Factorization

factorization class 8 rs aggarwal

Step by Step solutions of Exercise- 14D of RS Aggarwal ICSE Class-8 Maths chapter 14- Factorization by Goyal Brothers Prakashan is provided.

Exercise: 14D

Q1: x² + 18x + 81

Step 1: Identify first and last terms: x² and 81
x² = (x)², 81 = 9²
Step 2: Check the middle term matches 2ab: 2·x·9 = 18x
Step 3: Write as a square of a binomial:
x² + 18x + 81 = (x + 9)²
Answer: (x + 9)²


Q2: a² − 14a + 49

Step 1: Identify first and last terms: a² and 49
a² = (a)², 49 = 7²
Step 2: Check the middle term matches −2ab: −2·a·7 = −14a
Step 3: Write as a square of a binomial:
a² − 14a + 49 = (a − 7)²
Answer: (a − 7)²


Q3: 4x² + 12x + 9

Step 1: Identify first and last terms: 4x² and 9
4x² = (2x)², 9 = 3²
Step 2: Check the middle term matches 2ab: 2·(2x)·3 = 12x
Step 3: Write as a square of a binomial:
4x² + 12x + 9 = (2x + 3)²
Answer: (2x + 3)²


Q4: 9x² − 24x + 16

Step 1: Identify first and last terms: 9x² and 16
9x² = (3x)², 16 = 4²
Step 2: Check the middle term matches −2ab: −2·(3x)·4 = −24x
Step 3: Write as a square of a binomial:
9x² − 24x + 16 = (3x − 4)²
Answer: (3x − 4)²


Q5: 16 + 56x + 49x²

Step 1: Write in standard form: 49x² + 56x + 16
Step 2: Identify first and last terms: 49x² and 16
49x² = (7x)², 16 = 4²
Step 3: Check the middle term matches 2ab: 2·(7x)·4 = 56x
Step 4: Write as a square of a binomial:
49x² + 56x + 16 = (7x + 4)²
Answer: (7x + 4)²


Q6: 9 − 30z + 25z²

Step 1: Write in standard form: 25z² − 30z + 9
Step 2: Identify first and last terms: 25z² and 9
25z² = (5z)², 9 = 3²
Step 3: Check the middle term matches −2ab: −2·(5z)·3 = −30z
Step 4: Write as a square of a binomial:
25z² − 30z + 9 = (5z − 3)²
Answer: (5z − 3)²


Q7: 9q⁴r⁴ − 6p⁴q²r² + p⁸

Step 1: Identify first and last terms: 9q⁴r⁴ and p⁸
9q⁴r⁴ = (3q²r²)², p⁸ = (p⁴)²
Step 2: Check the middle term matches −2ab: −2·(3q²r²)·(p⁴) = −6p⁴q²r²
Step 3: Write as a square of a binomial:
9q⁴r⁴ − 6p⁴q²r² + p⁸ = (3q²r² − p⁴)²
Answer: (3q²r² − p⁴)²


Q8: \(\frac{9p^2}{q^2} + \frac{16r^2}{m^2} + \frac{24pr}{qm}\)

Step 1: Observe that this is a perfect square trinomial of the form \(a^2 + 2ab + b^2 = (a + b)^2\).
Step 2: Identify \(a\) and \(b\): \[ a = \frac{3p}{q}, \quad b = \frac{4r}{m} \] Check middle term: \[ 2ab = 2 \times \frac{3p}{q} \times \frac{4r}{m} = \frac{24pr}{qm} \] This matches the middle term.
Step 3: Write as square of sum: \[ = \left( \frac{3p}{q} + \frac{4r}{m} \right)^2 \]Answer: \(\left( \frac{3p}{q} + \frac{4r}{m} \right)^2\)


Q9: \(\frac{1}{4}z^6 + 9a^2 – 3az^3\)

Step 1: Rearrange the terms for clarity: \[ = \frac{1}{4}z^6 – 3az^3 + 9a^2 \]Step 2: Observe that this is a perfect square trinomial of the form \(a^2 – 2ab + b^2 = (a – b)^2\).
Step 3: Identify \(a\) and \(b\): \[ a = \frac{1}{2}z^3, \quad b = 3a \] Check middle term: \[ -2ab = -2 \times \frac{1}{2}z^3 \times 3a = -3az^3 \] This matches the middle term.
Step 4: Write as square of difference: \[ = \left( \frac{1}{2}z^3 – 3a \right)^2 \]Answer: \(\left( \frac{1}{2}z^3 – 3a \right)^2\)


Q10: \(\frac{9}{4}a^2 + \frac{49}{9}p^2 – 7ap\)

Step 1: Observe that this is a perfect square trinomial of the form \(a^2 – 2ab + b^2 = (a – b)^2\).
Step 2: Identify \(a\) and \(b\): \[ a = \frac{3}{2}a, \quad b = \frac{7}{3}p \] Check middle term: \[ -2ab = -2 \times \frac{3}{2}a \times \frac{7}{3}p = -7ap \] This matches the middle term.
Step 3: Write as square of difference: \[ = \left(\frac{3}{2}a – \frac{7}{3}p\right)^2 \]Answer: \(\left(\frac{3}{2}a – \frac{7}{3}p\right)^2\)


