Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

8th Standard Maths Exercise 3.4 Question 1.
Factorise the following by taking out the common factor
(i) 18xy – 12yz
(ii) 9x⁵y³ + 6x³y² – 18x²y
(iii) x(b – 2c) + y(b – 2c)
(iv) (ax + ay) + (bx + by)
(v) 2x²(4x – 1) – 4x + 1
(vi) 3y(x – 2)² – 2(2 – x)
(vii) 6xy – 4y² + 12xy – 2yzx
(viii) a³ – 3a² + a – 3
(ix) 3y³ – 48y
(x) ab² – bc² – ab + c²
Solution:
(i) 18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y (3x – 2z) = 6y (3x- 2z)

i) 9x⁵y³ + 6x³y² – 18x²y = (3 × 3 × x² × x³ × y × y²) + (2 × 3 × x² × x × y × y)
Taking out the common factors 3, x², y, we get
= 3 × x² × y (3x³ y² + 2xy – 6)
= 3x²y (3x³ y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)

(iv) (ax + ay) + (bx + by)
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay) + (bx + by) = a (x + y) + b (x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y)(a + b)

(v) 2x²(4x – 1) – 4x + 1
Taking out -1 from last two terms
2x² (4x – 1) – 4x + 1 = 2x² (4x – 1) – 1 (4x- 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1)(2x² – 1)

(vi) 3y(x – 2)² – 2(2 – x)
3y(x – 2)² – 2(2 – x) = 3y(x – 2)(x – 2)-2(-1) (x – 2) [∵ Taking out -1 from 2 – x]
= 3y (x – 2) (x – 2) + 2 (x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y (x – 2) + 2]

(vii) 6xy – 4y² + 12xy – 2yzx
= 6xy + 12xy – 4y² – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 × x × y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1)(2)y[2y + zx]
= 6xy (3) -2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy (2y + zx)
Taking out 2y from two terms
= 2y (9x – (2y + zx)) = 2y (9x – 2y – xz)

(viii) a² – 3a² + a – 3 = a² (a – 3) + 1 (a – 3) [∵ Grouping the terms suitably]
= (a – 3) (a² + 1)

(ix) 3y² – 48y = 3 × y × y² – 3 × 16 × y
Taking out 3 × y = 3y (y² – 16) = 3y (y² – 4²)
Comparing y² – 4² with a² – b²
a = y, b = 4
a² – b² = (a + b) (a – b)
y² – 4² = (y + 4) (y – 4)
∴ 3y (y² – 16) = 3y (y + 4) (y – 4)

(x) ab² – bc² – ab + c²
Grouping suitably
ab² – bc² – ab + c² = b ((ab – c²) – 1(ab – c²)
Taking out the binomial factor ab – c² = (ab – c²) (b – 1)

Samacheer Kalvi Guru 8th Maths Question 2.
Factorise the following expressions
(i) x² + 14x + 49
(ii) y² – 10y + 25
(iii) c² – 4c – 12
(iv) m² + m – 72
(v) 4x² – 8x + 3
Solution:
x² + 14x + 49 = x² + 14x + 72
Comparing with a² + 2ab + b² = (a + b)² we have a = x and b = 7
⇒ x² + 2(x) (7) + 7² = (x + 7)²
∴ x² + 14x + 49 = (x + 7)²

(ii) y² – 10y + 25 = y² – 10y + 5²
Comparing with a² – 2ab + b² = (a – b)² we get a = y ; b = 5
⇒ y² – 2(y) (5) + 5² = (y – 5)²
∴ y² – 10y + 25 = (y – 5)²

(iii) c² – 4c – 12
This is of the form ax² + bx + c
Where a = 1,b = – 4 c = – 12, x = c
Now the product ac = 1 × – 12 = – 12 and the sum b = -4

(iv) m² + m – 72

This is of the form ax² + bx + c
where a = 1, b = 1, c = -12
Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m² + m – 72 = m² + 9m – 8m – 12
= m (m + 9) – 8 (m + 9)
Taking out (m + 9)
= (m + 9) (m – 8)
∴ m² + m – 72 = (m + 9) (m – 8)

(v) 4x² – 8x + 3
This is of the form ax² + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
Sum b = -8

Samacheer Kalvi Guru Maths 8th Question 3.
Factorize the following expressions using a³ + b³ = (a + b)(a² – ab + b²) identity
(i) h³ + k²
(ii) 2a³ + 16
(iii) x³y³ + 27
(iv) 64m³ + n³
(v) r⁴ + 27p³r
Solution:
(i) h³ + k³
Comparing h³ + k³ with a³ + b³ = (a + b)(a² – ab + b²) we have a = h, b = k
∴ h³ + k³ = (h + k)(h² – hk + k²)

