find the degree and cofficients of -x² + 27
Answers
Step-by-step explanation:
Explanation:
The degree of reqd. polynomial, say
p
(
x
)
is
3
, and hence by the Fundamental Principle of Algebra, it must have
3
zeroes. These are given to be
−
2
,
1
and
4
.
As
−
2
is a zero of
p
(
x
)
,
x
−
(
−
2
)
=
x
+
2
must be a factor of
p
(
x
)
.
Similarly, other zeroes give us factors
(
x
−
1
)
and
(
x
−
4
)
Degree of
p
(
x
)
is
3
, so,
p
(
x
)
can not have any other factor except those described above. Of course,
p
(
x
)
can have a numerical factor, like
k
≠
0
.
In view of above, we can suppose that,
p
(
x
)
=
k
(
x
+
2
)
(
x
−
1
)
(
x
−
4
)
. Expanding the
R
.
H
.
S
.
, we have,
p
(
x
)
=
k
(
x
3
−
3
x
2
−
6
x
+
8
)
But this will give us the leading co-eff
=
k
, which is given to be
2
, so,
k
=
2
.
Hence the poly.
p
(
x
)
=
2
(
x
3
−
3
x
2
−
6
x
+
8
)
=
2
x
3
−
6
x
2
−
12
x
+
16
.
Answer:
Degree :- 2
coefficient :- -1
constant :- 27
The degree of the polynomial is to that's why it is Quadratic polynomial.