solve the equation x^4-2x^3-13x^2+38x-24=0 by newton raphson method
Answers
Step-by-step explanation:
Use sum of coefficients short cut, then rational roots theorem to find:
x
4
−
2
x
3
−
13
x
2
+
38
x
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
+
4
)
Explanation:
Let
f
(
x
)
=
x
4
−
2
x
3
−
13
x
2
+
38
x
−
24
First notice that the sum of the coefficients is
0
, so
f
(
1
)
=
0
and
(
x
−
1
)
is a factor of
f
(
x
)
.
x
4
−
2
x
3
−
13
x
2
+
38
x
−
24
=
(
x
4
−
x
3
)
−
(
x
3
−
x
2
)
−
(
14
x
2
−
14
x
)
+
(
24
x
−
24
)
=
x
3
(
x
−
1
)
−
x
2
(
x
−
1
)
−
14
x
(
x
−
1
)
+
24
(
x
−
1
)
=
(
x
3
−
x
2
−
14
x
+
24
)
(
x
−
1
)
Let
g
(
x
)
=
x
3
−
x
2
−
14
x
+
24
By the rational roots theorem, any rational root of
g
(
x
)
=
0
must be a factor of
24
and hence one of:
±
1
,
±
2
,
±
3
,
±
4
,
±
6
,
±
12
,
±
24
We soon find
g
(
2
)
=
8
−
4
−
28
+
24
=
0
,
so
(
x
−
2
)
is a factor of
g
(
x
)
x
3
−
x
2
−
14
x
+
24
=
(
x
3
−
2
x
2
)
+
(
x
2
−
2
x
)
−
(
12
x
−
24
)
=
x
2
(
x
−
2
)
+
x
(
x
−
2
)
−
12
(
x
−
2
)
=
(
x
2
+
x
−
12
)
(
x
−
2
)
Then
x
2
+
x
−
12
=
(
x
+
4
)
(
x
−
3
)
by finding two numbers whose product is
12
and whose difference is
1
Putting it all together:
x
4
−
2
x
3
−
13
x
2
+
38
x
−
24
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
(
x
+
4
)