Math 121 – Section 4.5 Solutions
12. The remainder when f(x) = −4x3+ 5x2+ 8 is divided by x+ 3 is:
f(−3) = −4(−3)3+ 5(−3)2+ 8 = 161
14. The remainder when f(x) = 4x4−15x2−4 is divided by x−2 is:
f(2) = 4(2)4−15(2)2−4 = 0
21. f(x) = −4x7+x3−x2+ 2
•f(x) has at most 7 real zeros since the degree is 7
•there are 3 sign changes in f(x)⇒f(x) has either 3 or 1 positive real zeros
•since f(−x) = 4x7−x3−x2+ 2, there are 2 sign changes in f(−x)⇒f(x) has either 2 or 0
negative real zeros
30. f(x) = x5−x4+x3−x2+x−1
•f(x) has at most 5 real zeros since the degree is 5
•there are 5 sign changes in f(x)⇒f(x) has either 5, 3, or 1 positive real zeros
•since f(−x) = −x5−x4−x3−x2−1, there are 0 sign changes in f(−x)⇒f(x) has 0 negative
real zeros
34. Given f(x) = x5−x4+ 2x2+ 3 we have a0= 3 and a5= 1. The factors of a0are p=±1,±3. The
factors of a5are q=±1. Therefore, the potential rational real zeros of f(x) are:
p
q=±1,±3
38. Given f(x) = 6x4−x2+ 2 we have a0= 2 and a4= 6. The factors of a0are p=±1,±2. The factors
of a4are q=±1,±2,±3,±6. Therefore, the potential rational real zeros of f(x) are:
p
q=±1,±1
2,±1
3,±1
6,±2,±2
3
45. Given f(x) = x3+ 2x2−5x−6 we have a0=−6 and a3= 1. The factors of a0are p=±1,±2,±3,±6.
The factors of a3are q=±1. Therefore, the potential rational real zeros of f(x) are:
p
q=±1,±2,±3,±6
1
