Posted by Suzie on Tuesday, September 27, 2011 at 11:27pm.
how to find all real zeros of "4x^5+8x^415x^323x^2+11x+15

calculus  Steve, Wednesday, September 28, 2011 at 5:43am
Since there is no general method to solve a quintic equation, we must assume that there is some lowhanging fruit, rootwise.
y(1) = 0, so
y = (x1)(4x^4 + 12x^3  3x^2  26x  15)
Try 1 again: no joy
Try 1. Yay and we have
y = (x1)(x+1)(4x^3 + 8x^2  11x  15)
Try 1 again: yay
y = (x1)(x+1)(x+1)(4x^2 + 4x  15)
y = (x1)(x+1)(x+1)(2x3)(2x+5)
I assume you can find the roots at this point...
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