(ii) 2a² + 16
(2 × a³) + (2 × 8) = 2(a³ + 8) = 2 (a³ + 2³)
∴ 2a³ + 16 = 2 (a³ + 2³)
Comparing with a³ + b³ we have a = a and b = 2
a³ + b³ = (a + b)(a² – ab + b²)
2(a³ + 2³) = 2[(a + 2) (a² – (a) (2) + 2²)] = 2[(a + 2) (a² – 2a + 4)]
2a³ +16 = 2 (a + 2) (a² – 2a + 4)

(iii) x³ y³ + 27 = (xy)³ + 3³
Comparing with a³ + b³ we have a = xy ;b = 3
a³ + b³ = (a + b) (a² – ab + b²)
(xy)³ + 3³ = (xy + 3) ((xy)² – (xy) (3) + 3²) = (xy + 3) (x²y² – 3xy + 9)
∴ x³y³ + 27 = (xy + 3)(x²y² – 3xy + 9)

(iv) 64m³ + n³ = (4³m³) + n³
= (4m)³ + n³
Comparing this with a³ + b³ we have a = 4m; b = n
a³ + b³ = (a + b) (a² – ab + b²)
(4m)³ + n³ = (4m + n) [(4m)2 – (4m) (n) + n2]
= (4m + n) [4²m² – 4mn + n²]
= (4m + n) [ 16m² – 4mn + n²]
64m³ + n³ = (4m + n) (16m² – 4mn + n²)

(v) r³ + 27p³r = r (r³ + 27p³) = r [r³ + 3³p³]
Comparing r³ + (3p)³ with a³ + b³ we have a = r; b = 3p
a³ + b³ = (a + b)(a² – ab + b²)
r[r³ + (3p)³] = r[(r + 3p)(r² – r(3p) + (3p)²)]
= r[(r+ 3p) (r² – 3rp + 3²p²]
= r(r + 3p) (r² – 3rp + 9p²)
r⁴ + 27p³r = r(r + 3p) (r² – 3rp + 9p²)

11th Maths Exercise 3.4 Samacheer Kalvi Question 4.
Factorize the following expressions using a³ – b³ = (a – b)(a² + ab + b²) identity
(i) y³ – 27
(ii) 3b³ – 192c³
(iii) -16y³ + 2x³
(iv) x³ y³ – 7³
(v) c³ – 27b³ a³
Solution:
(i) y³ – 27 = y³ – 3³
Comparing this with a³ – b³ , we have a = y and b = 3
a³ – b³ = (a + b) (a² + ab + b² )
y³ – 3³ = (y – 3)(y² + (y) (3) + 3² ) = (y – 3) (y² + 3y + 9)
y³ – 27 = (y – 3) (y² + 3y + 9)

(ii) 3b³ + 192c³ = (3 × b³) – (3 × 4 × 4 × 4 × c³) = 3(b³ – 4³ c³)
= 3(b³ – (4c)³ )
Comparing b³ – (4c)³ with a³ – b³ we have a = b and b = 4c
a³ – b³ = (a – b) (a² + ab + b²)
3(b³ – (4c)³) = 3[(b – 4c)(b² + (b)(4c) + (4c)²)]
= 3[(b – 4c)(b² + 4bc + 4² c²)]
3b³ – 192c³ = 3 [(b – 4c) (b² + 4bc + 16c²)]

(iii) -16y³ + 2x³ = 2x³ – 16y³ [∵ Addition is commatative]
= 2(x³ – 8y³) = 2(x³ – 23y³)
= 2(x³ – (2y)³)
Comparing x³ – (2y)³ with a³ – b³ we have a = x and b = 2y
a³ – b³ = (a – b)(a² + ab + b²)
2[x³ – (2y)³] = 2[(x – 2y) (x² + (x) (2y) + (2y)²)]
= 2[(x – 2y) (x² + 2xy + 2² y²)]
-16y³ + 2x³ = 2[(x – 2y)(x² + 2xy + 4y²)]

(iv) x³y³ – 7³ = (xy)³ – 7³
Comparing with a³ – b³ we have a = xy and b = 7
a³ – b³ = (a – b)(a² + ab + b²)
(xy)³ – 7³ = (xy – 7) ((xy)² + (xy) (7) + 7²)
x³y³ – 7³ = (xy – 7) (x²y² + 7xy + 49)

(v) c³ – 27 b³ a³ = c³ – 3³b³ a³ = c³ – (3ba)³
Comparing this with a³ – b³ we have a = x and b = 3ba
a³ – b³ = (a – b)(a² + ab + b²)
∴ c³ – (3ba)³ = (c – 3ba) (c² + (c) (3ba) + (3ba)²)
= (c – 3ba) (c² + 3bac + 3² b²a²)
c³ – 27b³a³ = (c – 3ab) (c² + 3bac + 9a²b²)