Roots of Polynomials
1. Use the Factor Theorem to prove: (x + 1) is a factor of x5 + 1
The root in (x+1) is -1. What is the remainder f(-1)? Zero, hence -1 is a root
2. Use synthetic division to find another factor of the polynomial in problem 1.
-1| 1 0 0 0 0 1
__-1_ 1_-1 1 -1___
1 -1 1 -1 1 0
x4-x3+x2-x+1
3. Give the roots of the following polynomial and their multiplicity. What is the degree of
the polynomial?
y = (x + 2)2(x − 1)
-2, multiplicity 2
1, multiplicity 1
4.
Is x = 2 a root of this polynomial: x6 − 3x5 + 3x4 − 3x3 + 3x2 −3x + 2?
Use synthetic division, since 2 is a factor of 2
2|1 -3 3 -3 3 -3 2
____2_-2_2_-2_2 -2______
1 -1 1 -1 1 -1 0
remainder is zero, so it is a root
5. What are the possible rational roots of x3 − 4x2 + 2x + 4?
± 1, ± 2, ± 4
Are any of the possible roots you found in problem 5 actual roots?
1 and -1 are easy to check: 1-4+2+4, 1 is not, -1 – 4 -2 + 4, -1 is not
2|1 -4 2 4
____2_-4_-4_
1 -2 -2 0 so 2 is a root, look at x2 – 2x – 2; possible roots are ± 1, ± 2, know ±
1don’t work, plug in ± 2, don’t work either. Discriminant is 12. √12 isn’t rational
6. What are the possible rational roots of x3 − 2x2 − 3x + 1?
±1
7. Are any of the possible roots you found in problem 7 actual roots? If so, factor the
polynomial as best you can.
These we can plug in: 1 – 2 – 3 + 1 so 1 is not, -1 -2 +3 +1 so -1 is not
8. What are the possible rational roots of x3 + 2x2 − 5x – 6?
±1, ±2, ±3
9. Are any of the possible roots you found in problem 9 actual roots? If so, factor the
polynomial as best you can.
try 1 and -1: 1 + 2 – 5 – 6 =0, so 1 isn’t, -1 + 2 + 5 – 6 = 0, so -1 is a root
to check ±2 can plug in or use synthetic division: -8 + 4 + 10 – 6 = 0, so -2 is a root
for 2, 8 + 4 – 10 – 6 ≠ 0, so 2 is not
check 3: 27 + 18 -15 -6 ≠ 0, -27 + 18 + 15 – 6 = 0, so -3 is
roots are -1, -2, -3, (x+1)(x+2)(x+3) is factorization
10. Let f(x) = x5 + x4 + x3 + x2 − 12x − 12. One root is
and another is −2i. At most, how
many rational roots can it have? List all the possible rational roots.
if √3 is a root, so is -√3, is -2i is a root, so is -2i. That is 4 roots, so it can have at most
one more, or one possible rational root