Prove that the polynomial $p(x)=\sqrt{2}x^3+4x^2+\frac{1}{7}x+10$ has a root in $\mathbb{R}$

Question

Prove that the polynomial $p(x)=\sqrt{2}x^3+4x^2+\frac{1}{7}x+10$ is continuous on $\mathbb{R}$ and has a root in $\mathbb{R}$

Since $p(x)$ is polynomial by using algebraic properties of continuous the function is continue

how to find it has root in $\mathbb{R}$ I am also trying to find using mean value theorem but not get it

Answer 1

With the new problem, the mean value theorem will tell you there is a root. The question does not ask to find it. If $x$ is huge and negative, the polynomial will be negative. Just evaluate it at $x=-100$ and see that $p(-100)\lt 0$. Then note $p(0)=10$ and you know there is a root in there somewhere.

Answer 2

There is an obvious algebraic answer, hinted to above. This one uses analytic geometry.

Recall that $p(x) = ax^2+bx+c$ is concave up iff $a>0$, and here $a=4$, so this is a parabola opening up, with a unique minimum point, which occurs at $$m = \frac{-b}{2a}=\frac{-1/7}{2\cdot 4} = -\frac{1}{56},$$ and $$ p(m) = 10 + \sqrt{2} + \frac{4}{56} - \frac{1}{7 \cdot 56} > 9+\sqrt{2}>0, $$ so the minimum point of $p$ is strictly above the $x$-axis, hence $p$ has no real roots.

Answer 3

Let $x_0>0$ be large enough such that $p(x_0)>0$ and $p(-x_0)<0$, and Intermediate value theorem implies that there exists a real root for the polynomial $p(x).$