Use the remainder and factor theorems to test polynomial roots
A zero remainder means x minus c is a factor
When a polynomial P(x) is divided by x−c, the remainder is P(c). Thus c is a zero, and x−c is a factor, exactly when P(c)=0. This works for real and complex values of c.
Problem
Example 1. For g(x)=2x3−x2−22x+11, determine whether −1 and −11 are zeros.
Example 2. For f(x)=3x3−25x2+37x+145, determine whether 4−i and 5−i are zeros.
Example 3. For h(x)=x4−6x3+3x2+4x+64, determine whether x−4 and x+2 are factors.
Answer
Example 1: −1 is not a zero; −11 is a zero.
Example 2: neither complex value is a zero.
Example 3: x−4 is a factor; x+2 is not.
For x+2, substitute c=-2. A complex result is zero only when both its real and imaginary parts are zero.
Step-by-step solution
1. Test the real and irrational values
Direct substitution gives
g(−1)=−2−1+22+11=30=0.
To test the radical, factor by grouping:
g(x)=x2(2x−1)−11(2x−1)=(2x−1)(x2−11).
At x=−11, the factor x2−11 is zero. Thus g(−11)=0, so this value is a zero.
2. Evaluate at four minus i
Use i2=−1:
(4−i)2=15−8i,(4−i)3=52−47i.
Substitute into the given polynomial:
f(4−i)=3(52−47i)−25(15−8i)+37(4−i)+145.
Combining real and imaginary parts gives 74+22i, which is nonzero. Hence 4−i is not a zero.
3. Evaluate at five minus i
Similarly,
(5−i)2=24−10i,(5−i)3=110−74i.
Therefore
f(5−i)=3(110−74i)−25(24−10i)+37(5−i)+145=60−9i.
This is also nonzero, so 5−i is not a zero. Test the stated coefficients directly rather than assuming that a listed value must be a root.
4. Test the proposed linear factors
For x−4, substitute x=4:
h(4)=256−384+48+16+64=0.
Thus x−4 is a factor. For x+2=x−(−2), substitute x=−2:
h(−2)=16+48+12−8+64=132=0.
Thus x+2 is not a factor.
5. Interpret a nonzero result
A nonzero value is the remainder, not a new root. No complete factorization is required merely to test a proposed zero or linear factor.
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