Chapter 4: Algebraic Expressions

Exercise 4.2 Solved Reference

Complete step-by-step solutions for Exercise 4.2, covering factorization of algebraic expressions, grouped substitution, perfect cubes, and sum/difference of cubes. Cleaned of all watermarks and optimized for presentation.

Question 1 Quadratic-Like Factorization
Factorize each of the following expressions using algebraic identity completeness.

(i) 4x4 + 81y4

Solution
Write expression: 4x4 + 81y4
Rewrite as perfect squares: (2x2)2 + (9y2)2
Add and subtract 2(2x2)(9y2) = 36x2y2 to complete the square:
=(2x2)2 + (9y2)2 + 2(2x2)(9y2) - 36x2y2 =(2x2 + 9y2)2 - (6xy)2
Apply difference of squares identity a2 - b2 = (a + b)(a - b):
=(2x2 + 9y2 + 6xy)(2x2 + 9y2 - 6xy) =(2x2 + 6xy + 9y2)(2x2 - 6xy + 9y2)
Answer: (2x2 + 6xy + 9y2)(2x2 - 6xy + 9y2)

(ii) a4 + 64b4

Solution
Write expression: a4 + 64b4
Rewrite as perfect squares: (a2)2 + (8b2)2
Add and subtract 2(a2)(8b2) = 16a2b2 to complete the square:
=(a2)2 + (8b2)2 + 2(a2)(8b2) - 16a2b2 =(a2 + 8b2)2 - (4ab)2
Apply x2 - y2 = (x + y)(x - y):
=(a2 + 8b2 + 4ab)(a2 + 8b2 - 4ab) =(a2 + 4ab + 8b2)(a2 - 4ab + 8b2)
Answer: (a2 + 4ab + 8b2)(a2 - 4ab + 8b2)

(iii) x4 + 4x2 + 16

Solution
Rearrange terms: x4 + 16 + 4x2
Complete the square for (x2)2 + 42 by adding and subtracting 2(x2)(4) = 8x2:
=(x2)2 + (4)2 + 2(x2)(4) - 8x2 + 4x2 =(x2 + 4)2 - 4x2 =(x2 + 4)2 - (2x)2
Apply a2 - b2 = (a + b)(a - b):
=(x2 + 4 + 2x)(x2 + 4 - 2x) =(x2 + 2x + 4)(x2 - 2x + 4)
Answer: (x2 + 2x + 4)(x2 - 2x + 4)

(iv) x4 - 14x2 + 1

Solution
Rearrange terms: x4 + 1 - 14x2
Complete the square for (x2)2 + 12 by adding and subtracting 2(x2)(1) = 2x^2:
=(x2)2 + 12 + 2(x2)(1) - 2x2 - 14x2 =(x2 + 1)2 - 16x2 =(x2 + 1)2 - (4x)2
Apply difference of squares identity:
=(x2 + 1 + 4x)(x2 + 1 - 4x) =(x2 + 4x + 1)(x2 - 4x + 1)
Answer: (x2 + 4x + 1)(x2 - 4x + 1)

(v) x4 - 30x2y2 + 9y4

Solution
Rearrange terms: x4 + 9y4 - 30x2y2
Rewrite as squares: (x2)2 + (3y2)2 - 30x2y2
Complete the square by adding and subtracting 2(x2)(3y2) = 6x2y2:
=(x2)2 + (3y2)2 + 2(x2)(3y2) - 6x2y2 - 30x2y2 =(x2 + 3y2)2 - 36x2y2 =(x2 + 3y2)2 - (6xy)2
Factor using difference of squares:
=(x2 + 3y2 + 6xy)(x2 + 3y2 - 6xy) =(x2 + 6xy + 3y2)(x2 - 6xy + 3y2)
Answer: (x2 + 6xy + 3y2)(x2 - 6xy + 3y2)

(vi) x4 - 7x2y2 + y4

Solution
Rearrange terms: x4 + y4 - 7x2y2
Complete the square for (x2)2 + (y2)2 by adding and subtracting 2x2y2:
=(x2)2 + (y2)2 + 2x2y2 - 2x2y2 - 7x2y2 =(x2 + y2)2 - 9x2y2 =(x2 + y2)2 - (3xy)2
Factor:
=(x2 + y2 + 3xy)(x2 + y2 - 3xy) =(x2 + 3xy + y2)(x2 - 3xy + y2)
Answer: (x2 + 3xy + y2)(x2 - 3xy + y2)
Question 2 Linear Groups & Substitutions
Factorize each of the following expressions using grouping and variable substitution.

