Question

1 .On factorizing (3a²bc + 9ab²c + 21abc?), we get
a) 3abc(3a+3b+7c)
b) 3abc(a+b+c)
c)abc(a+3b+7c)
d)3abc(a+3b+7c)​

Answer 1

Answer:

3abc(a+3b+7c)

Step-by-step explanation:

Just take the common out of the three terms

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