You should remember that you may find a root of a polynomial among the fractions that you may form using the divisors of constant term and divisors of leading coefficient such that:

`D_2 = {+-1 ; +-2} `

`D_6 = {+-1 ; +-2 ; +-3 ; +-6}`

Hence, evaluating the set...

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You should remember that you may find a root of a polynomial among the fractions that you may form using the divisors of constant term and divisors of leading coefficient such that:

`D_2 = {+-1 ; +-2} `

`D_6 = {+-1 ; +-2 ; +-3 ; +-6}`

Hence, evaluating the set of fractions that could be a root for the given polynomial yields:

`{+-1/2 ; +-1/3 ; +- 1/6 ; +- 2/3}`

You need to try `x = 1/2` to see if it verifies the polynomial such that:

`f(1/2) = 6(1/2)^4+5(1/2)^3-11(1/2)^2-10(1/2)-2`

`f(1/2) = 6/16 + 5/8 - 11/4 - 10/2 - 2`

`f(1/2) = (6 + 10 - 44 - 80 - 32)/16 != 0`

Trying `x = -1/2` yields:

`f(-1/2) = 6/16- 5/8 - 11/4+ 10/2 - 2`

`f(-1/2) = (6- 10 - 44+ 80 - 32)/16 = 0`

Notice that `x = -1/2` verifies the polynomial, hence, `x = -1/2` is a root for the given polynomial.

Considering `x = -1/3` yields:

`f(-1/3) = 6/81 - 5/27 - 11/9 + 10/3 - 2`

`f(-1/3) = (6 - 15 -99 +270 - 162)/16 = 0`

Notice that `x = -1/3` verifies the polynomial, hence, x = -1/3 is a root for the given polynomial.

You need to write the factored form of polynomial such that:

`6x^4+5x^3-11x^2-10x-2 = 6(x - x_1)(x - x_2)(x - x_3)(x - x_4)`

You may substitute -`1/2 ` for `x_1` and -`1/3 ` for `x_2` such that:

`6x^4+5x^3-11x^2-10x-2 = 6(x+1/2)(x+1/3)(x - x_3)(x - x_4)`

`6x^4+5x^3-11x^2-10x-2 = (2x+1)(3x+1)(x - x_3)(x - x_4)`

Notice that you may open the brackets such that:

`6x^4+5x^3-11x^2-10x-2 = (6x^2 + 5x + 1)(x^2- x(x_4 + x_3)` `+ x_3x_4) `

`6x^4+5x^3-11x^2-10x-2 = 6x^4 - 6x^3(x_4 + x_3) + 6x^2x_3x_4 + 5x^3 - 5x^2(x_4 + x_3) + 5xx_3x_4 + x^2 - x(x_4 + x_3) + x_3x_4`

`6x^4+5x^3-11x^2-10x-2 = 6x^4 + x^3(5-6(x_4 + x_3)) + x^2(6x_3x_4 - 5(x_3+x_4) + 1) + x(5x_3x_4 - x_3 - x_4) + x_3x_4`

Equating coefficients of like powers yields:

`x_3x_4 = -2`

`6x_3x_4 - 5(x_3+x_4) + 1 = -11 => 6x_3x_4 - 5(x_3+x_4) = -12`

`6*(-2) - 5(x_3+x_4) = -12 => 5(x_3+x_4) = 0 => x_3+x_4 = 0`

`5x_3x_4 - x_3 - x_4 = -10 => 5x_3x_4 = -10 => x_3x_4 = -2`

Since `x_3 = -x_4 => x_3x_4 = -x^2_3 = -x^2_4`

`x_3x_4 = -2 => -x^2_3 = -2 => x^2_3 = 2 => x_3 = +-sqrt2`

`x_4 = +-sqrt2`

Hence, you may write the complete factored form of the given polynomial such that:

`f(x) = (2x + 1)(3x + 1)(x - sqrt2)(x +sqrt2)`

**Hence, evaluating the factored form of the given polynomial yields `f(x) = (2x + 1)(3x + 1)(x - sqrt2)(x + sqrt2).` **