Problem: If 3x3−9x2+kx−12 is divisible by x−3, then it is also divisible by:
Answer Choices:
A. 3x2−x+4
B. 3x2−4
C. 3x2+4
D. 3x−4
E. 3x+4
Solution:
Since 3x3−9x2+kx−12 is divisible by x−3, 3⋅33−9⋅32+k⋅3−12=0∴k=4 Divide 3x9−9x2+4x−12 by x−3 and obtain 3x2+4.
or
3x3−9x2+kx−12=(x−3)(Ax2+Bx+C) =Ax3+(B−3A)x2+(C−3B)x−3C ∴A=3,C=4 and B−3A=−9∴B=0. The other factor is, therefore, 3x2+4.