Show that the polynomial $P(x)=x^4-x^2-x+2$ has no real roots

Question

Using clever algebra show that the polynomial $$P(x)=x^4-x^2-x+2$$ has no real roots.

Obviously, we can not use the derivative.

Using the general quartic formula is terrible.

I tried

$$(x^2+1)^2-3x^2-x+1=(x^2-x\sqrt 3+1)(x^2+1+x\sqrt 3)-x+1$$

But this didn't work.

Also factorisation doesn't work.

Finally I solved my question.

Note that, $P(-x)\ge P(x)$ holds for all $x\ge 0$.

So it is enough to show that $P(x)\ge 0$, for all $x\ge 0$.

$$P(x)=(x^2-1)^2+(x-1)^2+x>0.$$

So there are no real roots.

Answer 1

Why factorisation doesn't work?

Can you turn P(x) into sums of two square + positive constant form?

$P(x)=x^4-2x^2+x^2-x+2=....$

Answer 2

Alternative approach:

$f(x) = x^4 - x^2 - x + 2 \implies $

• $[E_1] ~: ~f(x) = x^4 - [(x + 2)(x-1)].$

• $[E_2] ~: ~f'(x) = 4x^3 - 2x - 1.$

Examining $E_1,~$ you have that for $~-2 \leq x \leq 1,~$
the term $~[(x + 2)(x-1)]~$ is non-positive.

Further, $f(0) > 0.$
So, you can immediately conclude that for $~-2 \leq x \leq 1,$
$f(x)~$ must be positive.

Now, use $E_2$ to consider the derivative in the two ranges $x \geq 1$ and $x \leq -2$.

For $~x \geq 1,~f'(x)~$ is clearly positive.
So, $f(1) > 0$ and $~f(x)~$ is strictly increasing for $~x \geq 1.$

For $x \leq -2$, you have the mirror analysis :
$f'(x)$ is clearly negative, for all $x \leq -2$ and $f(-2) > 0.$

So, for $~x < -2,~$ you must have that $~f(x)~$ is positive and is strictly decreasing down to $f(-2)$, which is also positive.

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The above analysis shows that it is impossible for $f(x)$ to equal $(0)$ for:

• $-2 \leq x \leq 1.$

• $1 < x.$

• $x < -2.$