³ + y³ = (x + y)(x² − xy + y²)
Similarly, the sum of two cubes factorises with a minus sign in the middle.
11. Identity: x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)
This long identity is useful when you know the sum of three numbers and the sum of their products taken two at a time.
12. How to Factorise Using Identities
Factorisation means breaking a big expression into simpler pieces that multiply back to give the original.
Step-by-step method
- Check for a common factor in all terms. If there is one, pull it out first.
- Look for patterns:
- Perfect square: a² + 2ab + b² = (a + b)²
- Difference of squares: a² − b² = (a + b)(a − b)
- Perfect square with minus: a² − 2ab + b² = (a − b)²
- For x² + px + q, find two numbers that multiply to give q and add to give p. Then split the middle term.
To factor x² + 7x + 12, find two numbers that multiply to 12 and add to 7. They are 3 and 4.
x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)
13. How to Simplify Rational Expressions
A rational expression is a fraction that has algebraic expressions on top and bottom. To simplify it:
- Fully factorise the numerator.
- Fully factorise the denominator.
- Cancel any common factors that appear on both top and bottom (only if they are not zero).
14. Exercise Set 4.1 — Solutions
Using (a + b)² = a² + 2ab + b²
| Question | Solution |
|---|---|
| (i) (7x + 4y)² | 49x² + 56xy + 16y² |
| (ii) (7x/5 + 3y/2)² | 49x²/25 + 21xy/5 + 9y²/4 |
| (iii) (2.5p + 1.5q)² | 6.25p² + 7.5pq + 2.25q² |
| (iv) (3s/4 + 8t)² | 9s²/16 + 12st + 64t² |
| (v) (x + 1/2y)² | x² + x/y + 1/4y² |
| (vi) (1/x + 1/y)² | 1/x² + 2/xy + 1/y² |
Finding values:
| Question | Working | Answer |
|---|---|---|
| (i) 64² | (60 + 4)² = 3600 + 480 + 16 | 4096 |
| (ii) 105² | (100 + 5)² = 10000 + 1000 + 25 | 11025 |
| (iii) 205² | (200 + 5)² = 40000 + 2000 + 25 | 42025 |
15. Exercise Set 4.2 — Solutions
Factor completely:
| Question | Solution |
|---|---|
| (i) 9x² + 24xy + 16y² | (3x + 4y)² |
| (ii) 4s² + 20st + 25t² | (2s + 5t)² |
| (iii) 49x² + 28xy + 4y² | (7x + 2y)² |
| (iv) 64p² + 32pq/3 + 4q²/9 | (8p + 2q/3)² |
| (v) 3a² + 4ab + 4b²/3 | (1/3)(3a + 2b)² |
| (vi) 9s²/5 + 6sv + 5v² | (1/5)(3s + 5v)² |
Finding values using (a − b)²:
| Question | Working | Answer |
|---|---|---|
| (i) 79² | (80 − 1)² = 6400 − 160 + 1 | 6241 |
| (ii) 193² | (200 − 7)² = 40000 − 2800 + 49 | 37249 |
| (iii) 299² | (300 − 1)² = 90000 − 600 + 1 | 89401 |
16. Exercise Set 4.3 — Solutions
Finding squares:
| Question | Working | Answer |
|---|---|---|
| (i) 117² | (100 + 17)² = 10000 + 3400 + 289 | 13689 |
| (ii) 78² | (80 − 2)² = 6400 − 320 + 4 | 6084 |
| (iii) 198² | (200 − 2)² = 40000 − 800 + 4 | 39204 |
| (iv) 214² | (200 + 14)² = 40000 + 5600 + 196 | 45796 |
| (v) 1104² | (1100 + 4)² = 1210000 + 8800 + 16 | 1218816 |
| (vi) 1120² | (1000 + 120)² = 1000000 + 240000 + 14400 | 1254400 |
Factor using suitable identities:
| Question | Solution |
|---|---|
| (i) 16y² − 24y + 9 | (4y − 3)² |
| (ii) 9s²/4 + 6st + 4t² | (3s/2 + 2t)² |
| (iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n² | (m/3 + k/2 + 3n)² |
| (iv) p²/16 − 2 + 16/p² | (p/4 − 4/p)² |
| (v) 9a² + 4b² + c² − 12ab + 6ac − 4bc | (3a − 2b + c)² |
Expand using (a + b + c)²:
| Question | Solution |
|---|---|
| (i) (p + 3q + 7r)² | p² + 9q² + 49r² + 6pq + 42qr + 14rp |
| (ii) (3x − 2y + 4z)² | 9x² + 4y² + 16z² − 12xy − 16yz + 24xz |
Is this an identity?
