## 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) 9x^{5}y^{3} + 6x^{3}y^{2} – 18x^{2}y

(iii) x(b – 2c) + y(b – 2c)

(iv) (ax + ay) + (bx + by)

(v) 2x^{2}(4x – 1) – 4x + 1

(vi) 3y(x – 2)^{2} – 2(2 – x)

(vii) 6xy – 4y^{2} + 12xy – 2yzx

(viii) a^{3} – 3a^{2} + a – 3

(ix) 3y^{3} – 48y

(x) ab^{2} – bc^{2} – ab + c^{2}

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) 9x^{5}y^{3} + 6x^{3}y^{2} – 18x^{2}y = (3 × 3 × x^{2} × x^{3} × y × y^{2}) + (2 × 3 × x^{2} × x × y × y)

Taking out the common factors 3, x^{2}, y, we get

= 3 × x^{2} × y (3x^{3} y^{2} + 2xy – 6)

= 3x^{2}y (3x^{3} y^{2} + 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) 2x^{2}(4x – 1) – 4x + 1

Taking out -1 from last two terms

2x^{2} (4x – 1) – 4x + 1 = 2x^{2} (4x – 1) – 1 (4x- 1)

Taking out the binomial factor 4x – 1, we get

= (4x – 1)(2x^{2} – 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 – 4y^{2} + 12xy – 2yzx

= 6xy + 12xy – 4y^{2} – 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) a^{2} – 3a^{2} + a – 3 = a^{2} (a – 3) + 1 (a – 3) [∵ Grouping the terms suitably]

= (a – 3) (a^{2} + 1)

(ix) 3y^{2} – 48y = 3 × y × y^{2} – 3 × 16 × y

Taking out 3 × y = 3y (y^{2} – 16) = 3y (y^{2} – 4^{2})

Comparing y^{2} – 4^{2} with a^{2} – b^{2}

a = y, b = 4

a^{2} – b^{2} = (a + b) (a – b)

y^{2} – 4^{2} = (y + 4) (y – 4)

∴ 3y (y^{2} – 16) = 3y (y + 4) (y – 4)

(x) ab^{2} – bc^{2} – ab + c^{2}

Grouping suitably

ab^{2} – bc^{2} – ab + c^{2} = b ((ab – c^{2}) – 1(ab – c^{2})

Taking out the binomial factor ab – c^{2} = (ab – c^{2}) (b – 1)

Question 2.

Factorise the following expressions

(i) x^{2} + 14x + 49

(ii) y^{2} – 10y + 25

(iii) c^{2} – 4c – 12

(iv) m^{2} + m – 72

(v) 4x^{2} – 8x + 3

Solution:

x^{2} + 14x + 49 = x^{2} + 14x + 72

Comparing with a^{2} + 2ab + b^{2} = (a + b)^{2} we have a = x and b = 7

⇒ x^{2} + 2(x) (7) + 7^{2} = (x + 7)^{2}

∴ x^{2} + 14x + 49 = (x + 7)^{2}

(ii) y^{2} – 10y + 25 = y^{2} – 10y + 5^{2}

Comparing with a^{2} – 2ab + b^{2} = (a – b)^{2} we get a = y ; b = 5

⇒ y^{2} – 2(y) (5) + 5^{2} = (y – 5)^{2}

∴ y^{2} – 10y + 25 = (y – 5)^{2}

(iii) c^{2} – 4c – 12

This is of the form ax^{2} + bx + c

Where a = 1,b = – 4 c = – 12, x = c

Now the product ac = 1 × – 12 = – 12 and the sum b = -4

(iv) m^{2} + m – 72

This is of the form ax^{2} + 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

m^{2} + m – 72 = m^{2} + 9m – 8m – 12

= m (m + 9) – 8 (m + 9)

Taking out (m + 9)

= (m + 9) (m – 8)

∴ m^{2} + m – 72 = (m + 9) (m – 8)

(v) 4x^{2} – 8x + 3

This is of the form ax^{2} + bx + c with a = 4 b = -8 c = 3

Product ac = 4 × 3 = 12

Sum b = -8

Question 3.

Factorize the following expressions using a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2}) identity

(i) h^{3} + k^{2}

(ii) 2a^{3} + 16

(iii) x^{3}y^{3} + 27

(iv) 64m^{3} + n^{3}

(v) r^{4} + 27p^{3}r

Solution:

(i) h^{3} + k^{3}

Comparing h^{3} + k^{3} with a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2}) we have a = h, b = k

∴ h^{3} + k^{3} = (h + k)(h^{2} – hk + k^{2})

(ii) 2a^{2} + 16

(2 × a^{3}) + (2 × 8) = 2(a^{3} + 8) = 2 (a^{3} + 2^{3})

∴ 2a^{3} + 16 = 2 (a^{3} + 2^{3})

Comparing with a^{3} + b^{3} we have a = a and b = 2

a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

2(a^{3} + 2^{3}) = 2[(a + 2) (a^{2} – (a) (2) + 2^{2})] = 2[(a + 2) (a^{2} – 2a + 4)]

