Cho hàm số y=f(x)=mx^4+nx^3+px^2+qx+r trong đó

Ta đặt $y=f^{\text{'}}\left(x\right)=k\left(x+2\right)\left(x-\frac{7}{6}\right)\left(x-3\right)$ .

Xét $\left\{\begin{array}{l}{S}_1=\left|k\underset{0}{\overset{\frac{7}{6}}{\int }}\left(x+2\right)\left(x-\frac{7}{6}\right)\left(x-3\right)d\text{x}\right|=\frac{65219}{1552}k\\ S_2=\left|k\underset{\frac{7}{6}}{\overset{3}{\int }}\left(x+2\right)\left(x-\frac{7}{6}\right)\left(x-3\right)d\text{x}\right|=\frac{65219}{1552}k\end{array}\right.$ .

Do đó:$S_1=S_2=\underset{0}{\overset{\frac{7}{6}}{\int }}{f}^{\text{'}}\left(x\right)d\text{x}=-\underset{\frac{7}{6}}{\overset{4}{\int }}{f}^{\text{'}}\left(x\right)d\text{x}\iff f\left(0\right)=f\left(3\right)$ .

Lập bảng biến thiên ta suy ra phương trình $f\left(x\right)=r=f\left(0\right)$  có tất cả 3 nghiệm.