Factorise : 2ab2c - 2a + 3b3c - 3b - 4b2c2 + 4c

Factorise : 2ab²c - 2a + 3b³c - 3b - 4b²c² + 4c

Sum

2ab²c - 2a + 3b³c - 3b - 4b²c² + 4c

= 2a (b²c - 1) + 3b (b²c - 1) - 4c (b²c - 1)

= (b²c - 1) (2a + 3b - 4c)

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Factorisation by Grouping

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Chapter 13: Factorisation - Exercise 13 (B) [Page 158]

Chapter 13 Factorisation
Exercise 13 (B) | Q 20 | Page 158

Factorise by the grouping method : a⁴ - 2a³ - 4a + 8

Factorise by the grouping method :
`a^2 + 1/a^2 - 2 - 3a + 3/a`

factorise: x² + 6x + 8

factorise: 6 + 7b - 3b²

factorise: 5 + 3a - 14a²

factorise: a² - 23a -108

factorise: b(c - d)² + a(d - c) + 3(c - d)

factorise: ax² + b²y - ab² - x²y

Factorise the following by grouping the terms:
3ax² - 5bx² + 9az² + 6ay² - 10by² - 15bz²

Factorise the following by grouping the terms:

`"p"^2 + (1)/"p"^2 - 2 - 5"p" + (5)/"p"`