2a^{3} +16 = 2 (a + 2) (a^{2} – 2a + 4)

(iii) x^{3} y^{3} + 27 = (xy)^{3} + 3^{3}

Comparing with a^{3} + b^{3} we have a = xy ;b = 3

a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})

(xy)^{3} + 3^{3} = (xy + 3) ((xy)^{2} – (xy) (3) + 3^{2}) = (xy + 3) (x^{2}y^{2} – 3xy + 9)

∴ x^{3}y^{3} + 27 = (xy + 3)(x^{2}y^{2} – 3xy + 9)

(iv) 64m^{3} + n^{3} = (4^{3}m^{3}) + n^{3}

= (4m)^{3} + n^{3}

Comparing this with a^{3} + b^{3} we have a = 4m; b = n

a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})

(4m)^{3} + n^{3} = (4m + n) [(4m)2 – (4m) (n) + n2]

= (4m + n) [4^{2}m^{2} – 4mn + n^{2}]

= (4m + n) [ 16m^{2} – 4mn + n^{2}]

64m^{3} + n^{3} = (4m + n) (16m^{2} – 4mn + n^{2})

(v) r^{3} + 27p^{3}r = r (r^{3} + 27p^{3}) = r [r^{3} + 3^{3}p^{3}]

Comparing r^{3} + (3p)^{3} with a^{3} + b^{3} we have a = r; b = 3p

a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

r[r^{3} + (3p)^{3}] = r[(r + 3p)(r^{2} – r(3p) + (3p)^{2})]

= r[(r+ 3p) (r^{2} – 3rp + 3^{2}p^{2}]

= r(r + 3p) (r^{2} – 3rp + 9p^{2})

r^{4} + 27p^{3}r = r(r + 3p) (r^{2} – 3rp + 9p^{2})

Question 4.

Factorize the following expressions using a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2}) identity

(i) y^{3} – 27

(ii) 3b^{3} – 192c^{3}

(iii) -16y^{3} + 2x^{3}

(iv) x^{3} y^{3} – 7^{3}

(v) c^{3} – 27b^{3} a^{3}

Solution:

(i) y^{3} – 27 = y^{3} – 3^{3}

Comparing this with a^{3} – b^{3} , we have a = y and b = 3

a^{3} – b^{3} = (a + b) (a^{2} + ab + b^{2} )

y^{3} – 3^{3} = (y – 3)(y^{2} + (y) (3) + 3^{2} ) = (y – 3) (y^{2} + 3y + 9)

y^{3} – 27 = (y – 3) (y^{2} + 3y + 9)

(ii) 3b^{3} + 192c^{3} = (3 × b^{3}) – (3 × 4 × 4 × 4 × c^{3}) = 3(b^{3} – 4^{3} c^{3})

= 3(b^{3} – (4c)^{3} )

Comparing b^{3} – (4c)^{3} with a^{3} – b^{3} we have a = b and b = 4c

a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

3(b^{3} – (4c)^{3}) = 3[(b – 4c)(b^{2} + (b)(4c) + (4c)^{2})]

= 3[(b – 4c)(b^{2} + 4bc + 4^{2} c^{2})]

3b^{3} – 192c^{3} = 3 [(b – 4c) (b^{2} + 4bc + 16c^{2})]

(iii) -16y^{3} + 2x^{3} = 2x^{3} – 16y^{3} [∵ Addition is commatative]

= 2(x^{3} – 8y^{3}) = 2(x^{3} – 23y^{3})

= 2(x^{3} – (2y)^{3})

Comparing x^{3} – (2y)^{3} with a^{3} – b^{3} we have a = x and b = 2y

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})

2[x^{3} – (2y)^{3}] = 2[(x – 2y) (x^{2} + (x) (2y) + (2y)^{2})]

= 2[(x – 2y) (x^{2} + 2xy + 2^{2} y^{2})]

-16y^{3} + 2x^{3} = 2[(x – 2y)(x^{2} + 2xy + 4y^{2})]

(iv) x^{3}y^{3} – 7^{3} = (xy)^{3} – 7^{3}

Comparing with a^{3} – b^{3} we have a = xy and b = 7

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})

(xy)^{3} – 7^{3} = (xy – 7) ((xy)^{2} + (xy) (7) + 7^{2})

x^{3}y^{3} – 7^{3} = (xy – 7) (x^{2}y^{2} + 7xy + 49)

(v) c^{3} – 27 b^{3} a^{3} = c^{3} – 3^{3}b^{3} a^{3} = c^{3} – (3ba)^{3}

Comparing this with a^{3} – b^{3} we have a = x and b = 3ba

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})

∴ c^{3} – (3ba)^{3} = (c – 3ba) (c^{2} + (c) (3ba) + (3ba)^{2})

= (c – 3ba) (c^{2} + 3bac + 3^{2} b^{2}a^{2})

c^{3} – 27b^{3}a^{3} = (c – 3ab) (c^{2} + 3bac + 9a^{2}b^{2})

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