JS-The-quartic-equation-x-4-2x-3-14x-15-0-has-one-root-equal-to-1-2i-Find-the-other-three-roots-

Question Number 106637 by john santu last updated on 06/Aug/20

@ JS @

The quartic equation x^{4} + {2 x}^{3} + 14 x + 15 = 0

has one root equal to 1 + 2 i . Find

the other three roots .

Answered by bemath last updated on 06/Aug/20

\underset{@ bemath @}{}

as the coefficient of the quartic are real; it follows

that 1 - 2 i is also a root . then

[x - (1 - 2 i)] [x - (1 + 2 i)] is a factor

of the quartic .

\Rightarrow x^{2} - 2 x + 5 . Therefore

x^{4} + {2 x}^{3} + 14 x + 15 = (x^{2} - 2 x + 5) (x^{2} + ax + b)

and we get a = 4 and b = 3 .

p (x) = (x^{2} - 2 x + 5) (x^{2} + 4 x + 3) = 0

\to {\begin{cases} x_{1, 2} = \frac{2 \pm 4 i}{2} = 1 \pm 2 i \\ x_{3, 4} = \frac{- 4 \pm 2}{2} = - 2 \pm 1 \end{cases}

Commented by john santu last updated on 06/Aug/20

good