Is my text book messing wih me again?

AI Thread Summary
The discussion focuses on determining the coefficient of x^6 in the expansion of the expression (2-3x^2)(1+2x)^8. The correct approach involves calculating contributions from both terms: the first term directly provides the coefficient of x^6, while the second term requires finding the coefficient of x^4 in the expansion due to the x^2 multiplier. The calculations yield a final coefficient of 244 for x^6. The confusion arises from understanding how the terms contribute to the overall coefficient. The clarification highlights the importance of correctly identifying the relevant terms in polynomial expansions.
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Here is an example in my book...

Determine the coefficient of x^6 in the expansion of

(2-3x^2)(1+2x)^8


(2-3x^2)(1+2x)^8 = 2(1+2x)^8 - 3x^2(1+2x)^8

The general term in the expansion of (1+2x)^8 is

C(8,k)(2x)^k

Therefore, the term containing x^6 in the expansion of (2-3x^2)(1+2x)^8 is

2(C(8,6))(2x)^6 - 3x^2(C(8,4))(2x)^4 = 2(2^6)(C(8,6))(x^6) - 3(2^4)(C(8,4))(x^6) = 224x^6

The coefficient is 244

Where the heck did these 4's come from? Shouldn't they be 6's or am I missing something?
 
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You're looking for the coefficient of x^6. In the first term, this will be equal to the coefficient of x^6, but the second term is premultiplied by x^2, thus to obtain the coefficient of x^6 in this term, you must look at the term x^4 in the expansion of (1+2x)^8 [since x^2.x^4=x^6]
 
Saved again, thanks Cristo
 
You're welcome.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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