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
x3−1039​x2+1029​x−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