Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5A Solutions
EXERCISE – 5A
(Q1) Factorise by taking out the common factors.
(1) 2 (2x – 5y) (3x + 4y) – 6 (2x – 5y) (x – y)
Solution:
2 (2x – 5y) (3x + 4y) – 6 (2x – 5y) (x – y)
=> Firstly common out the common factor.
=> (2x – 5y) [2 (3x + 4y) – 6 (x – y)]
=> (2x – 5y) [6x + 8y – 6x + 6y]
=> (2x – 5y) [14y]
=> 28xy – 70y
(2) xy(3x2 – 2y2) – yz (2y2 – 3x2) + zx(15x2 – 10y2)
Solution:
xy (3x2 – 2y2) – yz (2y2 – 3x2) + 2x (15x2 – 10y2)
Firstly common out the common factor
=> xy (3x2 – 2y2) – yz (- (3x2 – 2y2)) + 2x (15x2 – 10)
=> xy (3x2 – 2y2) + yz (3x2 – 2y2) + 5zx (3x2 – 2y2)
=> (3x2– 2y2) [xy + yz + 5zx]
(3) ab (a2 + b2 – c2) – bc (c2 – a2 – b2) + Ca (a2 + b2 – c2)
Firstly common out the common factors.
=> ab (a2 + b2 – c2) – bc (- (a2 + b2 – c2)) + ca (a2 + b2 – c2)
=> ab (a2 + b2 – c2) + bc (a2 + b2 – c2) + ca (a2 + b2 – c2)
=> (a2 + b2 – c2) [ab + bc + ca]
(4) 2x (a – b) + 3y (5a – 5b) + 4z(2b – 2a)
Solution:
2x (a – b) + 3y (5a – 5b) + 4z (2b – 2a)
Firstly common out the common factors
=> 2x (a – b) + 3y × 5 (a – b) + 4z × 2 (b-a)
=> 2x (a – b) + 15y (a – b) – 8z (a – b)
=> (a – b) [2a + 15y – 8z]
(5) Factorize by the grouping method:
a3 + a – 3a2 – 3
Solution:
a3 + a – 3a2 – 3
=> a (a2 + 1 – 3a – 3)
=> a (a2 – 3a – 2)
Factorize a2 – 3a – 2 = 0
a2 – 2a – a – 2 = 0
=> a (a – 2) + 1 (a – 2) = 0
= c
Solution:
a3 + a – 3a2 – 3
First we have to take common,
a (a2 + 1) – 3 (a2 + 1)
=> Again we have take common
(a2 + 1) (a – 3)
(6) 16 (a + b)2 – 4a – 4b
Solution:
16 (a + b)2 – 4a – 4b
=> 16 (a + b)2 – 4 (a + b)
We have to take common term
=> 16 (a + b)2 – 4 (a + b)
=> (a + b) {(a + b) – 4}
=> 4 (a + b) {4 (a + b) – 1}
(7) Factorize by the grouping method:
a4 – 2a3 – 4a + 8
Solution:
a4 – 2a3 – 4a + 8
First we have to take common
=> a (a – 2) – 4 (a – 2)
Again we have to take common
=> (a – 2) {a3 – 4}
(8) ab – 2b + a2 – 2a
Solution:
ab – 2b + a2 – 2a
First we have to take common.
=> b (a – 2) + a (a – 2)
=> Again we have to take common.
(9) ab (x2 + 1) + × (a2 + b2)
Solution:
ab (x2 + 1) + x (a2 + b2)
=> abx2 + ab + xa2 + ab2
=> abx2 + xa2 + xb2 + ab
First we have to take common term
=> ax (bx + a) + b (xb + a)
Again we have to take common term
=> (bx + a) (ax + b)
(10) a2 + b – ab – a
Solution:
a2 + b – ab – a
=> a2 – a + b – ab
First we have to take common
=> a (a – 1) – b (a – 1)
=> Again we have to take common (a – 1) (a – b)
(11) (ax + by)2 + (bx – ay)2
Solution:
(ax + by)2 + (bx – ay)2
First we have to expand –
Now, we have to use –
(a + b) = a2 + 2ab + b2
∴ (ax + by)2 = (ax)2 + 2 × (ax) × (by) + (by)2
= a2 x2 + 2abxy + b2y2
∴ (bx – ay)2 = (bx)2 – 2 × (bx) (ay) + (ay)2
= b2x2 – 2baxy + a2y2
=> a2x2 + 2abxy + b2y2 + b2x2 – 2abxy + a2y2
=> a2x2 + b2y2 + b2x2 + a2y2
=> a2x2 + a2y2+ b2y2 + b2x2
=> a2 (x2 + y2) + b2 (y2+ x2)
=> (x2 + y2) (a2 + b2)
(12) a2 x2 + (ax2 + 1) × a
Solution:
a2x2 + (ax2 + 1) x + a
=> a2x2 + ax3 + x + a
=> a2x2 + a + ax3 + x
= a (ax2 + 1) + x (ax2 + 1)
= (ax2 + 1) (a + x)
(14) a (a – 4) – a + 4
Solution:
a (a – 4) – a + 4
= a (a – 4) – 1 (a – 4)
= (a – 4) (a – 1)
(15) y2 – (a + b) y + ab
Solution:
y2 – (a + b) y + ab
=> y2 – (ay + by) + ab
=> y2 – ay – by + ab
=> y (y – a) – b (y – a)
=> (y – a) (y – b)
(17) x2 + y2 + x + y + 2xy
Solution:
x2 + y2 + x + y + 2xy
= (x2 + y2 + 2xy) + x + y
= (x + y)2 + (x + y)
= (x + y) (x + y + 1)
(18) a2 + 4b2 – 3a + 6b – 4ab
Solution:
a2 + 4b2 – 3a + 6b – 4ab
=> a2 + 4b2 – 4ab – 3a + 6b
=> (a – 2b)2 – 3a + 6b
=> (a – 2b)2 – 3(a – 2b)
=> (a – 2b) [a – 2b – 3]
(19) m (x – 3y)2 + n (3y – x) + 5x – 15y
Solution:
m (x – 3y)2 + n (3y – x) + 5x – 15y
= m (x – 3y)2 – n (x – 3y) + 5x – 15y
= m (x – 3y)2 – n (x – 3y) + 5 (x – 3y)
= (x – 3y) [m (x – 3y) – n + 5]
= (x – 3y) [mx – 3m – n + 5]
(20) x (6x – 5y) – 4 (6x – 5y)2
Solution:
x (6x – 5y) – 4 (6x – 5y)2
= (6x – 5y) [x – 4 (6x – 5y)]
= (6x – 5y) [x – 24x + 20y]
= (6x – 5y) [-20x + 20y]
Here is your solution of Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5A
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