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December 19, 2014

Posted by **edward** on Tuesday, December 20, 2011 at 8:07pm.

f(x) = 2x^4 + 19x^3 + 37x^2 - 55x - 75

- algebra -
**Damon**, Tuesday, December 20, 2011 at 8:34pmGraph this first to find the zeros, hopefully exactly.

- algebra -
**edward**, Tuesday, December 20, 2011 at 9:36pmThere is a process to this and it's why I am asking for help.

- algebra -
**Steve**, Wednesday, December 21, 2011 at 11:17amAll rational roots will have a numerator which is a factor of 75, and a denominator which is a factor of 2.

±1 ±3 ±5 ±15 ±25 ±75

±1/2 ±3/2 ±5/2 ±15/2 ±25/2 ±75/2

Descartes' Rule of Signs says that there are at most

1 positive real root

2x^4 + 19x^3 + 37x^2 - 55x - 75 has one change of sign

2 negative real roots

2x^4 - 19x^3 + 37x^2 + 55x - 75

So, we either have a double root, or some complex roots.

A little synthetic division reveals that there are no real roots greater than 2 or less than -6.

A little more produces the root -1

(x+1)(2x^3 + 17x^2 + 20x - 75)

(x+1)(2x-3)(x+5)^2

- algebra -
**DeAnn**, Thursday, April 26, 2012 at 8:45pmI have done synthetic divion and come up with 3/2 and -1 but still have not gotten it down to a trinomial.

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