Quadratic roots within integral

$\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$ Using the above property of definite integrals , $\int_{0}^{2}(1+\cos^8x)(ax^2+bx+c)dx=\int_{0}^{1}(1+\cos^8x)(ax^2+bx+c)dx+\int_{1}^{2}(1+\cos^8x)(ax^2+bx+c)dx$ $\implies \int_{1}^{2}(1+\cos^8x)(ax^2+bx+c)dx=0$ Since the area of the the region bounded by the curve $y=f(x)=(1+\cos^8x)(ax^2+bx+c)$ , $x=1$ , $x=2$ and $x$ -axis is $0$ , $f(x)$ takes both positive and negative values for $x\in [1,2]$ .

But, $1+\cos^8x>0, \forall x\in \mathbb{R} \implies ax^2+bx+c<0$ for some $x\in [1,2]$ and $ax^2+bx+c>0$ for some $x\in [1,2]$ . So the graph of $ax^2+bx+c$ will intersect the $x$ -axis at least once between $x=1$ and $x=2$ .

Therefore, $ax^2+bx+c=0$ must have at least one root in the interval $(1,2)$ .