Irrational Roots

Upon expansion, the equation becomes

$9x^{4} - 30x^{3} + 25x^{2} - 25 = 0 \Longrightarrow (3x^{2} - 5x)^{2} - 25 = 0 \Longrightarrow (3x^{2} - 5x + 5)(3x^{2} - 5x - 5) = 0.$

Now $3x^{2} - 5x + 5 = 0$ yields no real, (and hence no irrational), solutions since the discriminant is less than zero.

With $3x^{2} - 5x - 5 = 0$ we have solutions $x = \dfrac{5 \pm \sqrt{85}}{6},$ and so the answer is $\boxed{2}.$

$(x-1)(x-2)(3x-2)(3x+1) = 21$

$[(x-1)(3x-2)][(x-2)(3x+1)] = 21$

$[3x^2 - 5x + 2][3x^2 - 5x - 2] = 21$

Let $y = 3x^2 - 5x$

$[y + 2][y - 2] = 21$

$y^2 - 4 = 21$

$y^2 = 25$

$y = 5$ or $y = -5$

Thus: $3x^2 - 5x = 5$ or $3x^2 - 5x = -5$ . The first equation yields two real solutions: $\frac{5 + \sqrt{85}}{6}$ and $\frac{5 - \sqrt{85}}{6}$ . The second equation's discriminant is less than zero, so it has no real solutions.

Therefore, the equation $(x-1)(x-2)(3x-2)(3x+1) = 21$ yields 2 irrational roots altogether.