Find the rational zeros of the polynomial function. Then enter the function in factored form.
h(x)=6x^3-11x²-3x + 2.
The zeros are x __,__,__

The zeros are -1/2, 1/3 and 2.

The factors are (x - 2)(3x - 1)(2x + 1)

Step-by-step explanation:

h(x)= 6x^3 - 11x² - 3x + 2 = 0

As the last term is 2 we try  to see if +/-1 or +/- 2 are zeros

f(1) = -6, f(-1) = -18  so  they are not zeros.

f(2) = 6*8 - 11*4 - 3(2) + 2

= 48 - 44 - 6 + 2

=  4 - 6 + 2

= 0.  

So x = 2 is a zero and x - 2 is a factor.

Dividing by x - 2:

x - 2 ) 6x^3 - 11x² - 3x + 2 ( 6x^2 + x - 1 <Quotient

         6x^3  - 12x^2

                      x^2 - 3x

                      x^2 - 2x

                             - x+ 2

                             -x + 2

Factoring the quotient:

6x^2 + x - 1

= (3x - 1)(2x + 1) =  0

x = 1/3, x = -1/2..