(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 .