Jul 28, 2010 · Clearly the task at hand is to find the values of these constants. If we can convert (1) into (2) by determining quadratic factors that satisfies the equation, we can find the roots with the …

Mar 7, 2020 · Can you specify exactly how to determine the roots using this method? The form that I have is f(x) = a(x-h)^2 + k, not equivalent to zero.

Feb 6, 2021 · All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what …

Jan 4, 2016 · The only method that I know is to multiply and then find the derivative of function then apply Sturm's theorem but it seems vague when you have to solve the question in 3 to 5 minutes .

How would you find the roots of a function when: The function is a polynomial with a degree greater than 3 The polynomial is not factorable ? For example (let's just say): $2x^5-3x^3+13$ &nbsp...

Apr 19, 2011 · The question is already answered, but I wanted to add in my intuition for the why of the solution: The degree of a polynomial determines how many roots it can have.

Sep 17, 2016 · Generally, you need to have some intuition about the shapes of the functions. The first half of your function is a circle, and the last half is something that sort of looks like a parabola. You …

2 Bascially we are trying to find the roots of a quadratic equation, and 'apparently' there is a theorem for this, but every one that I have found so far mentions that the degree of the polynomial is directly …

Oct 7, 2014 · I've got the following polynomial $$ x^3-6x^2-2x+40 $$ and I want to find its roots. The only option I see at the moment is to compute all the divisors of $40$ and their inverse, and manually …

Nov 29, 2019 · Is it possible to find the complex roots of a function by using Newton's method? If the answer is yes, how to do it?