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

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)

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

(iv) m2 + m – 72
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.4 2
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 8th Maths Term 1 Chapter 3 Algebra Ex 3.4 3

SamacheerKalvi.Guru

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)

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)

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