How do I solve this Pre-Calculus problem?

Question

List all possible rational zeros of f (x)= 2x^3-4x^2+5x-3. Then determine the rational zeros.

Answer 1

There is a rational zero test.

All rational solutions to f(x) = 0 will be contained in the set of ± r / ± s where
r = the factors of the coefficient xn = the highest power term s = the factors of the coefficient of x0 = the constant term

 

For f (x)= 2x^3-4x^2+5x-3   

x(n) = 2   x(0) = - 3  

So the possible rational solutions are   1 , -1 , 2, -2 , 3, -3 , 2/3 , -2/3

 

Turns out x = 1 is a rational solution

 

Answer 2

By the Rational Zero Theorem if p/q is a rational zero of f(x), then p is a divisor of the constant term, -3, and q is a divisor of the leading coefficient, 2.

 

So, p/q = ±3, ±1, ±1/2, ±3/2

 

By inspection, we see that 1 is a root.  Divide synthetically by x-1:

 

1⌋    2      -4      5      -3

    _______2__-2____3

       2      -2     3        0

 

So, f(x) = (x-1)(2x² - 2x + 3)

 

  2x²-2x+3 has imaginary roots, so the only rational zero is 1