(a).
\Identify Possible Rational Zeros:
\Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.
\The function is
.
Because the leading coefficient is
, the possible rational zeros are the intezer factors of the constant term
.
Therefore the possible rational zeros of
are
.
Examine the variations in sign for
and
:
Consider
.
The original polynomial equation has
variation in sign.
.
.
There are no sign changes.so ,there are no negative roots.
\Therefore by Descartes
rule of signs
has either
or
positive real zeros and no negative real zeros.
Synthetic Division:
\Setup the synthetic division using a zero place for the missing terms
in the dividend.
.
By The synthetic division method it is determined that
has no rational zeros. 
To find the factors of
Substiute
.
.
.
.
To find the rational zeros use the quadratic formula
.
Consider
.
Where
.
Substiute the values in the quadratic formula
.

The rational roots are
and
.
Earlier we assume that
.
Now
and
.
Therefore
and
are the factors of
.
The factor
can be written as
.
The factor
has no real zeros and is therefore irreducable over the reals.
So
is written as a product of linear and irreducable quadratic factors is
.
(b).
\We can factor
by writing the expression first as a difference of squares
or
.Then factor the difference of squares as
and
.
written as a product of linear factors is
.
(c).
\Because the function has degree
, by the corollary of the fundemental therom of algebra
has excatly
zeros,including any that may be repeated.
The linear factorization yeilds
zeros:
.
(a).
\The function
is written as a product of linear and irreducable quadratic factors is
.
(b).
\
written as a product of linear factors is
.
(c).
\The zeros are
.