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

Latest revision as of 23:51, 11 October 2020

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

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

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 $\boxed{\mathrm{(B)}\ 5}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
-== Problem==
+== Problem ==
-The formula which expresses the relationship between <math>x</math> and <math>y</math> as shown in the accompanying table is:
+The number of terms in the expansion of <math> [(a+3b)^{2}(a-3b)^{2}]^{2} </math> when simplified is:
-<cmath> \begin{tabular}[t]{|c|c|c|c|c|c|}\hline x&0&1&2&3&4\\\hline y&100&90&70&40&0\\\hline\end{tabular} </cmath>
+<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
-<math> \textbf{(A)}\ y=100-10x\qquad\textbf{(B)}\ y=100-5x^{2}\qquad\textbf{(C)}\ y=100-5x-5x^{2}\qquad\\ \textbf{(D)}\ y=20-x-x^{2}\qquad\textbf{(E)}\ \text{None of these} </math>
+== Solution ==
-==Solution==
+Use properties of exponents to move the squares outside the brackets use difference of squares.
-Plug in the points <math>(0,100)</math> and <math>(4,0)</math> into each equation. The only one that works for both points is <math>\mathrm{(C)}.</math> Plug in the rest of the points to confirm the answer is indeed <math>\boxed{\mathrm{(C)}\ y=100-5x-5x^2.}</math>
+<cmath>[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4</cmath>
-==See Also==
+Using the binomial theorem, we can see that the number of terms is <math>\boxed{\mathrm{(B)}\ 5}</math>.
-{{AHSME box|year=1950|num-b=15|num-a=17}}
+== See Also ==
+{{AHSME 50p box|year=1950|num-b=15|num-a=17}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}