posted by Suzie on .
how to find all real zeros of "4x^5+8x^4-15x^3-23x^2+11x+15
Since there is no general method to solve a quintic equation, we must assume that there is some low-hanging fruit, root-wise.
y(1) = 0, so
y = (x-1)(4x^4 + 12x^3 - 3x^2 - 26x - 15)
Try 1 again: no joy
Try -1. Yay and we have
y = (x-1)(x+1)(4x^3 + 8x^2 - 11x - 15)
Try -1 again: yay
y = (x-1)(x+1)(x+1)(4x^2 + 4x - 15)
y = (x-1)(x+1)(x+1)(2x-3)(2x+5)
I assume you can find the roots at this point...