Problem:
For some real number r, the polynomial 8x3−4x2−42x+45 is divisible by (x−r)2. Which of the following numbers is closest to r?
Answer Choices:
A. 1.22
B. 1.32
C. 1.42
D. 1.52
E. 1.62
Solution:
Since the given polynomial is divisible by (x−r)2 the remainders when it is divided by (x−r) and (x−r)2 must be zero. Using long division the quotient when 8x3−4x2−42x+45 is divided by (x−r) is 8x2+ (8r−4)x+(8r2−4r−42), the remainder 8r3−4r2−42r+45 being zero. Since (x−r) also divides the above quadratic polynomial its remainder must be zero. Thus
24r2−8r−42=0
The roots of the last equation are 23 and −67. Since r=−67 does not make the first remainder vanish, r=23=1.5.