Exercise 4.1
Q1. Using identity (a + b)² = a² + 2ab + b²
(i) (7x + 4y)²
= (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²
(ii) (5/7 x + 2/3 y)²
= (5/7 x)² + 2(5/7 x)(2/3 y) + (2/3 y)²
= 25/49 x² + 20/21 xy + 4/9 y²
(iii) (2.5p + 1.5q)²
= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²
(iv) (4/3 s + 2/8 t)²
= (4/3 s)² + 2(4/3 s)(2/8 t) + (2/8 t)²
= 16/9 s² + 16/12 st + 4/64 t²
= 16/9 s² + 4/3 st + 1/16 t²
(v) (x + 2y)²
= x² + 2(x)(2y) + (2y)²
= x² + 4xy + 4y²
(vi) (2x + y)²
= (2x)² + 2(2x)(y) + y²
= 4x² + 4xy + y²
Q2. Find values using identity
(i) (64)²
= (60 + 4)²
= 60² + 2(60)(4) + 4²
= 3600 + 480 + 16
= 4096
(ii) (105)²
= (100 + 5)²
= 10000 + 1000 + 25
= 11025
(iii) (205)²
= (200 + 5)²
= 40000 + 2000 + 25
= 42025
Exercise 4.2
Q1. Factor completely
(i) 9x² + 24xy + 16y²
This is of the form (a + b)²
= (3x)² + 2(3x)(4y) + (4y)²
= (3x + 4y)²
(ii) 4s² + 20st + 25t²
= (2s)² + 2(2s)(5t) + (5t)²
= (2s + 5t)²
(iii) 49x² + 28xy + 4y²
= (7x)² + 2(7x)(2y) + (2y)²
= (7x + 2y)²
(iv) (9/64)p² + (3/32)pq + q²
= (3/8 p)² + 2(3/8 p)(q) + q²
= (3/8 p + q)²
(v) (4/3)a² + ab + (3/4)b²
= (2/√3 a)² + 2(2/√3 a)(√3/2 b) + (√3/2 b)²
= (2/√3 a + √3/2 b)²
(vi) (5/9)s² + 6sv + 5v²
Here middle term does not match perfect square identity, so factor normally
= (5/3 s + 3v)(1/3 s + ? v) → try splitting
6sv = 5sv + sv
= (5/9)s² + 5sv + sv + 5v²
= (5/9)s(s + 9v) + v(s + 9v)
= (5/9 s + v)(s + 9v)
Q2. Using identity (a – b)² = a² – 2ab + b²
(i) (79)²
= (80 – 1)²
= 80² – 2(80)(1) + 1²
= 6400 – 160 + 1
= 6241
(ii) (193)²
= (200 – 7)²
= 40000 – 2800 + 49
= 37249
(iii) (299)²
= (300 – 1)²
= 90000 – 600 + 1
= 89401
Exercise 4.3
Q1. Find squares using suitable identity
(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
= (1100 + 20)²
= 1210000 + 44000 + 400
= 1254400
Q2. Factor using identities
(i) 16y² – 24y + 9
= (4y)² – 2(4y)(3) + 3²
= (4y – 3)²
(ii) (4/9)s² + 6st + 4t²
Check form: (a + b)²
= (2/3 s)² + 2(2/3 s)(3t) + (3t)²
= (2/3 s + 3t)²
(iii) 9m² + 3mk + k² + n² + 2mn + 9
Group terms:
= (9m² + 3mk + k²) + (n² + 2mn + 9)
= (3m + k)² + (n + 3)²
(no further factorisation possible)
(iv) p² – (2/16)p + (4/16)
= p² – (1/8)p + 1/4
Not perfect square directly, factor normally
= (p – 1/2)(p – 1/4)
(v) 9a² + 4b² + 12ab + 6ac + 4bc
Group:
= (9a² + 12ab + 4b²) + (6ac + 4bc)
= (3a + 2b)² + 2c(3a + 2b)
= (3a + 2b)(3a + 2b + 2c)
Q3. Expand using (a + b + c)²
(i) (p + 3q + 7r)²
= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr
(ii) (3x – 2y + 4z)²
= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(3x)(4z)
= 9x² + 4y² + 16z² – 12xy – 16yz + 24xz
Q4. Is this an identity?
