All roots of the quartic equation $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real

Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But then the same is true of $f(x-\frac14) = x^4 + \frac58 x^2 + Bx + A$ for some $B$ and $A$ (we don't need the formula). If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$ then $a^2+b^2+c^2+d^2 = -2\frac58 < 0$, so $a,b,c,d$ cannot all be real, QED.

Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But $f''(x) = 12x^2 + 6x + 2 > 0$ for all $x$. Therefore $f$ is convex upwards and hence can have at most two real roots, QED.

Assume contrariwise that $f(x)=ax^4+bx^3+x^2+x+1$ has 4 distinct real zeros for some choice of $a,b$. Then $$ g(x)=f(x)(x-1)=ax^5+(b-a)x^4+(1-b)x^3-1 $$ has either five distinct zeros, or a double zero at $x=1$ together with three other zeros.

But $$ g'(x)=5ax^4+4(b-a)x^3+3(1-b)x^2 $$ has a double zero at $x=0$, and at most two other zeros.

Rolle's theorem tells us that the derivative of a differentiable function has a zero between any two of its zeros. This rules out the possibility of $g(x)$ having five distinct zeros.

The remaining possibility is that $g(x)$ has a double zero at $x=1$ and three other zeros. This means that $x=1$ is one of the zeros of $g'(x)$, so by Rolle's theorem $g'(x)$ should have three zeros different from $x=1$. We saw that this cannot be the case, so we have arrived at a contradiction.

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How to see this? The fixed part of the three lowest degree terms of the polynomial $f(x)$ is known to be a factor of $x^3-1$. With nothing else to go by, I decided to check out, where using that bit takes us.