$\begingroup$I am to find the solutions of $2x^3-3x^2+32x+17$.
My textbook says the solutions are $\frac{-1}{2}$, $1\pm4i$
I got $\frac{-1}{2}$ and $1\pm i \sqrt{17}$
First I used the fundamental theorem of algebra to find candidate zeros and verified using synthetic division that $\frac{-1}{2}$ is a zero.
I then had:
$(x+\frac{1}{2})(2x^2-4x+34)$
Then, using the quadratic formula with $(2x^2-4x+34)$ to find the zeros:
$a=2$
$b=-4$
$c=34$
$$\frac{4\pm\sqrt{4^2-4(2)(34)}}{4}$$
$$\frac{4\pm\sqrt{-272}}{4}$$
$$\frac{4\pm4i\sqrt{17}}{4}$$
$$1\pm\sqrt{17}i$$
Where did I go wrong and how can I arrive at $1\pm4i$?
$\endgroup$61 Answer
$\begingroup$Your problem is expression under root is $16-8*34 = -256 \ne -272$
$\endgroup$4
