given the real part of 1 root of a polynomial with degree 4, find all the other roots.
ive tried - long divsion and equating the remainder to zero,
and substituting 2 + ib in the equation,
but both are too complex and didnt give me the right answer, which is
$$z = frac{3 pm sqrt{5} }{2}$$ and 2 + i and 2 - i
)
complex-numbers
asked Nov 25 '18 at 11:05
Vanessa
656
add a comment |
ive tried - long divsion and equating the remainder to zero,
and substituting 2 + ib in the equation,
but both are too complex and didnt give me the right answer, which is
$$z = frac{3 pm sqrt{5} }{2}$$ and 2 + i and 2 - i
)
complex-numbers
asked Nov 25 '18 at 11:05
Vanessa
656
What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
1
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32
add a comment |
ive tried - long divsion and equating the remainder to zero,
and substituting 2 + ib in the equation,
but both are too complex and didnt give me the right answer, which is
$$z = frac{3 pm sqrt{5} }{2}$$ and 2 + i and 2 - i
)
complex-numbers
asked Nov 25 '18 at 11:05
Vanessa
656
ive tried - long divsion and equating the remainder to zero,
and substituting 2 + ib in the equation,
but both are too complex and didnt give me the right answer, which is
$$z = frac{3 pm sqrt{5} }{2}$$ and 2 + i and 2 - i
)
complex-numbers
complex-numbers
asked Nov 25 '18 at 11:05
Vanessa
656
asked Nov 25 '18 at 11:05
Vanessa
656
edited Nov 25 '18 at 11:32
asked Nov 25 '18 at 11:05
Vanessa
656
asked Nov 25 '18 at 11:05
Vanessa
656
asked Nov 25 '18 at 11:05
Vanessa
656
656
What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
1
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32
add a comment |
What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
1
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32
What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
1
1
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32
add a comment |
1 Answer
1
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If you substitute $2+bi$ into the equation you get
$$11+7ib-12b^2-7ib^3+b^4=0.$$ So $7ib(1-b^2)=0$ and since $b=0$ is no solution you
have $b=pm 1$. If you divide your original polynominal by
$(z-(2+i))(z-(2-i)) = z^2 - 4z +5$ you have to solve
$$z^2+3z+1 =0.$$
Can you continue?
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
add a comment |
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If you substitute $2+bi$ into the equation you get
$$11+7ib-12b^2-7ib^3+b^4=0.$$ So $7ib(1-b^2)=0$ and since $b=0$ is no solution you
have $b=pm 1$. If you divide your original polynominal by
$(z-(2+i))(z-(2-i)) = z^2 - 4z +5$ you have to solve
$$z^2+3z+1 =0.$$
Can you continue?
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
add a comment |
If you substitute $2+bi$ into the equation you get
$$11+7ib-12b^2-7ib^3+b^4=0.$$ So $7ib(1-b^2)=0$ and since $b=0$ is no solution you
have $b=pm 1$. If you divide your original polynominal by
$(z-(2+i))(z-(2-i)) = z^2 - 4z +5$ you have to solve
$$z^2+3z+1 =0.$$
Can you continue?
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
add a comment |
If you substitute $2+bi$ into the equation you get
$$11+7ib-12b^2-7ib^3+b^4=0.$$ So $7ib(1-b^2)=0$ and since $b=0$ is no solution you
have $b=pm 1$. If you divide your original polynominal by
$(z-(2+i))(z-(2-i)) = z^2 - 4z +5$ you have to solve
$$z^2+3z+1 =0.$$
Can you continue?
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
If you substitute $2+bi$ into the equation you get
$$11+7ib-12b^2-7ib^3+b^4=0.$$ So $7ib(1-b^2)=0$ and since $b=0$ is no solution you
have $b=pm 1$. If you divide your original polynominal by
$(z-(2+i))(z-(2-i)) = z^2 - 4z +5$ you have to solve
$$z^2+3z+1 =0.$$
Can you continue?
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
answered Nov 25 '18 at 11:31
gammatester
16.7k21632
16.7k21632
add a comment |
add a comment |
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What does $$(=0/5(3pmsqrt5)$$ mean? It has unbalanced parentheses, as written it's two ways of writing zero, and it's only two roots where the polynomial should have three more.
– Gerry Myerson
Nov 25 '18 at 11:10
I'm so sorry, i edited it now.
– Vanessa
Nov 25 '18 at 11:19
OK, that's better (though it would be better still if you would take a few minutes with the help menu to learn about formatting mathematics on this website). But I'm still worried that that equation has four solutions, while what you're calling the right answer only comes to two.
– Gerry Myerson
Nov 25 '18 at 11:24
1
if $2+ib$ is a root, so does $2-ib$, This means the polynomial contains a quadratic factor of the form $z^2-4z+A$. By matching the coeff of $z^3$, the polynomial also contains a factor of the form $z^2+3z+B$. Now compare the coefficients of $z^2, z^1$ and $z^0$ in $$z^4 - z^3 - 6z^2 + 11z + 5 = (z^2-4z+A)(z^2+3z+B)$$ You will get $A = 5,B = 1$ and hence obtain 4 roots, two of them $2pm i$ will be complex, the other two $frac{-3 pm sqrt{5}}{2}$ will be real.
– achille hui
Nov 25 '18 at 11:32