(a+b+c)(a+b−c)(a−b+c)(−a+b+c) = 2(a²b² + b²c² + c²a²) − (a⁴ + b⁴ + c⁴)
Answer: Yes. When you group and multiply the left side, it simplifies exactly to the right side.
17. Exercise Set 4.4 — Solutions
Fill in the blanks:
| Question | Completed Form |
|---|---|
| (i) s² − 11s + 24 | (s − 3)(s − 8) |
| (ii) (?)(x + 1) = 3x² − 4x − 7 | (3x − 7)(x + 1) |
| (iii) 10x² − 11x − 6 = (2x − ?)(? + 2) | (2x − 3)(5x + 2) |
| (iv) 6x² + 7x + 2 | (2x + 1)(3x + 2) |
Find products using identities:
| Question | Identity Used | Answer |
|---|---|---|
| (i) 41² | (40 + 1)² | 1681 |
| (ii) 27² | (30 − 3)² | 729 |
| (iii) 23 × 17 | (20 + 3)(20 − 3) | 391 |
| (iv) 135² | (130 + 5)² | 18225 |
| (v) 97² | (100 − 3)² | 9409 |
| (vi) 18 × 29 | 18(30 − 1) | 522 |
| (vii) 34 × 43 | Direct / distributive | 1462 |
| (viii) 205² | (200 + 5)² | 42025 |
Factor:
| Question | Solution |
|---|---|
| (i) 9a² + b² + 4c² − 6ab + 12ac − 4bc | (3a − b + 2c)² |
| (ii) 16s² + 25t² − 40st | (4s − 5t)² |
| (iii) r² − r − 42 | (r − 7)(r + 6) |
| (iv) 49g² + 14gh + h² | (7g + h)² |
| (v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw | (8u − 11v − 2w)² |
18. Exercise Set 4.5 — Solutions
Simplify the rational expressions:
| Question | Solution Steps | Final Answer |
|---|---|---|
| (i) (3p² + 3pq − 18q²) / (p² + 3pq − 10q²) | Num: 3(p+3q)(p−2q) Den: (p+5q)(p−2q) | 3(p + 3q) / (p + 5q) |
| (ii) (3m³ − 3m²n − 5mn² + 5n³) / (m³ − 3m²n + 5mn² − 15n³) | Factor by grouping | (3m² − 5n²) / (m² + 5n²) — check denominator grouping |
| (iii) (w³ + v³ + x³ − 3wvx) / (w² + v² + x² + wv + vx + wx) | Use x³+y³+z³−3xyz identity | (w + v + x) — after cancelling matching factor |
| (iv) (4y² − 25z²) / (2y − 5z) | Num: (2y+5z)(2y−5z) | 2y + 5z |
| (v) [(x²+7x+12)(x²−9)] / [(x²−6x+9)(x²+x−12)] | Factor all quadratics and cancel | (x+3)² / (x−3)² |
| (vi) (p⁴ − 16) / (p² − 4p + 4) | Num: (p²+4)(p+2)(p−2) Den: (p−2)² | (p²+4)(p+2) / (p−2) |
19. End-of-Chapter Exercises — Solutions
1. Products
| Question | Solution |
|---|---|
| (i) (−3x + 4)² | 9x² − 24x + 16 |
| (ii) (2s + 7)(2s − 7) | 4s² − 49 |
| (iii) (p² + 1/2)(p² − 1/2) | p⁴ − 1/4 |
| (iv) (2n + 7)(2n − 7) | 4n² − 49 |
| (v) (s − 2t)(s² + 2st + 4t²) | s³ − 8t³ |
| (vi) (1/2r − 4r)² | 1/4r² − 4 + 16r² |
| (vii) (−3m + 4k − l)² | 9m² + 16k² + l² − 24mk − 8kl + 6ml |
| (viii) (x − y/3)³ | x³ − x²y + xy²/3 − y³/27 |
| (ix) (7k/2 − 2m/3)³ | 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27 |
2. Values
| Question | Working | Answer |
|---|---|---|
| (i) 17 × 21 | (19−2)(19+2) = 361 − 4 | 357 |
| (ii) 104 × 96 | (100+4)(100−4) = 10000 − 16 | 9984 |
| (iii) 24 × 16 | (20+4)(20−4) = 400 − 16 | 384 |