Q11: t² + 22t + 85

Step 1: Check if it is a perfect square trinomial:
t² = (t)², middle term = 22t = 2·t·? → b = 11, last term = 85 ≠ 11² = 121
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of 85 that add up to 22 → 5 and 17
Split middle term: t² + 5t + 17t + 85
Step 3: Factor by grouping:
(t² + 5t) + (17t + 85) = t(t+5) + 17(t+5) = (t+5)(t+17)
Answer: (t+5)(t+17)


Q12: x² − 10x + 24

Step 1: Check if it is a perfect square trinomial:
x² = (x)², middle term = -10x = -2·x·? → b = 5, last term = 24 ≠ 5² = 25
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of 24 that add up to -10 → -4 and -6
Split middle term: x² – 4x – 6x + 24
Step 3: Factor by grouping:
(x² – 4x) + (-6x + 24) = x(x-4) – 6(x-4) = (x-4)(x-6)
Answer: (x-4)(x-6)


Q13: m² − 3m − 40

Step 1: Check if it is a perfect square trinomial:
m² = (m)², middle term = -3m = -2·m·? → b = 3/2, last term = -40 ≠ (3/2)² = 9/4
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of -40 that add up to -3 → 5 and -8
Split middle term: m² + 5m – 8m – 40
Step 3: Factor by grouping:
(m² + 5m) + (-8m – 40) = m(m+5) – 8(m+5) = (m+5)(m-8)
Answer: (m+5)(m-8)


Q14: x² + x − 72

Step 1: Check if it is a perfect square trinomial:
x² = (x)², middle term = x = 2·x·? → b = 1/2, last term = -72 ≠ (1/2)² = 1/4
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of -72 that add up to 1 → 9 and -8
Split middle term: x² + 9x – 8x – 72
Step 3: Factor by grouping:
(x² + 9x) – (8x + 72) = x(x+9) – 8(x+9) = (x-8)(x+9)
Answer: (x-8)(x+9)


Q15: p² − 7p − 120

Step 1: Check if it is a perfect square trinomial:
p² = (p)², middle term = -7p = -2·p·? → b = 7/2, last term = -120 ≠ (7/2)² = 49/4
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of -120 that add up to -7 → 8 and -15
Split middle term: p² + 8p – 15p – 120
Step 3: Factor by grouping:
(p² + 8p) – (15p + 120) = p(p+8) – 15(p+8) = (p-15)(p+8)
Answer: (p-15)(p+8)


Q16: 16 − 17z + z²

Step 1: Write in standard form:
z² – 17z + 16
Step 2: Check if it is a perfect square trinomial:
z² = (z)², middle term = -17z = -2·z·? → b = 17/2, last term = 16 ≠ (17/2)² = 289/4
→ Not a perfect square trinomial
Step 3: Factorization using splitting method:
Find factors of 16 that add up to -17 → -1 and -16
Split middle term: z² – z – 16z + 16
Step 4: Factor by grouping:
(z² – z) – (16z – 16) = z(z-1) – 16(z-1) = (z-1)(z-16)
Answer: (z-1)(z-16)


Q17: a² + 5a − 104

Step 1: Check if it is a perfect square trinomial:
a² = (a)², middle term = 5a = 2·a·? → b = 5/2, last term = -104 ≠ (5/2)² = 25/4
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Find factors of -104 that add up to 5 → 13 and -8
Split middle term: a² + 13a – 8a – 104
Step 3: Factor by grouping:
(a² + 13a) – (8a + 104) = a(a+13) – 8(a+13) = (a-8)(a+13)
Answer: (a-8)(a+13)


Q18: 3x² + 11x + 10

Step 1: Check if it is a perfect square trinomial:
3x² = (√3·x)², last term = 10 ≠ (√3·x·?)²
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Multiply first and last term: 3·10 = 30
Find factors of 30 that add up to 11 → 6 and 5
Split middle term: 3x² + 6x + 5x + 10
Step 3: Factor by grouping:
(3x² + 6x) + (5x + 10) = 3x(x+2) + 5(x+2) = (3x+5)(x+2)
Answer: (3x+5)(x+2)


Q19: 6x² + 7x − 3

Step 1: Check if it is a perfect square trinomial:
6x² = (√6·x)², last term = -3 ≠ (√6·x·?)²
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Multiply first and last term: 6·(-3) = -18
Find factors of -18 that add up to 7 → 9 and -2
Split middle term: 6x² + 9x – 2x – 3
Step 3: Factor by grouping:
(6x² + 9x) – (2x + 3) = 3x(2x+3) – 1(2x+3) = (3x-1)(2x+3)
Answer: (3x-1)(2x+3)


