I have to tackle a question related with roots of polynomial function.
The graph of the polynomial is shown below:
(A)explain why, of the four roots of the equation f(x)=o, two are real
and two are complex
I found I am a little bit messy with the concepts…..when f(x)=0, the two real roots are 2 and -4, how can I find the complex roots from the graph??
(B)The curve passes though the point (-1,-18). Find f(x) in the form
(x)=(x-a)(x-b)(x^2+cx+d), were a, b, c, d belongs to Z
This question is simple, I put in 2,-4, and -32 to be a,b,c, and then plug in (-1,-18), and found that c=-33
(C)Find the two complex roots of the equation f(x)=0 in Cartesian form
(D)express each of the four roots of the equation in the Euler’s form
Again, I know Cartesian form and euler’s form,but how can I find the complex roots?
Hope someone can help me ! Thanks!!
Best Answer
There's no easy answer to this question that I'm aware of. But fortunately that's not what you were asked. You were just asked to show that two are real and two are complex. You see the real ones with your eyes, and the other two must be complex. The fundamental theorem of algebra says there must be four, up to multiplicity, and since complex roots come in conjugate pairs, you know they are distinct, so there are two different ones.
That's not what I found. Also you didn't mention $d.$
Once you have found $c$ and $d$, the complex roots will just be the roots of $x^2+cx+d,$ since the two real roots are already factored out. You can find the roots with the quadratic formula.