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October 1, 2014

October 1, 2014

Posted by **JB** on Monday, January 2, 2012 at 11:39pm.

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

- Rational Zeros -
**Reiny**, Monday, January 2, 2012 at 11:47pmI tried ±1, ±3, ±5

found -1 and -5 to be zeros

So after synthetic division by x+1 and x+5

I got

2x^4+19x^3+37x^2-55x-75 = (x+1)x+5)(2x^2 + 7x - 15)

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

so we have a double zero at -5, and zeros at -1 and 3/2

- Rational Zeros -
**Steve**, Tuesday, January 3, 2012 at 12:15amany possible rational zeroes would have a numerator which divides 75 and a denominator which divides 2. So, the list would be

±1 ±3 ±5 ±15 ±25 ±75

±1/2 ±3/2 ±5/2 ±15/2 ±25/2 ±75/2

- Rational Zeros -
**mike**, Tuesday, April 24, 2012 at 12:55amDERRRR

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