$\text{ If }\mathrm{a},\mathrm{b},\mathrm{c}\text{ are non }-\text{ zero real numbers such that }$$\underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=$

${\int }_0^2 \left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=0,\text{ then the equation }a{x}^2+bx+c=0\text{ will have }$ 

$f\left(x\right)={\int }_0^x \left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$

$\begin{array}{r}\begin{array}{rl} & f\left(0\right)=0,f\left(1\right)=0,f\left(2\right)=0\\ & f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow \left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)=0\end{array}\\ \Rightarrow a{x}^2+bx+c=0\text{ it gives two roots }\end{array}$

$f\left(0\right)=0=f\left(1\right)\text{ and }{f}^{\mathrm{\prime }}\left(x\right)=0\Rightarrow \text{ at least one }\mathrm{x}\text{ between }0\text{ and }1$

$f\left(0\right)=0=f\left(2\right)\text{ and }{f}^{\mathrm{\prime }}\left(x\right)=0\Rightarrow \text{ at least one }\mathrm{x}\text{ between }1\text{ and }2$