Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

APP. B] ALGEBRAIC SYSTEMS 459

POLYNOMIALS OVER A FIELD

B.26.Supposef(t)= 2 t^3 − 3 t^2 − 6 t−2. Find all the roots off(t)knowing thatf(t)has a rational root.

The rational roots off(t)must be among±1,±2,±1/2. Testing each possible root, we get, by synthetic division

(or dividing by 2t+1),

−^12

∣

∣∣

∣

∣

2 − 3 − 6 − 2

− 1 + 2 + 2

2 − 4 − 4 + 0

Thereforet=− 1 /2 is a root and

f(t)=(t+ 1 / 2 )( 2 t^2 − 4 t− 4 )=( 2 t+ 1 )(t^2 − 2 t− 2 )

We can now use the quadratic formula ont^2 − 2 t−2 to obtain the following three roots off(t):

t=− 1 / 2 ,t= 1 +

√

3 ,t= 1 −

√

3

B.27.Letf(t)=t^4 − 3 t^3 + 3 t^2 + 3 t−20. Find all the roots off(t)given thatt= 1 + 2 iis a root.

Since 1+ 2 iis a root, then 1− 2 iis a root andc(t)=t^2 − 2 t+5 is a factor off(t). Dividingf(t)byc(t)we get

f(t)=(t^2 − 2 t+ 5 )(t^2 −t− 4 )

The quadratic formula witht^2 −t−4 gives us the other roots off(t). That is, the four roots off(t)follow:

t= 1 + 2 i, t= 1 − 2 i, t=( 1 +

√

17 )/ 2 ,t=( 1 −

√

17 )/ 2

B.28.LetK=Z 8. Find all roots off(t)=t^2 + 6 t.

HereZ 8 ={ 0 , 1 , 2 ,..., 7 }. Substitute each element ofZ 8 intof(t)to obtain:

f( 0 )= 0 ,f( 2 )= 0 ,f( 4 )= 0 ,f( 6 )= 0

Thenf(t)has four roots,t= 0 , 2 , 4 ,6. (Theorem B.21 does not hold here sinceKis not a field.)

B.29.Supposef(t)is a real polynomial with odd degreen. Show thatf(t)has a real root.

The complex (nonreal) roots come in pairs. Sincef(t)has an odd numbernof roots (counting multiplicity),

f(t)must have at least one real root.

B.30.Prove Theorem B.15 (Euclidean Division Algorithm): Letf(t)andg(t)be polynomials over a fieldK

withg(t)=0. Then there exist polynomialsq(t)andr(t)such that

f(t)=q(t)g(t)+r(t)

where eitherr(t)≡0 or deg(r)<deg(g).

Iff(t)=0 or if deg(f)<deg(g), then we have the required representationf(t)= 0 g(t)+f(t). Now suppose

deg(f)≥deg(g), say

f(t)=antn+···+a 1 t+a 0 and g(t)=bmtm+···+b 1 t+b 0

wherean,bm=0 andn>m. We form the polynomial

f 1 (t)=f(t)−

an

bm

tn−mg(t) (1)

(This is the first subtraction step in “long division.”) Then deg(f 1 )<deg(f). By induction, there exist polynomials

q 1 (t)andr(t)such thatf 1 (t)=q 1 (t)g(t)+r(t)where eitherr(t)≡0 or deg(r)<deg(g). Substituting this into (1)

and solving forf(t),weget

f(t)=

[

q 1 (t)+

an

bm

tn−m

]

g(t)+r(t)

which is the desired representation.