find a polynomial whose zeroes are 2,1 and -1. what is its degree.
Answers
Step-by-step explanation:
Question:
Use the given information about a polynomial whose coefficients are real numbers to find the remaining zeros.
degree: 6
Zeros:
−
8
+
11
i
,
−
7
+
17
i
,
16
−
i
√
2
Find the Conjugate Complex Zeros of a Polynomial
Fundamental Theorem of Algebra
The fundamental theorem of algebra states that every non-constant, single-variable polynomial of degree
n
has exactly
n
complex zeros.
Complex Conjugates
The conjugate of a complex number is the number having an equivalent real value and an imaginary value equal in magnitude and opposite in sign.
Therefore, if
z
=
a
+
b
i
is a complex number, then the conjugate of
z
, denoted by
¯¯¯
z
, is equal to
¯¯¯
z
=
a
−
b
i
.
Complex Conjugate Root Theorem
The complex conjugate root theorem says that, if
P
(
x
)
is a single-variable polynomial with real coefficients, and if
a
+
b
i
, with
a
,
b
∈
Z
, is a root of
P
(
x
)
, then
a
−
b
i
is also a root of
P
(
x
)
.
In other words, for polynomials with real coefficients, complex zeros appear in conjugate pairs.
Suppose we are asked to find all of the zeros of a polynomial of degree 6 and we are give three complex zeros.
If no pair of the given zeros are conjugates, then we can find the conjugate of each given zero.
The conjugates are themselves zeros of the polynomial, giving us six zeros in total.
Answer and Explanation:
Let
P
(
x
)
be the polynomial of degree
6
.
We are given that
{
−
8
+
11
i
,
−
7
+
17
i
,
16
−
i
√
2
}
are zeros of
P
(
x
)
.
{
−
8
+
11
i
,
−
7
+
17
i
,
16
−
i
√
2
}
are not pairwise conjugate, therefore, by the complex conjugate root theorem,
{
−
8
−
11
i
,
−
7
−
17
i
,
16
+
i
√
2
}
are also zeros of
P
(
x
)
.
By the fundamental theorem of algebra, a polynomial of degree
6
has exactly
6
complex roots. Therefore we have identified all of the roots of
P
(
x
)
.
Step-by-step explanation:
may beee 6
hiiii dear friend!!!!