Problem: If x4+4x3+6px2+4qx+r is exactly divisible by x3+3x2+9x+3, the value of (p+q)r is:
Answer Choices:
A. −18
B. 12
C. 15
D. 27
E. 45
Solution:
Let x4+4x3+6px2+4qx+r=(x+a)(x3+3x2+9x+3)=x4+(a+3)x2+(3a+9)x2+(9a+3)x+3a. Equating the coefficients of like powers of x, we have a=1,p=2,q=3,r=3 so that (p+q)r =(2+3)(3)=15
or
Divide x4+4x3+6px2+4qx+r by x3+3x2+9x+3; the quotient is x+1 and the remainder is the second-degree polynomial (6p−12)x2+(4q−12)x+r−3. This remainder equals zero since the division is exact. Therefore p=2,q=3,r=3 and (p+q)r=15