List all possible rational zeros for the polynomial below. Find all real zeros and factor completely. Please show all your work.
f(x) = 2x^4 + 19x^3 + 37x^2 - 55x - 75
4 answers
Graph this first to find the zeros, hopefully exactly.
There is a process to this and it's why I am asking for help.
All 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
±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
I have done synthetic divion and come up with 3/2 and -1 but still have not gotten it down to a trinomial.
3 answers