# Find all complex solutions (real and non real) of $2x^3-3x^2+32x+17$

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.

$$(x+\frac{1}{2})(2x^2-4x+34)$$

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 $$1\pm4i$$?

• $$2x^3-3x^2+32x+17 = (2 x+1) \left(x^2-2 x+17\right)$$ See your issue now?
– Moo
Aug 21 '20 at 15:24
• $4^2-4\cdot2\cdot34=-256=-16^2$. Aug 21 '20 at 15:26
• $-272$ is wrong here Aug 21 '20 at 15:26
• $4^2 - 4*2*34= 16 - 272$ which not $-272$. Some how the $4^2$ that came before the $-4*2*34$ just got up and walked off the paper!.... $12-272 = -256$ and $\sqrt {-256} = \pm 16i$. Aug 21 '20 at 15:30
• To the person who gives the downvote, let me tell you that you are wrong. The question is not school homework and he knows what to apply to solve the problem, he is asking for the part of calculations where his error lives. Aug 21 '20 at 15:34

Your problem is expression under root is $$16-8*34 = -256 \ne -272$$