(i) (x + 1)(x + 2)(x + 3)(x + 4) + 1

Solution
Group factors by matching sums of constants: 1 + 4 = 5 and 2 + 3 = 5.
=[(x + 1)(x + 4)][(x + 2)(x + 3)] + 1 =[x2 + 5x + 4][x2 + 5x + 6] + 1
Let y = x2 + 5x. Substitute y into the expression:
=(y + 4)(y + 6) + 1 =y2 + 10y + 24 + 1 =y2 + 10y + 25
Complete the perfect square: y2 + 2(y)(5) + 52 = (y + 5)2.
Substitute back y = x2 + 5x:
=(x2 + 5x + 5)2
Answer: (x2 + 5x + 5)2

(ii) (x + 2)(x - 7)(x - 4)(x - 1) + 17

Solution
Group factors by matching constant sums: 2 - 7 = -5 and -4 - 1 = -5:
=[(x + 2)(x - 7)][(x - 4)(x - 1)] + 17 =[x2 - 5x - 14][x2 - 5x + 4] + 17
Let y = x2 - 5x. Substitute:
=(y - 14)(y + 4) + 17 =y2 - 10y - 56 + 17 =y2 - 10y - 39
Factorize the quadratic in y by splitting the middle term:
=y2 - 13y + 3y - 39 =y(y - 13) + 3(y - 13) =(y - 13)(y + 3)
Substitute back y = x2 - 5x:
=(x2 - 5x - 13)(x2 - 5x + 3)
Answer: (x2 - 5x - 13)(x2 - 5x + 3)

(iii) (2x2 + 7x + 3)(2x2 + 7x + 5) + 1

Solution
Identify the repeating algebraic expression: 2x2 + 7x.
Let y = 2x2 + 7x. Substitute into expression:
=(y + 3)(y + 5) + 1 =y2 + 8y + 15 + 1 =y2 + 8y + 16
Factor as a perfect square: (y + 4)2.
Substitute back y = 2x2 + 7x:
=(2x2 + 7x + 4)2
Answer: (2x2 + 7x + 4)2

(iv) (3x2 + 5x + 3)(3x2 + 5x + 5) - 3

Solution
Let y = 3x2 + 5x. Substitute:
=(y + 3)(y + 5) - 3 =y2 + 8y + 15 - 3 =y2 + 8y + 12
Split the middle term:
=y2 + 6y + 2y + 12 =y(y + 6) + 2(y + 6) =(y + 6)(y + 2)
Substitute back y = 3x2 + 5x:
=(3x2 + 5x + 6)(3x2 + 5x + 2)
Answer: (3x2 + 5x + 6)(3x2 + 5x + 2)

(v) (x + 1)(x + 2)(x + 3)(x + 6) - 3x2

Solution
Group terms by matching product of constants: 1 × 6 = 6 and 2 × 3 = 6.
=[(x + 1)(x + 6)][(x + 2)(x + 3)] - 3x2 =[x2 + 7x + 6][x2 + 5x + 6] - 3x2
Rearrange to group x2 + 6: [(x2 + 6) + 7x][(x2 + 6) + 5x] - 3x2.
Let y = x2 + 6. Substitute:
=(y + 7x)(y + 5x) - 3x2 =y2 + 12xy + 35x2 - 3x2 =y2 + 12xy + 32x2
Split the middle term for y2 + 12xy + 32x2 using 8x and 4x:
=y2 + 8xy + 4xy + 32x2 =y(y + 8x) + 4x(y + 8x) =(y + 8x)(y + 4x)
Substitute back y = x2 + 6:
=(x2 + 8x + 6)(x2 + 4x + 6)
Answer: (x2 + 8x + 6)(x2 + 4x + 6)

(vi) (x + 1)(x - 1)(x + 2)(x - 2) + 5x2

Solution
Multiply the pairs:
=(x2 - 1)(x2 - 4) + 5x2 =x4 - 5x2 + 4 + 5x2 =x4 + 4
Complete the square for (x2)2 + 22 by adding and subtracting 4x2:
=(x2)2 + 22 + 4x2 - 4x2 =(x2 + 2)2 - (2x)2
Apply difference of squares identity:
=(x2 + 2 + 2x)(x2 + 2 - 2x) =(x2 + 2x + 2)(x2 - 2x + 2)
Answer: (x2 + 2x + 2)(x2 - 2x + 2)
Question 3 Perfect Cubes
Factorize each of the following expressions using perfect cube identities:
(a ± b)3 = a3 ± 3a2b + 3ab2 ± b3.

