Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6
Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6
Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6
9th Maths Exercise 3.6 Question 1.
Factorise the following:
(i) x2 + 10x + 24
(ii) z2 + 4z – 12
(iii) p2 – 6p – 16
(iv) t2 + 72 – 17t
(v) y2 – 16y – 80
(vi) a2 + 10a – 600
Solution:
(i) x2 + 10x + 24
x2 + 10x + 24 = x2 + 6x + 4x + 24
= x(x + 6) + 4 (x + 6)
= (x + 6) (x + 4)
(ii) z2 + 4z – 12
z2 + 4z – 12 = z2 + 6z – 2z- 12
= z (z + 6) – 2 (z + 6)
= (z + 6) (z – 2)
(iii) p2 – 6p – 16
p2 – 6p – 16 = p2 – 8p + 2p – 16
= p(p – 8) + 2(p – 8)
= (p – 8)(p + 2)
(iv) t2 + 72 – 17t
t2 + 72 – 17t = t2 – 17t + 72
= t2 – 9t – 8t + 72
= t(t – 9) – 8 (t – 9)
= (t – 9) (t – 8)
(v) y2 – 16y – 80
y2 – 16y – 80 = y2 – 20y + 4y – 80
= y(y – 20) + 4 (y – 20)
= (y – 20) (y + 4)
(vi) a2 + 10a – 600
a2 + 10a – 600 = a2 + 30a – 20a – 600
= a(a + 30) -20 (a + 30)
= (a + 30) (a – 20)
Exercise 3.6 Class 9 Question 2.
Factorise the following
(i) 2a2 + 9a + 10
(ii) 5x2 – 29xy – 42y2
(iii) 9 – 18x + 18x2
(iv) 6x2 + 16xy + 8y2
(v) 12x2 + 36x2y + 27y2x2
(vi) (a + b)2 + 9 (a + b) + 18
Solution:
(i) 2a2 + 9a + 10
2a2 + 9a + 10 = 2a2 + 4a + 5a + 10
= 2a(a + 2) + 5 (a + 2)
= (a+ 2) (2a+ 5)
(ii) 5x2 – 29xy – 42y2
5x2 – 35xy + 6xy – 42y2
= 5x (x – 7) + 6y (x – 7)
= (x – 7) (5x + 6y)
(iii) 9 – 18x + 8x2
= 8x2 – 18x + 9
= 8x2 – 6x – 12x + 9
= 2x (4 x – 3) – 3 (4x – 3)
= (4x – 3) (2x – 3)
(iv) 6x2 + 16xy + 8y2
= 2 (3x2 + 8xy + 4y2)
= 2 (3x2 + 8xy + 4y2)
= 2 (3x2 + 6xy + 2xy + 4y2)
= 2 (3x (x + 2y) + 2y (x + 2y))
= 2 (x + 2y) (3x + 2y)
(v) 12x2 + 36x2y + 27y2x2
= 27y2x2 + 36x2y + 12x2 = 3x2(9y2 + 12y + 4)
= 3x2 (9y2 + 6y + 6y + 4) = 3x2 (3y (3y + 2) + 2 (3y + 2))
= 3x2 (3y + 2) (3y + 2) = 3x2 (3y + 2) (3y + 2)
(vi) (a + b)2 + 9 (a + 6) + 18
= (a + b)2 + 6 (a + b) + 3 (a + b) + 18
= (a + b) ((a + b) + 6) + 3 ((a + b) + 6)
= ((a + 6) + 6) ((a + b) + 3) = (a + b + 6) (a + b + 3)
Ex 3.6 Class 9 Question 3.
Factorise the following:
(i) (p – q)2 – 6(p – q) – 16
(ii) m2 + 2mn – 24n2
(iii) √5a2 + 2a – 3√5
(iv) a4 – 3a2 + 2
(v) 8m3 – 2m2n – 15mn2
(vi) 1/x2+1y2+2/xy
Solution:
(i) (p – q)2 – 6 (p – q) – 16
= (p – q)2 – 8(p – q) + 2(p – q) – 16
= (p – q)((p – q) – 8) + 2((p – q) – 8)
= (p – q – 8)(p – q + 2)
(ii) m2 + 2mn – 24n2
= m2 + 6mn – 4mn – 24n2
= m(m + 6n) – 4n(m + 6n)
= (m + 6n)(m – 4n)
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