Posted by **Suzie** on Tuesday, September 27, 2011 at 11:27pm.

how to find all real zeros of "4x^5+8x^4-15x^3-23x^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 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...

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