Try This By Hand

We claim that $P(x)>0$ for all real $x$ .

• If $x\leq 0$ , then it's obvious that $P(x)>0$ .

• If $x\geq1$ , then $x^4-x \geq0$ . It immediately follows that $P(x)>0$ .

• If $0<x<1$ , then $-x+1>0$ and that implies that $P(x)>0$ .

This proves our claim. $_\square$

And we can say that $P(x)$ has $\boxed{0}$ real roots.

$P(x) = x^4 + 2x^2 - x + 1 = x^4 + 2 \left( x - \frac{1}{4} \right)^2 + \frac{7}{8} > 0$ for all $x$ , so no real roots.

Note that $P(x)=(x^2+1)^2-x$ . Thus, if we prove that $(x^2+1)^2-x\ge 0 \iff (x^2+1)^2\ge x$ then there are no real roots.

If $x < 0$ , then this is trivial because the $LHS$ is positive and the $RHS$ is negative.

If $x\ge 0$ , then we can take the square root and get $x^2+1\ge \sqrt{x}$

Note that $\left(x-\dfrac{1}{2}\right)^2\ge 0 \iff x^2-x+\dfrac{1}{4}\ge 0\iff x^2+1\ge x+\dfrac{3}{4}$

Also, $\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge 0\iff x-\sqrt{x}+\dfrac{1}{4}\ge 0\iff x+\dfrac{1}{4}\ge \sqrt{x}$

Putting these two inequalities together, we get $x^2+1\ge x+\dfrac{3}{4} > x+\dfrac{1}{4}\ge \sqrt{x}$ so $(x^2+1)^2 > x$ so there are $\boxed{0}$ real solutions to the original polynomial.

$P(x) = (x^2 + 1)^2 - x > 0$ for all $x<0$ . For $x>0$ , $f'(x) > 0 , f(0) > 0$ so no real solution.

$P(x)=x^4+x^2+(x-\frac 1 2 )^2+\frac 3 4 >0 \text{ since all first three terms} \geq 0.$

$P(x)=x^4 + 2(x - \frac {1} {4})^2 + \frac {7} {8}$

We have $x^4 > 0$ and $2(x-\frac {1} {4})^2 > 0$ for all real x, hence $P(x) > 0$ for all real x.

Thus $P(x)$ has no real root.