Difference between revisions of "1953 AHSME Problems/Problem 4"

Latest revision as of 18:58, 20 April 2020

The roots of $x(x^2+8x+16)(4-x)=0$ are:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}$

We factor the middle part into $(x+4)^2$ .

The equation now becomes $x(x+4)^2(4-x)=0$

The solutions are then $0, -4, 4$ . So the answer is $\boxed{\text{C}}$

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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 4: / Line 4: @@
 ==Solution==
-We can divide both sides by <math>-1</math> to get the equation:
-<math>x(x^2+8x+16)(x-4)=0</math>
 We factor the middle part into <math>(x+4)^2</math>.
-The equation now becomes <math>x(x+4)^2(x-4)=0</math>
+The equation now becomes <math>x(x+4)^2(4-x)=0</math>
 The solutions are then <math>0, -4, 4</math>.  So the answer is <math>\boxed{\text{C}}</math>