Question

Use the rational root theorem with polynomial division if needed to find all the zeroes of:  
  P(x)=2x^4-3x^3+3x^2+5x-3.

Answer 1

P(x) = 2x^4 - 3x^3 + 3x^2 + 5x - 3
To use rational root theorem, we get all the factors of constant divided by the factors of the leading coefficient (or the term with variable of highest degree). In this case, the constant is -3, and the leading coefficient is 2. Thus,
+/- 1,3 / 1,2
= 1, -1, 3, -3, 1/2 , -1/2 , 3/2 , -3/2
To check which of these are the possible roots, substitute each value of x to the function, and if the answer you got is zero, then that value of x is a root.
After trial and error, the values of x that you'll get are -1 and 1/2. Checking:
x = -1:
P(-1) = 2(-1)^4 - 3(-1)^3 + 3(-1)^2 + 5(-1) - 3
P(-1) = 2 + 3 + 3 - 5 - 3
P(-1) = 0

x = 1/2:
P(1/2) = 2(1/2)^4 - 3(1/2)^3 + 3(1/2)^2 + 5(1/2) - 3
P(1/2) = 1/8 - 3/8 + 3/4 + 5/2 - 3
P(1/2) = -2/8 + 3/4 + 10/4 - 12/4
P(1/2) = -1/4 + 13/4 - 12/4
P(1/2) = 0

Since the highest degree of x in the function is 4, there should be 4 roots. Divide the two factors (from the roots you found) by the function. You should get 2(x^2 - 2x + 3).
From here, you can say that the roots of this quadratic equation are imaginary. Use the quadratic formula to get the remaining roots:
Quadratic formula:
x = (-b +/- sqrt(b^2 - 4ac)) / 2a

hope this helps~ `u`