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Bunuel
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x + x^2 + x^3+1 = 28
X( x + x^2 + x^3+1) = 28x
(x + x^2 + x^3+x^4)4 = 28x/4=7x
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I did it this way:
x + x^2 + x^3 + x^4 = x + (x + x^2 + x^3) * x = x + 27x = 28x
the arithm mean is 28x/4, ie 7x.

Answer is C.
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Bunuel
If x + x^2 + x^3 = 27, then the average (arithmetic mean) of x + x^2 + x^3 + x^4 is equal to which of the following?

A. 3
B. 3x
C. 7x
D. 9
E. 9x

We need to determine the value of (x + x^2 + x^3 + x^4)/4.

We can manipulate x + x^2 + x^3 = 27 by multiplying the entire equation by x and we have:

x^2 + x^3 + x^4 = 27x

Now we have:

(x + x^2 + x^3 + x^4)/4 = (x + 27x)/4 = 28x/4 = 7x

Answer: C
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Look at the question, answer is hidden.
x + x^2 + x^3 + x^4
= x(1+x+x^2+x^3)
= x(1+27) [it is given that x+x^2+x^3 = 27]
=28x
Sum is 28x of 4 numbers. So average is 28x/4 = 7x.
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x + \(x^{2}\) + \(x^{3}\) = 27

Average of = x , \(x^{2}\) , \(x^{3}\) ,\(x^{3}\) , \(x^{4}\)

==>(x + \(x^{2}\) + \(x^{3}\) +\(x^{3}\) + \(x^{4}\) ) /4

==> (x + x(x + \(x^{2}\) + \(x^{3}\) )) / 4

==> \(\frac{x + 27x}{4}\)

==> \(\frac{28x}{4}\) = 7x
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I don't find the question very well formulated. The average of x + x^2 + x^3 + x^4 is just 28x (because it is only 1 number, the average is itself). The average of x, x^2, x^3 and x^4 is 7x.
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x+x^2+x^3 = 27

(x+x^2+x^3+x^4)/4 = y
x(1+x+x^2+x^3)/4 = y
x28/4 = y
7x=y
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The explanation was good here
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The question language is a bit wrong. It should say the arithmetic mean of x, x^2, x^3 and x^4, not the arithmetic mean of x + x^2 + x^3 + x^4. The latter term means just one term, not 4.
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