The roots are where the function crosses the xaxis. So, you can just look for where it crosses and those would be your roots.
http://answers.yahoo.com/question/index?qid=200812...

It's not just quartics. Polynomial equations that has real coefficients must have an even number of imaginary solutions. Of course, you can make a Quartic polynomial that has only
http://au.answers.yahoo.com/question/index?qid=201...

y = 6x^5+5x^450x^365x^2+20x+12 = (x+2)(6x^4 7x^336x^2+7x+6) =(x+2)(x3)(6x^3+11x^23x2) =(x+2)(x3)(2x1)(3x^2+7x+2) 3x^2+7x+2 = 3[(x+7/6)^2+2/349/36] = 3[(x+7/6)^225/36] y
http://answers.yahoo.com/question/index?qid=200704...

You don't happen to go to UHS do you XD Take home test?
http://answers.yahoo.com/question/index?qid=201001...
