Question

What are the possible rational zeros of f(x) = x4 + 6x3 − 3x2 + 17x − 15?

Answer 1

$f(x)=x^4+6x^3-3x^2+17x-15\\\\15:\{\pm1;\ \pm3;\ \pm5;\ \pm15\}\\1:\{\pm1\}\\\\Answer:\boxed{\{\pm1;\ \pm3;\ \pm5;\ \pm15\}}$

Answer 2

Consider the polynomial $f(x) = x^4 + 6x^3 - 3x^2 + 17x- 15.$

The rational zeros could be only of form c/d, where c is integer number among divisors of the last term (-15) and d is integer number among divisors of the first term (1).

The divisors of -15 are: $\pm 1, \pm 3, \pm 5, \pm 15.$

The divisors of 1 are: $\pm 1.$

Possible rational zeros: $\pm 1, \pm 3, \pm 5, \pm 15.$

Check them:

$f(1)=1^4 + 6\cdot 1^3 - 3\cdot 1^2 + 17\cdot 1- 15=1+6-3+17-15=6\neq 0;$

$f(-1)=(-1)^4 + 6\cdot (-1)^3 - 3\cdot (-1)^2 + 17\cdot (-1)- 15=1-6-3-17-15=-40\neq 0;$

$f(3)=3^4 + 6\cdot 3^3 - 3\cdot 3^2 + 17\cdot 3- 15=81+162-27+51-15=252\neq 0;$

$f(-3)=(-3)^4 + 6\cdot (-3)^3 - 3\cdot (-3)^2 + 17\cdot (-3)- 15=81-162-27-51-15=-174\neq 0;$

$f(5)=5^4 + 6\cdot 5^3 - 3\cdot 5^2 + 17\cdot 5- 15=625+750-75+85-15=1370\neq 0;$

$f(-5)=(-5)^4 + 6\cdot (-5)^3 - 3\cdot (-5)^2 + 17\cdot (-5)- 15=625-750-75-85-15=-300\neq 0;$

$f(15)=15^4 + 6\cdot 15^3 - 3\cdot 15^2 + 17\cdot 15- 15=50625+20250-675+255-15=70440\neq 0;$

$f(-15)=(-15)^4 + 6\cdot (-15)^3 - 3\cdot (-15)^2 + 17\cdot (-15)- 15=50625-20250-675-255-15=29430\neq 0.$

There are no rational zeros.