Stuck while seeking $a$ with which the difference between quadratic roots is above 3

Question

The problem is:

Find the values of a with which the roots of the following inequality will form an interval longer than 3:

$$x^2-(a^2+3a+1)x+a^2+3a^3\le0$$

From Vieta's formulas, the square of the difference of the roots of a reduced quadratic ($x^2+px+q=0$) is $$(x_1-x_2)^2=p^2-4q$$

Thus, I decided for starters to derive $a$ from

$$(x_1-x_2)^2=9$$

$$p^2-4q=(a^2+3a+1)^2-4(a^2+3a^3)=a^4-6a^3+7a^2+6a+1$$

But I've no idea how to solve the

$$a^4-6a^3+7a^2+6a+1=9$$

Maybe there is a simpler way?

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P.S.
$\checkmark$ The textbook's answer is

Answer 1

You are trying to solve the equation

$$a^4 - 6a^3 + 7a^2+6a-8 = 0$$

My advice is you first try to guess at least one root of the polynomial.

Hints:

1. There exists a rational root of the above polynomial.
2. There is a very handy theorem for finding rational roots of polynomials.
3. Try to plug in the simplest possible values for $a$ that you can think of...