Advertisements
Advertisements
प्रश्न
Factorise : 2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c
योग
Advertisements
उत्तर
2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c
= 2a (b2c - 1) + 3b (b2c - 1) - 4c (b2c - 1)
= (b2c - 1) (2a + 3b - 4c)
shaalaa.com
Factorisation by Grouping
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Factorise by the grouping method : ab - 2b + a2 - 2a
Factorise : a3 -a2 +a -1
Factorise : x2 - (b - 2)x - 2b
factorise: x2 + 4x + 3
factorise: x2 + 5xy + 4y2
factorise: 2x2 + xy - 6y2
factorise: x2y2 - 3xy - 40
factorise: 3a2x - bx + 3a2 - b
factorise: b(c - d)2 + a(d - c) + 3(c - d)
Factorise the following by grouping the terms:
8(2a + b)2 - 8a -4b
