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Question

Question 31

The value of (3x3+9x2+27x)÷3x is
a) x2+9+27x
b) 3x3+3x2+27x
c) 3x3+9x2+9
d) x2+3x+9

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Solution

d) x2+3x+9

​​​​​​​We have,
(3x3+9x2+27x)÷3x=3x3+9x2+27x3x=3x33x+9x23x+27x3x=x2+3x+9

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