How do I find the equation of a polynomial with real cofficients having $2$ and $5-2i$ as zeroes?
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0
down vote
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How do I find the equation of a polynomial with real coefficients having $2$ and $5-2i$ as zeroes?
Do I set the two equal to zero to help me find the equation? Would anyone be able to show the work?
algebra-precalculus
edited Nov 23 at 3:51
Rócherz
2,7012721
asked Nov 23 at 3:44
User231
165
add a comment |
up vote
0
down vote
favorite
How do I find the equation of a polynomial with real coefficients having $2$ and $5-2i$ as zeroes?
Do I set the two equal to zero to help me find the equation? Would anyone be able to show the work?
algebra-precalculus
edited Nov 23 at 3:51
Rócherz
2,7012721
asked Nov 23 at 3:44
User231
165
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I find the equation of a polynomial with real coefficients having $2$ and $5-2i$ as zeroes?
Do I set the two equal to zero to help me find the equation? Would anyone be able to show the work?
algebra-precalculus
edited Nov 23 at 3:51
Rócherz
2,7012721
asked Nov 23 at 3:44
User231
165
How do I find the equation of a polynomial with real coefficients having $2$ and $5-2i$ as zeroes?
Do I set the two equal to zero to help me find the equation? Would anyone be able to show the work?
algebra-precalculus
algebra-precalculus
edited Nov 23 at 3:51
Rócherz
2,7012721
asked Nov 23 at 3:44
User231
165
edited Nov 23 at 3:51
Rócherz
2,7012721
asked Nov 23 at 3:44
User231
165
edited Nov 23 at 3:51
Rócherz
2,7012721
edited Nov 23 at 3:51
Rócherz
2,7012721
edited Nov 23 at 3:51
Rócherz
2,7012721
2,7012721
asked Nov 23 at 3:44
User231
165
asked Nov 23 at 3:44
User231
165
asked Nov 23 at 3:44
User231
165
165
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
If it has real coefficients and
$a+bi$ as a root
then it also has
$a-bi$ as a root.
So a factor is
$(x-(a+bi))(x-(a-bi))
=x^2-2ax+a^2+b^2
$.
So if roots are
$2$ and $5-2i$
the polynomial is
$(x-2)(x^2-10x+29)
=x^3 - 12 x^2 + 49 x - 58
$.
answered Nov 23 at 3:49
marty cohen
72k547126
add a comment |
up vote
0
down vote
Observe that $5+2i$ is also a root. Thus your polynomial is: $P(x)=(x-2)(x-5+2i)(x-5-2i)=(x-2)(x^2-10x+29)$
answered Nov 23 at 3:50
DeepSea
70.8k54487
add a comment |
up vote
0
down vote
You have roots $x = 2$ and $x = 5 - 2i$. Now work backwards.
begin{align*}
x & = 2 & x & = 5 - 2i\
x - 2 & = 0 & x - 5 & = 2i\
& & (x - 5)^2 & = (2i)^2\
& & x^2 - 10x + 25 & = -4\
& & x^2 - 10x + 29 & = 0
end{align*}
One solution is
begin{align*}
(x - 2)(x^2 - 10x + 29) & = x(x^2 - 10x + 29) - 2(x^2 - 10x + 29)\
& = x^3 - 10x^2 + 29x - 2x^2 + 20x - 58\
& = x^3 - 12x^2 + 49x - 58
end{align*}
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If it has real coefficients and
$a+bi$ as a root
then it also has
$a-bi$ as a root.
So a factor is
$(x-(a+bi))(x-(a-bi))
=x^2-2ax+a^2+b^2
$.
So if roots are
$2$ and $5-2i$
the polynomial is
$(x-2)(x^2-10x+29)
=x^3 - 12 x^2 + 49 x - 58
$.
answered Nov 23 at 3:49
marty cohen
72k547126
add a comment |
up vote
0
down vote
If it has real coefficients and
$a+bi$ as a root
then it also has
$a-bi$ as a root.
So a factor is
$(x-(a+bi))(x-(a-bi))
=x^2-2ax+a^2+b^2
$.
So if roots are
$2$ and $5-2i$
the polynomial is
$(x-2)(x^2-10x+29)
=x^3 - 12 x^2 + 49 x - 58
$.
answered Nov 23 at 3:49
marty cohen
72k547126
add a comment |
up vote
0
down vote
up vote
0
down vote
If it has real coefficients and
$a+bi$ as a root
then it also has
$a-bi$ as a root.
So a factor is
$(x-(a+bi))(x-(a-bi))
=x^2-2ax+a^2+b^2
$.
So if roots are
$2$ and $5-2i$
the polynomial is
$(x-2)(x^2-10x+29)
=x^3 - 12 x^2 + 49 x - 58
$.
answered Nov 23 at 3:49
marty cohen
72k547126
If it has real coefficients and
$a+bi$ as a root
then it also has
$a-bi$ as a root.