LHS = (a+b+c)² + (a+b−c)² + (a−b+c)² + (a−b−c)²
Expand all terms:
Each gives: a² + b² + c² ± 2ab ± 2bc ± 2ca
When all 4 are added:
All mixed terms cancel
So, LHS = 4a² + 4b² + 4c²
RHS = 2(a² + b² + c²) = 2a² + 2b² + 2c²
Since LHS ≠ RHS
So, NOT an identity
Exercise 4.4
Q1. Fill in the blanks
(i) s² – 11s + 24
Find two numbers whose product = 24 and sum = –11
→ –3 and –8
So,
= (s – 3)(s – 8)
(ii) (____)(x + 1) = 3x² – 4x – 7
Factor RHS:
3x² – 4x – 7
= 3x² + 3x – 7x – 7
= 3x(x + 1) –7(x + 1)
= (3x – 7)(x + 1)
So blank = (3x – 7)
(iii) 10x² – 11x – 6 = (2x – )( + 2)
Factor:
10x² – 11x – 6
= 10x² – 15x + 4x – 6
= 5x(2x – 3) + 2(2x – 3)
= (2x – 3)(5x + 2)
So blanks = 3 and 5x
(iv) 6x² + 7x + 2
= 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1)
Q2. Use identities
(i) (41)²
= (40 + 1)²
= 1600 + 80 + 1 = 1681
(ii) (27)²
= (30 – 3)²
= 900 – 180 + 9 = 729
(iii) (23 × 17)
= (20 + 3)(20 – 3)
= 20² – 3²
= 400 – 9 = 391
(iv) (135)²
= (100 + 35)²
= 10000 + 7000 + 1225
= 18225
(v) (97)²
= (100 – 3)²
= 10000 – 600 + 9
= 9409
(vi) (18 × 29)
= ( (23 – 5)(23 + 6) ) → better method:
= ( (23)² – (5×6?) not suitable )
Better:
= (20 – 2)(20 + 9)
= 400 + 180 – 40 – 18
= 522
(vii) (34 × 43)
= ( (38 – 4)(38 + 5) ) not ideal
Better:
= ( (34)(43) )
= ( (34)(40 + 3) )
= 1360 + 102 = 1462
(viii) (205)²
= (200 + 5)²
= 40000 + 2000 + 25
= 42025
Q3. Factor
(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc
= (3a – b + 2c)²
(ii) 16s² + 25t² – 40st
= (4s – 5t)²
(iii) r² – r – 42
Find numbers: product = –42, sum = –1
→ –7 and 6
= (r – 7)(r + 6)
(iv) 49g² + 14gh + h²
= (7g + h)²
(v) 64u² + 121v² + 4w² – 176uv – 32uw + 44vw
= (8u – 11v – 2w)²
Exercise 4.5
Q1. Simplify rational expressions
(i) (p² + pq − 10q²) / (3p² − 3pq − 18q²)
Factor numerator:
p² + pq − 10q²
= p² + 5pq − 4pq − 10q²
= p(p + 5q) −2q(2p + 5q)
= (p + 5q)(p − 2q)
Factor denominator:
3p² − 3pq − 18q²
= 3(p² − pq − 6q²)
= 3(p − 3q)(p + 2q)
Final answer:
= (p + 5q)(p − 2q) / [3(p − 3q)(p + 2q)]
(ii) (m²n + mn²) / (n³ − n²m + nm² − m³)
Numerator:
= mn(m + n)
Denominator:
Group terms:
= (n³ − m³) + (nm² − n²m)
= (n − m)(n² + nm + m²) + nm(m − n)
= (n − m)(n² + nm + m² − nm)
= (n − m)(n² + m²)
Final answer:
= mn(m + n) / [(n − m)(n² + m²)]
(iii) (w² + v² + x² − wv − vx − wx) / (w³ + v³ + x³ − wvx)
Use identity:
w³ + v³ + x³ − 3wvx = (w + v + x)(w² + v² + x² − wv − vx − wx)
So denominator = (w + v + x)(numerator)
Final answer:
= 1 / (w + v + x)
(iv) (25z² − y²) / (4y² − 20yz + 25z²)
Numerator:
= (5z − y)(5z + y)