Q20: 3z² − 4z − 4

Step 1: Check if it is a perfect square trinomial:
3z² = (√3·z)², last term = -4 ≠ (√3·z·?)²
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Multiply first and last term: 3·(-4) = -12
Find factors of -12 that add up to -4 → 2 and -6
Split middle term: 3z² + 2z – 6z – 4
Step 3: Factor by grouping:
(3z² + 2z) – (6z + 4) = z(3z+2) – 2(3z+2) = (z-2)(3z+2)
Answer: (z-2)(3z+2)


Q21: 72 − x − x²

Step 1: Rearrange in standard form:
72 − x − x² = -x² − x + 72
Factor out -1: -(x² + x − 72)
Step 2: Factorization using splitting method:
Multiply first and last term: 1·(-72) = -72
Find factors of -72 that add up to 1 → 9 and -8
Split middle term: x² + 9x – 8x – 72
Step 3: Factor by grouping:
(x² + 9x) – (8x + 72) = x(x+9) – 8(x+9) = (x-8)(x+9)
Include -1 factor: -(x-8)(x+9) = (8-x)(x+9)
Answer: (8-x)(x+9)


Q22: x² − 3xy − 40y²

Step 1: Check if it is a perfect square trinomial:
x² = (x)², last term = -40y² ≠ (something)²
→ Not a perfect square trinomial
Step 2: Factorization using splitting method:
Multiply first and last term: 1·(-40) = -40
Find factors of -40 that add up to -3 → 5 and -8
Split middle term: x² + 5xy – 8xy – 40y²
Step 3: Factor by grouping:
(x² + 5xy) – (8xy + 40y²) = x(x+5y) – 8y(x+5y) = (x-8y)(x+5y)
Answer: (x-8y)(x+5y)


Q23: 3x²y + 11xy + 6y

Step 1: Factor out the common factor:
3x²y + 11xy + 6y = y(3x² + 11x + 6)
Step 2: Factorize quadratic 3x² + 11x + 6 using splitting method:
Multiply first and last term: 3·6 = 18
Find factors of 18 that add up to 11 → 9 and 2
Split middle term: 3x² + 9x + 2x + 6
Step 3: Factor by grouping:
(3x² + 9x) + (2x + 6) = 3x(x+3) + 2(x+3) = (3x+2)(x+3)
Include y: y(3x+2)(x+3)
Answer: y(3x+2)(x+3)


Q24: (a-b)² – 5(a-b) + 6

Step 1: Substitute x = a-b
Then expression becomes: x² – 5x + 6
Step 2: Factorize quadratic x² – 5x + 6
Find factors of 6 that add up to -5 → -2 and -3
Split middle term: x² – 2x – 3x + 6
Step 3: Factor by grouping:
(x² – 2x) – (3x – 6) = x(x-2) – 3(x-2) = (x-3)(x-2)
Step 4: Replace x = a-b
(x-3)(x-2) = (a-b-3)(a-b-2)
Answer: (a-b-3)(a-b-2)


Q25: (a-3b)² – 4(a-3b) – 21

Step 1: Substitute x = a-3b
Then expression becomes: x² – 4x – 21
Step 2: Factorize quadratic x² – 4x – 21
Find factors of -21 that add up to -4 → 3 and -7
Split middle term: x² + 3x – 7x – 21
Step 3: Factor by grouping:
(x² + 3x) – (7x + 21) = x(x+3) – 7(x+3) = (x-7)(x+3)
Step 4: Replace x = a-3b
(x-7)(x+3) = (a-3b-7)(a-3b+3)
Answer: (a-3b-7)(a-3b+3)


Q26: 3(y-2)² – (y-2) – 44

Step 1: Substitute x = y-2
Then expression becomes: 3x² – x – 44
Step 2: Factorize quadratic 3x² – x – 44
Multiply first and last term: 3 * -44 = -132
Find factors of -132 that add up to -1 → 11 and -12
Split middle term: 3x² + 11x – 12x – 44
Step 3: Factor by grouping:
(3x² + 11x) – (12x + 44) = x(3x + 11) – 4(3x + 11) = (x-4)(3x+11)
Step 4: Replace x = y-2
(x-4)(3x+11) = (y-6)(3y+5)
Answer: (y-6)(3y+5)


Q27: 7 + 10(x+y) – 8(x+y)²

Step 1: Substitute z = x+y
Expression becomes: 7 + 10z – 8z²
Rewrite in standard form: -8z² + 10z + 7
Factor out -1: -(8z² – 10z – 7)
Step 2: Factorize quadratic 8z² – 10z – 7
Multiply first and last term: 8 * -7 = -56
Find factors of -56 that add up to -10 → -14 and 4
Split middle term: 8z² – 14z + 4z – 7
Step 3: Factor by grouping:
(8z² – 14z) + (4z – 7) = 2z(4z – 7) + 1(4z – 7) = (2z + 1)(4z – 7)
Include the negative sign: -(2z + 1)(4z – 7)
Step 4: Replace z = x+y
-(2(x+y)+1)(4(x+y)-7) = (2x + 2y + 1)(7 – 4x – 4y)
Answer: -(2x + 2y + 1)(7 – 4x – 4y)


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