Need this solutions this morning be helpful

Dec 2019
18
0
Abeokuta
One root of the equation z^3 - 6z + 13z + k = 0, where is real, is 2 + I. Find the other roots and the value of k.
 

romsek

Math Team
Sep 2015
2,776
1,552
USA
I assume that you mean $z^3 -6z^2 + 13z + k = 0$

I also assume you mean $k$ is real. Since all the coefficients are real you know that roots come in conjugate pairs. Or you would if you read your textbook.

Thus you have two of the roots, $2+i, ~ 2-i$.

Divide your polynomial by $(x-(2+i))(x-(2-i))$ and you'll end up with a first degree polynomial quotient with an obvious zero and a remainder that must be zeroed out.
Zeroing out the remainder allows you to solve for $k$
 
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Dec 2019
18
0
Abeokuta
I assume that you mean $z^3 -6z^2 + 13z + k = 0$

I also assume you mean $k$ is real. Since all the coefficients are real you know that roots come in conjugate pairs. Or you would if you read your textbook.

Thus you have two of the roots, $2+i, ~ 2-i$.

Divide your polynomial by $(x-(2+i))(x-(2-i))$ and you'll end up with a first degree polynomial quotient with an obvious zero and a remainder that must be zeroed out.
Zeroing out the remainder allows you to solve for $k$
Which textbook is day please .. can you send it
 

skipjack

Forum Staff
Dec 2006
21,321
2,390
As $(2 + i)^3 - 6(2 + i)^2 + 13(2 + i) + k = 10 + k$, $k = -10$.

$z^3 - 6z^2 + 13z - 10 = (z - 2)(z^2 - 4z + 5) = (z - 2)(z - 2 + i)(z - 2 - i)$
 
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