Denominator:
= (2y − 5z)²
Final answer:
= (5z + y) / (2y − 5z)
(v) (x² − 6x + 8 − 9) / (x² + 6x − 7 − (x² + 12x))
Simplify numerator:
= x² − 6x − 1
Denominator:
= x² + 6x − 7 − x² − 12x
= −6x − 7
Final answer:
= (x² − 6x − 1) / (−6x − 7)
(vi) (p² − 4p + 4) / (p⁴ − 16)
Numerator:
= (p − 2)²
Denominator:
= (p² − 4)(p² + 4)
= (p − 2)(p + 2)(p² + 4)
Cancel (p − 2):
Final answer:
= (p − 2) / [(p + 2)(p² + 4)]
End-of-Chapter Exercises
Q1. Use suitable identities
(i) (–3x + 4)²
Use (a + b)²
= (−3x)² + 2(−3x)(4) + 4²
= 9x² − 24x + 16
(ii) (2s + 7)(2s − 7)
Use (a + b)(a − b) = a² − b²
= (2s)² − 7²
= 4s² − 49
(iii) (p + 1/2)(p − 1/2)
= p² − (1/2)²
= p² − 1/4
(iv) (2n + 7)(2n − 7)
= (2n)² − 7²
= 4n² − 49
(v) (s − 2t)(s² + 2st + 4t²)
Use (a − b)(a² + ab + b²) = a³ − b³
= s³ − (2t)³
= s³ − 8t³
(vi) (1/2 r − 2/4 r)²
Simplify first:
2/4 r = 1/2 r
So expression = (1/2 r − 1/2 r)² = 0
(vii) (−3m + 4k − l)²
Use (a + b + c)²
= (−3m)² + (4k)² + (−l)²
- 2(−3m)(4k) + 2(4k)(−l) + 2(−3m)(−l)
= 9m² + 16k² + l² − 24mk − 8kl + 6ml
(viii) (x − y)³
Use (a − b)³
= x³ − 3x²y + 3xy² − y³
(ix) (2/7 k − 3/3 m)³
Simplify: 3/3 m = m
= (2/7 k − m)³
Use (a − b)³
= (2/7 k)³ − 3(2/7 k)²(m) + 3(2/7 k)(m²) − m³
= 8/343 k³ − 12/49 k²m + 6/7 km² − m³
Q2. Find values
(i) 17 × 21
= (19 − 2)(19 + 2)
= 19² − 2²
= 361 − 4 = 357
(ii) 104 × 96
= (100 + 4)(100 − 4)
= 100² − 4²
= 10000 − 16 = 9984
(iii) 24 × 16
= (20 + 4)(20 − 4)
= 400 − 16 = 384
(iv) 147³
= (150 − 3)³
= 150³ − 3(150)²(3) + 3(150)(3²) − 3³
= 3375000 − 202500 + 4050 − 27
= 3176523
(v) 199³
= (200 − 1)³
= 8000000 − 120000 + 600 − 1
= 7880599
(vi) 127³
= (100 + 27)³
= 1000000 + 3(10000)(27) + 3(100)(729) + 19683
= 1000000 + 810000 + 218700 + 19683
= 2048383
(vii) (−107)³
= −(107³)
107³ = (100 + 7)³
= 1000000 + 210000 + 14700 + 343
= 1225043
So = −1225043
(viii) (−299)³
= −(299³)
299³ = (300 − 1)³
= 27000000 − 270000 + 900 − 1
= 26730900 − 1
= 26730899
So = −26730899
Q3. Factor the following
(i) 4y² + y²/1 + 16
This is written incorrectly, correct form is likely:
4y² + 4y + 1 + 16 → but assuming identity form
If intended:
4y² + 4y + 1 = (2y + 1)²
(ii) 9m² − n²/1 + 25
Assuming form:
= 9m² + 25 − n²
No perfect identity, so write as:
= (3m)² + 5² − n²
(no further factorisation)
(iii) 27b³ − b³/1 + 64
Use a³ − b³ + c³ type → not standard
If taken as:
27b³ + 64 = (3b + 4)(9b² − 12b + 16)
(iv) x² + 5/6 x + 1/6
Find numbers: product = 1/6, sum = 5/6
→ 1/2 and 1/3
= (x + 1/2)(x + 1/3)
(v) 27u³ − 125/1 + (5/27)u² + (25/9)u
Group:
= (3u − 5)³ (after rearranging identity form)
(vi) 64y³ + 125z³
= (4y)³ + (5z)³
= (4y + 5z)(16y² − 20yz + 25z²)
(vii) p³ + q³ + r³ − 3pqr
= (p + q + r)(p² + q² + r² − pq − qr − rp)
(viii) 9m² – 12m + 4
= (3m − 2)²
(ix) (8/9)x³ − 6y³ + xyz
Rearrange: not exact identity, so no standard factorisation
(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= (2x + 3y + 6z)²
(xi) 27u³ − 216/1 + (2/9)u² + 4u
Rearrange into cube form:
= (3u − 6)³
Q4. Simplify
(i) (4x² − 1) / (4x² + 4x + 1)
= (2x − 1)(2x + 1) / (2x + 1)²
Cancel:
= (2x − 1)/(2x + 1)
(ii) (9a² − 36b²) / (9a² − 3ab + 24b²)
Numerator:
= (3a − 6b)(3a + 6b)
Denominator: not perfect factor match
So final:
= (3a − 6b)(3a + 6b) / (9a² − 3ab + 24b²)
(iii) (s² − 2st + t² − 35) / (s³ + t³ + 125)
Numerator:
= (s − t)² − 35
Denominator:
= s³ + t³ + 5³
= (s + t + 5)(s² + t² + 25 − st − 5t − 5s)
No cancellation
Q5. Rectangle dimensions
(i) 25a² – 30ab + 9b²
= (5a − 3b)²
So length = (5a − 3b), breadth = (5a − 3b)
(ii) 36s² – 49t²
= (6s − 7t)(6s + 7t)
Length = (6s + 7t), breadth = (6s − 7t)
Q6. Cuboid dimensions
(i) 6a² − 24b²
= 6(a² − 4b²)
= 6(a − 2b)(a + 2b)
Dimensions: 6, (a − 2b), (a + 2b)
(ii) 3ps² − 15ps + 12p
= 3p(s² − 5s + 4)
= 3p(s − 1)(s − 4)
Dimensions: 3p, (s − 1), (s − 4)
Q7. Area of path
Side = 40
New side = 40 + 2s
Area of path = (40 + 2s)² − 40²
= 1600 + 160s + 4s² − 1600
= 160s + 4s²
Q8. Number + reciprocal = 3/10
Let number = x
x + 1/x = 3/10
Multiply by x:
x² + 1 = 3x/10
10x² + 10 = 3x
10x² − 3x + 10 = 0
Solve → no real solution
Q9. Pool length
Area = 2x² + 7x + 3
Width = 2x + 1
Factor area:
= (2x + 1)(x + 3)
So length = (x + 3)
Q10. Show p = r
Given factors: (x − 2), (x − 1/2)
So polynomial = p(x − 2)(x − 1/2)
Expand:
= p[x² − (5/2)x + 1]
= px² − (5p/2)x + p
Compare with px² + 5x + r
So:
−5p/2 = 5 → p = −2
and r = p
So p = r
Q11. Prove identity
Given: a + b + c = 5
ab + bc + ca = 10
Use identity:
a³ + b³ + c³ − 3abc
= (a + b + c)(a² + b² + c² − ab − bc − ca)
Now:
a² + b² + c² = (a+b+c)² − 2(ab+bc+ca)
= 25 − 20 = 5
So:
= 5(5 − 10)
= 5(−5) = −25
Q12. Show n³ − n divisible by 6
n³ − n = n(n² − 1)
= n(n − 1)(n + 1)
Product of 3 consecutive numbers → divisible by 2 and 3
So divisible by 6
Q13. Find value
(i) x³ + y³ − 12xy + 64, given x + y = −4
Use:
x³ + y³ + 64 − 12xy
= (x + y + 4)(…)
Since x + y = −4 → first factor = 0
Answer = 0
(ii) x³ − 8y³ − 36xy − 216, x = 2y + 6
Substitute:
x³ − 8y³ − 36xy − 216
= (2y + 6)³ − 8y³ − 36y(2y + 6) − 216
Expand:
= (8y³ + 72y² + 216y + 216) − 8y³ − 72y² − 216y − 216
= 0