<p>I have a math competition and one of the areas I am not sure about. Could finding a root of a function be explained.
sample problem:
f(x)=5x^(5)+x^(4)+21x^(3)+15x^(2)-98x+56
A. 7i
B. -7i
C. (7i)^(.5)
D. +(7i)^(.5) and -(7i)^(.5)
E. none of the above</p>
<p>If someone could show me how to find the root, you could make up an easier more basic problem if you don’t want to explain how to find the one above?</p>
<p>I don’t think there’s a simple way to find the roots of a quintic, as there is for the simpler polynomials (e.g., nothing in the way of a quadratic equation.)</p>
<p>the roots of an equation, if that is your question, is the number that makes the function output zero. Since this is multiple choice you could input the choices you are given, and see if they result in zero.</p>
<p>If it weren’t, I would try to factor the polynomial, although this looks like it may not be very easy to do that to.</p>
<p>Another rule you may find useful is that if a polynomial has a complex or imaginary root, it will always be a double root.</p>
<p>I’ve factored it, and I get</p>
<p>(.2)*(m-1)(5m-4)(5m+10)(m^2+7)</p>
<p>real roots include {1, 4/5, -2}, and the roots from the final m^2 term will be your imaginary one. I conclude that it is {sqrt(7)*i}</p>
<p>Thank you, my competition is for algebra 2, but that wasn’t covered in the class.</p>