Answer:
f(x) = 4x^5-2x^4+30x^3-15x^2+50x-25
Step-by-step explanation:
By the rational roots theorem, any rational zeros of
f
(
x
)
are expressible in the form
p
q
for integers
p
,
q
with
p
a divisor of the constant term
25
and
q
a divisor of the coefficient
4
of the leading term.
That means that the only possible rational zeros are:
±
1
4
,
±
1
2
,
±
1
,
±
5
4
,
±
5
2
,
±
5
,
±
25
4
,
±
25
2
,
±
25
In fact, note that the signs of the coefficients alternate, with no powers of
x
missing. Hence (by Descartes rule of signs) there are
1
,
3
or
5
positive Real zeros and no negative Real zeros.
So the only possible rational zeros are:
1
4
,
1
2
,
1
,
5
4
,
5
2
,
5
,
25
4
,
25
2
,
25
Trying each in turn, we find:
f
(
1
2
)
=
4
(
1
32
)
−
2
(
1
16
)
+
30
(
1
8
)
−
15
(
1
4
)
+
50
(
1
2
)
−
25
f
(
1
2
)
=
1
−
1
+
30
−
30
+
200
−
200
8
=
0
So
x
=
1
2
is a zero and
(
2
x
−
1
)
a factor:
4
x
5
−
2
x
4
+
30
x
3
−
15
x
2
+
50
x
−
25
=
(
2
x
−
1
)
(
2
x
4
+
15
x
2
+
25
)
We can factor the remaining quartic as a quadratic in
x
2
2
x
4
+
15
x
2
+
25
=
2
x
4
+
10
x
2
+
5
x
2
+
25
2
x
4
+
15
x
2
+
25
=
2
x
2
(
x
2
+
5
)
+
5
(
x
2
+
5
)
2
x
4
+
15
x
2
+
25
=
(
2
x
2
+
5
)
(
x
2
+
5
)
Hence zeros:
x
=
±
√
10
2
i
and
x
=
±
√
5
i
