Question

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 __,__,__

Answer 1

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

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

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..