Advertisements
Advertisements
Question
Factorise the following by grouping the terms:
2a + b + 3c - d + (2a + b)3 + (2a + b)2(3c - d)
Advertisements
Solution
2a + b + 3c - d + (2a + b)3 + (2a + b)2(3c - d)
= (2a + b + 3c - d) + [(2a + b)3 + (2a + b)2(3c - d)]
= 1(2a + b + 3c - d) + (2a + b)2(2a + b + 3c - d)
= (2a + b + 3c - d)[1 + (2a + b)2].
APPEARS IN
RELATED QUESTIONS
Factorise by the grouping method : a2 + b - ab - a
Factorise by the grouping method : (ax + by)2 + (bx - ay)2
Factorise by the grouping method : (2a-b)2 -10a + 5b
Factorise by the grouping method : y2 - (a + b) y + ab
Factorise using the grouping method :
x (6x - 5y) - 4 (6x - 5y)2
factorise: x2 + 4x + 3
factorise: 2a2 - 17ab + 26b2
Factorise: 2(x + 2y)2 − 5(x + 2y) + 2
Factorise the following by grouping the terms:
4abx2 + 49aby2 + 14xy(a2 + b2)
Factorise the following by grouping the terms:
9x3 + 6x2y2 - 4y3 - 6xy
