Advertisements
Advertisements
प्रश्न
Factorise: 5 - (3a2 - 2a) (6 - 3a2 + 2a)
Advertisements
उत्तर
5 - (3a2 - 2a) (6 - 3a2 + 2a)
= 5 - (3a2 - 2a)[6 - (3a2 - 2a)]
Assume that 3a2 - 2a = x
Therefore,
5 - (3a2 - 2a)(6 - 3a2 + 2a)
= 5 - x(6 - x)
= 5 - 6x + x2
= x2 - (5 + 1)x + 5
= x2 - 5x - 1x + 5
= x(x - 5) -1 (x - 5)
= (x - 5) (x - 1)
put x = 3a2 - 2a
= (3a2 - 2a - 5) (3a2 - 2a - 1)
= (3a2 - 5a + 3a - 5) (3a2 - 3a + a - 1)
= [a(3a - 5) + 1(3a - 5)] [3a(a - 1) + 1(a - 1)]
= (3a - 5) (a + 1) (3a + 1) (a - 1)
APPEARS IN
संबंधित प्रश्न
Factorise.
p2 − 7p − 44
Factorise.
2x2 + x − 45
Factorise.
44x2 − x − 3
Factorise:
6a2 - a - 15
Factorise : 3 - a (4 + 7a)
Factorise the following by splitting the middle term:
x2 - 11x + 24
Factorise the following:
y2 + 3y + 2 + by + 2b
Factorise the following:
5(3x + y)2 + 6(3x + y) - 8
Factorise the following:
12 - (y + y2)(8 - y - y2)
Factorise the following:
6 - 5x + 5y + (x - y)2
