ANSWER TO QUESTION 1
Check attachment for long division.
From our long division we can write the following;
[tex]\frac{8x^4-6x^3+7x^2-11x+10}{2x-1} =4x^3-x^2+3x-4+\frac{6}{2x-1}[/tex]
Since the polynomial
[tex]8x^4-6x^3+7x^2-11x+10[/tex] leaves a non zero  remainder of [tex]6[/tex] when divided by [tex]2x-1[/tex], we conclude that [tex]2x-1[/tex] not a factor of the dividend,
ANSWER TO QUESTION 2
We want to factor
[tex]f(x)=x^4+x^3-8x^2+6x+36[/tex]
The possible rational roots are;
[tex]\pm1, \pm 2,\pm 3, \pm 4,\pm 6,\pm 9, \pm18, \pm 36[/tex]
We found
[tex]f(-2)=(-2)^4+(-2)^3-8(-2)^2+6(-2)+36[/tex]
[tex]f(-2)=16+-8-32+-12+36[/tex]
[tex]f(-2)=-36+36[/tex]
[tex]f(-2)=0[/tex]
and
[tex]f(-3)=(-3)^4+(-3)^3-8(-3)^2+6(-3)+36[/tex]
[tex]f(-3)=81-27-72-18+36[/tex]
[tex]f(-3)=-36+36[/tex]
[tex]f(-3)=0[/tex].
This means that [tex](x+2)[/tex] and [tex](x+3)[/tex] are factors of the polynomial.
This also means that
[tex](x+2)(x+3)=x^2+5x+6[/tex] is also a factor of the polynomial
So we apply long division to obtain the remaining factors as shown in the attachment.
[tex]\Rightarrow f(x)=x^4+x^3-8x^2+6x+36=(x+2)(x+3)(x^2-4x+6)[/tex]
We factor further to obtain;
[tex]\Rightarrow f(x)=x^4+x^3-8x^2+6x+36=(x+2)(x+3)(x-(2-\sqrt{2}i))(x-(2+\sqrt{2}i))[/tex]