So a factor is
$(x-(a+bi))(x-(a-bi))
=x^2-2ax+a^2+b^2
$.
So if roots are
$2$ and $5-2i$
the polynomial is
$(x-2)(x^2-10x+29)
=x^3 - 12 x^2 + 49 x - 58
$.
answered Nov 23 at 3:49
marty cohen
72k547126
answered Nov 23 at 3:49
marty cohen
72k547126
answered Nov 23 at 3:49
marty cohen
72k547126
answered Nov 23 at 3:49
marty cohen
72k547126
72k547126
add a comment |
add a comment |
up vote
0
down vote
Observe that $5+2i$ is also a root. Thus your polynomial is: $P(x)=(x-2)(x-5+2i)(x-5-2i)=(x-2)(x^2-10x+29)$
answered Nov 23 at 3:50
DeepSea
70.8k54487
add a comment |
up vote
0
down vote
Observe that $5+2i$ is also a root. Thus your polynomial is: $P(x)=(x-2)(x-5+2i)(x-5-2i)=(x-2)(x^2-10x+29)$
answered Nov 23 at 3:50
DeepSea
70.8k54487
add a comment |
up vote
0
down vote
up vote
0
down vote
Observe that $5+2i$ is also a root. Thus your polynomial is: $P(x)=(x-2)(x-5+2i)(x-5-2i)=(x-2)(x^2-10x+29)$
answered Nov 23 at 3:50
DeepSea
70.8k54487
Observe that $5+2i$ is also a root. Thus your polynomial is: $P(x)=(x-2)(x-5+2i)(x-5-2i)=(x-2)(x^2-10x+29)$
answered Nov 23 at 3:50
DeepSea
70.8k54487
answered Nov 23 at 3:50
DeepSea
70.8k54487
answered Nov 23 at 3:50
DeepSea
70.8k54487
answered Nov 23 at 3:50
DeepSea
70.8k54487
70.8k54487
add a comment |
add a comment |
up vote
0
down vote
You have roots $x = 2$ and $x = 5 - 2i$. Now work backwards.
begin{align*}
x & = 2 & x & = 5 - 2i\
x - 2 & = 0 & x - 5 & = 2i\
& & (x - 5)^2 & = (2i)^2\
& & x^2 - 10x + 25 & = -4\
& & x^2 - 10x + 29 & = 0
end{align*}
One solution is
begin{align*}
(x - 2)(x^2 - 10x + 29) & = x(x^2 - 10x + 29) - 2(x^2 - 10x + 29)\
& = x^3 - 10x^2 + 29x - 2x^2 + 20x - 58\
& = x^3 - 12x^2 + 49x - 58
end{align*}
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
add a comment |
up vote
0
down vote
You have roots $x = 2$ and $x = 5 - 2i$. Now work backwards.
begin{align*}
x & = 2 & x & = 5 - 2i\
x - 2 & = 0 & x - 5 & = 2i\
& & (x - 5)^2 & = (2i)^2\
& & x^2 - 10x + 25 & = -4\
& & x^2 - 10x + 29 & = 0
end{align*}
One solution is
begin{align*}
(x - 2)(x^2 - 10x + 29) & = x(x^2 - 10x + 29) - 2(x^2 - 10x + 29)\
& = x^3 - 10x^2 + 29x - 2x^2 + 20x - 58\
& = x^3 - 12x^2 + 49x - 58
end{align*}
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
add a comment |
up vote
0
down vote
up vote
0
down vote
You have roots $x = 2$ and $x = 5 - 2i$. Now work backwards.
begin{align*}
x & = 2 & x & = 5 - 2i\
x - 2 & = 0 & x - 5 & = 2i\
& & (x - 5)^2 & = (2i)^2\
& & x^2 - 10x + 25 & = -4\
& & x^2 - 10x + 29 & = 0
end{align*}
One solution is
begin{align*}
(x - 2)(x^2 - 10x + 29) & = x(x^2 - 10x + 29) - 2(x^2 - 10x + 29)\
& = x^3 - 10x^2 + 29x - 2x^2 + 20x - 58\
& = x^3 - 12x^2 + 49x - 58
end{align*}
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
You have roots $x = 2$ and $x = 5 - 2i$. Now work backwards.
begin{align*}
x & = 2 & x & = 5 - 2i\
x - 2 & = 0 & x - 5 & = 2i\
& & (x - 5)^2 & = (2i)^2\
& & x^2 - 10x + 25 & = -4\
& & x^2 - 10x + 29 & = 0
end{align*}
One solution is
begin{align*}
(x - 2)(x^2 - 10x + 29) & = x(x^2 - 10x + 29) - 2(x^2 - 10x + 29)\
& = x^3 - 10x^2 + 29x - 2x^2 + 20x - 58\
& = x^3 - 12x^2 + 49x - 58
end{align*}
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
answered Nov 23 at 12:34
N. F. Taussig
43.2k93254
43.2k93254
add a comment |
add a comment |
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