What are all the possible rational zeros for #f(x)=3x^3-8x^2-2x-3# and how do you find all zeros?
1 Answer
Explanation:
#f(x) = 3x^3-8x^2-2x-3#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/3, +-1, +-3#
In addition note that the signs of the coefficients of
So check the possible positive rational zeros first:
#f(1/3) = 3(1/27)-8(1/9)-2(1/3)-3 = (1-8-6-27)/9= -40/9#
#f(1) = 3-8-2-3 = -10#
#f(3) = 3(27)-8(9)-2(3)-3 = 81-72-6-3 = 0#
So
#3x^3-8x^2-2x-3 = (x-3)(3x^2+x+1)#
Use the quadratic formula to solve
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-1+-sqrt(1^2-4(3)(1)))/(2*3)#
#color(white)(x) = (-1+-sqrt(1-12))/6#
#color(white)(x) = (-1+-sqrt(-11))/6#
#color(white)(x) = -1/6+-sqrt(11)/6i#