Problem:
The roots of x3+2x2−x+3 are p,q, and r. What is the value of
(p2+4)(q2+4)(r2+4)?
Answer Choices:
A. 64
B. 75
C. 100
D. 125
E. 144
Solution:
By Vieta's Formulas,
p+q+rpq+pr+qrpqr=−2=−1, and =−3
The expression to be evaluated is
(p2+4)(q2+4)(r2+4)=p2q2r2+4(p2q2+p2r2+q2r2)+16(p2+q2+r2)+64
The first term is
p2q2r2=(pqr)2=(−3)2=9
To find the second term, note that
(pq+pr+qr)2=p2q2+p2r2+q2r2+2pqr(p+q+r)
Rearranging and substituting yields
p2q2+p2r2+q2r2=(−1)2−2(−3)(−2)=−11
For the third term,
(p+q+r)2=p2+q2+r2+2(pq+pr+qr)
so
p2+q2+r2=(−2)2−2(−1)=6
Therefore
(p2+4)(q2+4)(r2+4)=9+4⋅(−11)+16⋅6+64=125
With i=−1, the given expression can be factored as
(p2+4)(q2+4)(r2+4)=(p−2i)(p+2i)(q−2i)(q+2i)(r−2i)(r+2i)=[(p−2i)(q−2i)(r−2i)]⋅[(p+2i)(q+2i)(r+2i)]
Let f(x) be the given polynomial. Then f(x)=−(p−x)(q−x)(r−x) and
−f(2i)=(p−2i)(q−2i)(r−2i)
which is the first bracketed expression above. Furthermore
f(−x)=(−x−p)(−x−q)(−x−r)=−(p+x)(q+x)(r+x)
Therefore
−f(−x)=(p+x)(q+x)(r+x)
It follows that
−f(−2i)=(p+2i)(q+2i)(r+2i)
the second bracketed expression above. Hence
(p2+4)(q2+4)(r2+4)=−f(2i)⋅(−f(−2i))=f(2i)⋅f(−2i).
Because
f(2i)=(2i)3+2(2i)2−(2i)+3=−8i−8−2i+3=−10i−5
and
f(−2i)=(−2i)3+2(−2i)2−(−2i)+3=8i−8+2i+3=10i−5,
the requested product is
f(2i)⋅f(−2i)=(−10i−5)(10i−5)=(D)125.
Note: The roots of the polynomial are approximately −2.757,0.379+0.972i, and 0.379−0.972i.
The problems on this page are the property of the MAA's American Mathematics Competitions