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1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get : (a) 3abc (3a + 3b + 7c) (b) 3abc (a + b + c) (c) abc (a + 3b + 7c) (d) 3abc (a + 3b + 7c)

1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get : (a) 3abc (3a + 3b + 7c) (b) 3abc (a + b + c) (c) abc (a + 3b + 7c) (d) 3abc (a + 3b + 7c)
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1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get :

(a) 3abc (3a + 3b + 7c)
(b) 3abc (a + b + c)
(c) abc (a + 3b + 7c)
(d) 3abc (a + 3b + 7c)

asked
User Fsbflavio
by
8.1k points

2 Answers

2 votes

Answer:

Fast

Explanation:

You told me "Please answer fast." ;-;

answered
User Erick R Soto
by
7.9k points
6 votes
actually it’s none of these.

24a^2bc+9ab2c
48abc+9ab2c

answer: 3abc ^ (16+3b)
answered
User GangstaGraham
by
7.9k points
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