You know that a polynomial can be writen in function of it's roots like:
[tex]ax^4+bx^3+cx^2+dx+e=(x-x_1)(x-x_2)(x-x_3)(x-x_4)[/tex]
x1, x2, x3 and x4 are the roots. So you have that 1 and 2 are roots, so part of the polynomial is:
[tex](x-1)\cdot(x-2)=x^2-3x+2[/tex]
If this 2 degree polynomial represent's 2 of the 4 roots of the original polynomial, it means that the original one can be divided by this one. Notice that if we do that, we will get a 2 degree polynomial, which roots are very easy to find. Let's divide:
So, the new polynomial you have to find the roots of is:
[tex]12x^2+x-1[/tex]
Now for this one is very easy. We just use the formula:
[tex]\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where a, b and c are the coefficients of the polynomial:
[tex]ax^2+bx+c[/tex]
For our polynomial, a = 12, b = 1 and c = -1. Let's call this new roots x3 and x4:
[tex]x_3=\frac{-1+\sqrt[]{1^2-(4\cdot12\cdot(-1))}}{2\cdot12}=\frac{-1+\sqrt[]{1+48}}{24}=\frac{-1+\sqrt[]{49}}{24}=\frac{-1+7}{24}=\frac{1}{4}=0.25[/tex][tex]x_3=\frac{-1-\sqrt[]{1^2-(4\cdot12\cdot(-1))}}{2\cdot12}=\frac{-1-\sqrt[]{1+48}}{24}=\frac{-1-\sqrt[]{49}}{24}=\frac{-1-7}{24}=-\frac{1}{3}\cong0.333[/tex]
So, the roots of the given polynomial, other than 1 and 2, are 1/4 and 1/3
