Difference between revisions of "1950 AHSME Problems/Problem 16"

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== Problem==
 
== Problem==
  
The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{4} </math> when simplified is:
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The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:
  
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>

Revision as of 20:59, 26 September 2019

Problem

The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

Use properties of exponents to move the squares outside the brackets use difference of squares.

\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]

Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5.}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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