I 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
List all possible rational zeros for the polynomial below. Find all real zeros of the polynomial and factor
f(x)=2x^4+19x^3+37x^2-55x-75
3 answers
any 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
±1 ±3 ±5 ±15 ±25 ±75
±1/2 ±3/2 ±5/2 ±15/2 ±25/2 ±75/2
DERRRR