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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6

Question 1.

Factorise the following.

(i) x² + 10x + 24

Solution:

Product = 24, sum = 10

Split the middle term as 6x and 4x

x² + 10x + 24 = x² + 6x + 4x + 24

= x(x + 6) + 4 (x + 6)

= (x + 6) (x + 4)

(ii) z² + 4z – 12

Solution:

Product = -12, sum = 4

Split the middle term as 6z and -2z

z² + 4z – 12 = z² + 6z – 2z – 12

= z(z + 6) – 2 (z + 6)

= (z + 6) (z – 2)

(iii) p² – 6p – 16

Solution:

Product = -16, sum = -6

Split the middle term as – 8p and 2p

p² – 6p – 16 = p² – 8p + 2p – 16

= p(p – 8) + 2 (p – 8)

= (p – 8) (p + 2)

(iv) t² + 72 – 17t

Solution:

Product = +72, sum = -17

Split the middle term as -9t and -8t

t² – 17t + 72 = t² – 91 – 8t + 72

= t(t – 9) – 8 (l – 9)

= (t – 9) (t – 8)

(v) y² – 16y – 80

Solution:

Product = -80, sum = -16

Split the middle term as -20y and 4y

y² – 16y – 80 = y² – 20y + Ay – 80

= y(y – 20) + 4 (y – 20)

= (y – 20) (y + 4)

(vi) a² + 10a – 600

Solution:

Product = -600, sum =10

Split the middle term as 30a and -20a

a² + 10a – 600 = a² + 30a – 20a – 600

= a(a + 30) – 20 (a + 30)

= (a + 30) (a – 20)

Question 2.

Factorise the following.

(i) 2a² + 9a + 10

Solution:

Product = 2 × 10 = 20, sum = 9

Split the middle term as 5a and 4a

2a² + 9a + 10 = 2a² + 5a + 4a + 10

= a(2a + 5) + 2 (2a + 5)

= (2a+ 5) (a+ 2)

(ii) 5x² – 29xy – 42y²

Solution:

Product = 5 × -42 = -210, sum = -29

Split the middle term as -35x and 6x

5x² – 29xy – 42y² = 5x² – 35xy + 6xy – 42y²

= 5x (x – 7y) + 6y (x – 7y)

= (x – 7y) (5x + 6y)

(iii) 9 – 18x + 8x²

Solution:

Product = 9 × 8 = 72, sum = -18

Split the middle term as -12x and -6x

9 – 18x + 8x² = 8x² – 18x + 9

= 8x² – 12x – 6x + 9

= 4x (2x – 3) – 3 (2x – 3)

= (2x – 3) (4x – 3)

(iv) 6x² + 16xy + 8y²

Solution:

Product = 6 × 8 = 48, sum = 16

Split the middle term as 4xy and 12xy

6x² + 16xy + 8y² = 6x² + 12xy + 4xy + 8y²

= 6x (x + 2y) + 4y(x + 2y)

= (x + 2y) (6x + 4y)

= 2(x + 2y) (3x + 2y)

(v) 12x² + 36x²y + 27y²x²

Solution:

3x²2 [4 + 12y + 9y²]

= 3x² [9y² + 12y + 4]

Product = 9 x 4 = 36, sum =12

Split the middle term as 6y and 6y

12x² + 36x²y + 21y²x² = 3x² [9y² + 12y + 4]

= 3x² [9y² + 6y + 6y + 4]

= 3x² [3y(3y + 2) + 2(3y + 2)]

= 3x² (3y + 2) (3y + 2)

= 3x² (3y + 2)2

(vi) (a + b)² + 9 (a + b) + 18

Solution:

Let (a + b) = x

x² + 9x + 18

Product =18, sum = 9

Split the middle term as 6x and 3x

x² + 9x + 18 = x² + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

But x = a + b

(a + b)² + 9(a + b) + 18 = (a + b + 6) (a + b + 3)

Question 3.

Factorise the following.

(i) (p – q)² – 6(p – q) – 16

Solution:

Let (p – q) = x

(p – q)² – 6 (p – q) – 16 = x² – 6x – 16

Product = -16, sum = -6

Split the middle term as -8x and 2x

x² – 6x – 16 = x² – 8x + 2x – 16

= x(x – 8) + 2(x – 8)

= (x – 8) (x + 2)

(But x = p – q)

= (p – q – 8) (p – q + 2)

(ii) m² + 2mn – 24n²

Solution:

Product = -24, sum = 2

Split the middle term as 6mn and -4mn

m² + 2mn – 24m² = m² + 6mn – 4mn – 24n²

= m(m + 6n) – 4n (m + 6n)

= (m + 6n) (m – 4n)

(iii) √5 a² + 2a – 3√5?

Solution:

Product = √5 × – 3√5 = -15, sum = 2

Split the middle term as 5x and -3x

√5 a² + 2a – 3√5 = √5a² + 5a – 3a – 3√5

= √5 a(a + √5) – 3(a + √5)

= (a + √5) (√5a – 3)

(iv) a^{4} – 3a² + 2

Solution:

Let a² = x

a^{4} – 3a² + 2 = (a²)² – 3a² + 2

= x² – 3x + 2

Product = 2 and sum = -3

Split the middle term as -x and -2x

x² – 3x + 2 = x² – x – 2x + 2

= x(x – 1) – 2(x – 1)

= (x – 1) (x – 2)

a^{4} – 3a² + 2 = (a^{2} – 1)(a^{2} – 2) [But a^{2} = x]

= (a + 1) (a – 1) (a^{2} – 2)

(v) 8m^{3} – 2m^{2}n – 15mn^{2}

Solution:

8m^{3} – 2m^{2}n – 15mn^{2} = m(8m^{2} – 2mn – 15n^{2})

Product = 8(-15) = -120 and sum = -2

Split the middle term as -12mn and 10mn

8m^{3} – 2m^{2}n – 15mn^{2} = m[8m^{2} – 2mn – 15n^{2}]

= m[8m^{2}– 12mn + 10mn- 15n^{2}]

= m[4m (2m – 3n) + 5n(2m – 3n)]

= m(2m – 3n) (4m + 5n)

(vi) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}\)

Solution: