Question

Chứng minh đa thức \(x^4+x^3+4x^2+3x+3\) vô nghiệm

Answer 1

\(=\left(x^4+x^3+x^2\right)+\left(3x^2+3x+3\right)=x^2\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)

\(=\left(x^2+3\right)\left(x^2+x+1\right)=\left(x^2+3\right)\left(x^2+2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

\(=\left(x^2+3\right)\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

vì \(x^2>=0;3>0\Rightarrow x^2+3>0\)

\(\left(x+\frac{1}{2}\right)^2>=0;\frac{3}{4}>0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\left(x^2+3\right)\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)>0\Rightarrow\)đa thức trên vô nghiệm