If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$...
$begingroup$
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
linear-algebra
$endgroup$
add a comment |
$begingroup$
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
linear-algebra
$endgroup$
add a comment |
$begingroup$
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
linear-algebra
$endgroup$
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:
I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.
Does a better method than the lackluster substitution, exist?
The answer is:
$x^2-3x+2$
linear-algebra
linear-algebra
asked Dec 24 '18 at 16:12
CaptainQuestionCaptainQuestion
1337
1337
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\
&=1
end{align}
$$
and
$$
begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\
&=2
end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
$endgroup$
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
add a comment |
$begingroup$
Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
$endgroup$
add a comment |
$begingroup$
Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$begin{align}
& 1 {-}3 , 5, {-}2\
&color{#c00}{{-}1, 2 {-}3}\
& {-}1 2 {-}2\
&color{#c00}{ 1 {-}2 3}\
&qquadqquadquad color{#0a0}1
end{align}qquadqquadquad$$
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\
&=1
end{align}
$$
and
$$
begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\
&=2
end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
$endgroup$
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
add a comment |
$begingroup$
Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\
&=1
end{align}
$$
and
$$
begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\
&=2
end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
$endgroup$
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
add a comment |
$begingroup$
Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\
&=1
end{align}
$$
and
$$
begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\
&=2
end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
$endgroup$
Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\
&=1
end{align}
$$
and
$$
begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\
&=2
end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
answered Dec 24 '18 at 16:29
robjohn♦robjohn
269k27311638
269k27311638
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
add a comment |
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
$begingroup$
Never thought about using the initial polynomial to create a reduction formula absolutely radical!
$endgroup$
– CaptainQuestion
Dec 24 '18 at 16:43
1
1
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
$begingroup$
@CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
$endgroup$
– Bill Dubuque
Dec 24 '18 at 16:44
add a comment |
$begingroup$
Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
$endgroup$
add a comment |
$begingroup$
Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
$endgroup$
add a comment |
$begingroup$
Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
$endgroup$
Hint:
As $a,b$ are the roots of $x^2-2x+3=0$
$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$
Similarly for $b$
answered Dec 24 '18 at 16:20
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$begin{align}
& 1 {-}3 , 5, {-}2\
&color{#c00}{{-}1, 2 {-}3}\
& {-}1 2 {-}2\
&color{#c00}{ 1 {-}2 3}\
&qquadqquadquad color{#0a0}1
end{align}qquadqquadquad$$
$endgroup$
add a comment |
$begingroup$
Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$begin{align}
& 1 {-}3 , 5, {-}2\
&color{#c00}{{-}1, 2 {-}3}\
& {-}1 2 {-}2\
&color{#c00}{ 1 {-}2 3}\
&qquadqquadquad color{#0a0}1
end{align}qquadqquadquad$$
$endgroup$
add a comment |
$begingroup$
Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$begin{align}
& 1 {-}3 , 5, {-}2\
&color{#c00}{{-}1, 2 {-}3}\
& {-}1 2 {-}2\
&color{#c00}{ 1 {-}2 3}\
&qquadqquadquad color{#0a0}1
end{align}qquadqquadquad$$
$endgroup$
Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$begin{align}
& 1 {-}3 , 5, {-}2\
&color{#c00}{{-}1, 2 {-}3}\
& {-}1 2 {-}2\
&color{#c00}{ 1 {-}2 3}\
&qquadqquadquad color{#0a0}1
end{align}qquadqquadquad$$
edited Dec 24 '18 at 16:44
answered Dec 24 '18 at 16:19
Bill DubuqueBill Dubuque
212k29195651
212k29195651
add a comment |
add a comment |
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