| (iv) 147³ | (150−3)³ | 3176523 |
| (v) 199³ | (200−1)³ = 8000000 − 120000 + 600 − 1 | 7880599 |
| (vi) 127³ | (130−3)³ | 2048383 |
| (vii) (−107)³ | −(100+7)³ | −1225043 |
| (viii) (−299)³ | −(300−1)³ | −26730899 |
3. Factorisation
| Question | Solution |
|---|---|
| (i) 4y² + 1 + 1/16y² | (2y + 1/4y)² |
| (ii) 9m² − 1/25n² | (3m + 1/5n)(3m − 1/5n) |
| (iii) 27b³ − 1/64b³ | (3b − 1/4b)(9b² + 3/4 + 1/16b²) |
| (iv) x² + 5x/6 + 1/6 | (x + 1/2)(x + 1/3) |
| (v) 27u³ − 1/125 − 27u²/5 + 9u/25 | (3u − 1/5)³ |
| (vi) 64y³ + 1/125 | (4y + 1/5)(16y² − 4y/5 + 1/25) |
| (vii) p³ + 27q³ + r³ − 9pqr | (p + 3q + r)(p² + 9q² + r² − 3pq − pr − 3qr) |
| (viii) 9m² − 24m + 16 | (3m − 4)² |
| (ix) 9x³ − 8y³/3 + z³/3 + 6xyz | Use x³+y³+z³−3xyz identity after adjusting fractions |
| (x) 4x² + 9y² + z² + 12xy + 4xz + 6yz | (2x + 3y + z)² |
| (xi) 27u³ − 1/216 − 9u²/2 + u/4 | (3u − 1/6)³ |
4. Simplification
| Question | Solution |
|---|---|
| (i) (4x² + 4x + 1) / (4x² − 1) | (2x + 1) / (2x − 1) |
| (ii) (9a² − 24ab + 16b²) / (9a² − 16b²) | (3a − 4b) / (3a + 4b) |
| (iii) (s³ + 125t³) / (s² − 2st − 35t²) | (s² − 5st + 25t²) / (s − 7t) |
5. Rectangle Dimensions
| Area | Length | Breadth |
|---|---|---|
| (i) 25a² − 30ab + 9b² | 5a − 3b | 5a − 3b |
| (ii) 36s² − 49t² | 6s + 7t | 6s − 7t |
6. Cuboid Dimensions
| Volume | Dimensions |
|---|---|
| (i) 6a² − 24b² | 6, (a + 2b), (a − 2b) |
| (ii) 3ps² − 15ps + 12p | 3p, (s − 1), (s − 4) |
7. Playground Path Area
Inner side = 40 m, outer side = (40 + 2s) m.
8. Number and Reciprocal
Let the number be x.
Answer: The number is 3 or 1/3.
9. Pool Length
Area = 2x² + 7x + 3, Width = 2x + 1
10. Proving p = r
Since (x − 2) is a factor: p(2)² + 5(2) + r = 0 → 4p + r = −10
Since (x − 1/2) is a factor: p(1/4) + 5/2 + r = 0 → p + 4r = −10
Solving these two equations gives p = −2 and r = −2. Hence p = r.
11. Proving a³ + b³ + c³ − 3abc = −25
Given: a + b + c = 5 and ab + bc + ca = 10.
First find a² + b² + c² using (a+b+c)² = a²+b²+c² + 2(ab+bc+ca):
Now use the identity:
12. Divisibility by 6
n³ − n = n(n² − 1) = (n − 1)n(n + 1). This is a product of three consecutive integers.
- At least one of them is even, so it is divisible by 2.
- Exactly one of them is a multiple of 3, so it is divisible by 3.
Since it is divisible by both 2 and 3, it is divisible by 6.
13. Find Values
(i) x³ + y³ − 12xy + 64 when x + y = −4
Rewrite as x³ + y³ + 4³ − 3(x)(y)(4). Using the identity with z = 4:
(ii) x³ − 8y³ − 36xy − 216 when x = 2y + 6
Rewrite as x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6).
Here a + b + c = x − 2y − 6 = (2y + 6) − 2y − 6 = 0.