(i) 8x3 + 12x2 + 6x + 1

Solution
Rewrite the terms as cubes and multiples:
=(2x)3 + 3(2x)2(1) + 3(2x)(1)2 + (1)3
Compare with identity a3 + 3a2b + 3ab2 + b3 = (a + b)3, where a = 2x and b = 1.
Factor as: (2x + 1)3.
Answer: (2x + 1)3

(ii) 27a3 + 108a2b + 144ab2 + 64b3

Solution
Rewrite the terms as cubes and products:
=(3a)3 + 3(3a)2(4b) + 3(3a)(4b)2 + (4b)3
Check coefficients: 3(9)(4) = 108 and 3(3)(16) = 144. This matches exactly.
Factor as (a + b)3 with a = 3a and b = 4b:
=(3a + 4b)3
Answer: (3a + 4b)3

(iii) x3 + 18x2y + 108xy2 + 216y3

Solution
Rewrite the expression:
=(x)3 + 3(x)2(6y) + 3(x)(6y)2 + (6y)3
Check coefficients: 3(6) = 18 and 3(36) = 108. This matches exactly.
Factor as (x + 6y)3:
=(x + 6y)3
Answer: (x + 6y)3

(iv) 8x3 - 125y3 + 150xy2 - 60x2y

Solution
Rearrange terms in descending order of x: 8x3 - 60x2y + 150xy2 - 125y3.
Rewrite the terms:
=(2x)3 - 3(2x)2(5y) + 3(2x)(5y)2 - (5y)3
Check coefficients: -3(4)(5) = -60 and 3(2)(25) = 150. This matches exactly.
Factor using (a - b)3 with a = 2x and b = 5y:
=(2x - 5y)3
Answer: (2x - 5y)3
Question 4 Sum or Difference of Cubes
Factorize each of the following expressions using sum or difference of cubes identities:
a3 ± b3 = (a ± b)(a2 ∓ ab + b2).

(i) 125a3 - 1

Solution
Rewrite as difference of cubes: (5a)3 - (1)3.
Apply identity x3 - y3 = (x - y)(x2 + xy + y2):
=(5a - 1)[(5a)2 + (5a)(1) + (1)2] =(5a - 1)(25a2 + 5a + 1)
Answer: (5a - 1)(25a2 + 5a + 1)

(ii) 64x3 + 125

Solution
Rewrite as sum of cubes: (4x)3 + (5)3.
Apply identity x3 + y3 = (x + y)(x2 - xy + y2):
=(4x + 5)[(4x)2 - (4x)(5) + (5)2] =(4x + 5)(16x2 - 20x + 25)
Answer: (4x + 5)(16x2 - 20x + 25)

(iii) x6 - 27

Solution
Rewrite as difference of cubes: (x2)3 - (3)3.
Apply identity a3 - b3 = (a - b)(a2 + ab + b2):
=(x2 - 3)[(x2)2 + (x2)(3) + (3)2] =(x2 - 3)(x4 + 3x2 + 9)
Answer: (x2 - 3)(x4 + 3x2 + 9)

(iv) 1000a3 + 1

Solution
Rewrite as sum of cubes: (10a)3 + (1)3.
Apply identity:
=(10a + 1)[(10a)2 - (10a)(1) + (1)2] =(10a + 1)(100a2 - 10a + 1)
Answer: (10a + 1)(100a2 - 10a + 1)

(v) 343x3 + 216

Solution
Rewrite as sum of cubes: (7x)3 + (6)3.
Apply identity:
=(7x + 6)[(7x)2 - (7x)(6) + (6)2] =(7x + 6)(49x2 - 42x + 36)
Answer: (7x + 6)(49x2 - 42x + 36)

(vi) 27 - 512y3

Solution
Rewrite as difference of cubes: (3)3 - (8y)3.
Apply identity:
=(3 - 8y)[(3)2 + (3)(8y) + (8y)2] =(3 - 8y)(9 + 24y + 64y2)
Answer: (3 - 8y)(9 + 24y + 64y2)
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