Problem:
The roots of the polynomial 10x3−39x2+29x−6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 2 units. What is the volume of the new box?
Answer Choices:
A. 524
B. 542
C. 581
D. 30
E. 48
Solution:
Let f(x)=10x3−39x2+29x−6 and let a,b, and c be the roots of f(x). Hence f(x)=10(x−a)(x−b)(x−c). Then the volume of the new box is
(2+a)(2+b)(2+c)=−(−2−a)(−2−b)(−2−c)=−10f(−2)=−101(−80−156−58−6)=(D)30
OR
Dividing by 10 gives the polynomial
x3−1039x2+1029x−53
whose roots are the same. If a,b, and c are the roots, then Vieta's Formulas give
abc=53,ab+bc+ac=1029, and a+b+c=1039
The volume of the new box is
(a+2)(b+2)(c+2)=abc+2(ab+bc+ac)+4(a+b+c)+8
Substituting gives
53+2⋅1029+4⋅1039+8=(D)30
Note: The solutions presented here did not require finding the roots. In fact, the roots of f(x) are 3,21, and 52. The dimensions of the new box are 5,25, and 512, which gives a volume of 5⋅25⋅512=30.
The problems on this page are the property of the MAA's American Mathematics Competitions