\(B\left(x\right)=\left(x+2\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(B\left(x\right)=\left(x+2\right)\left(2x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\2x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Ta có:
\(B\left(x\right)=\left(x+2\right)\left(2x+5\right)\)
\(B\left(x\right)=0\Leftrightarrow\left(x+2\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy đa thức đã cho có nghiệm \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
B(x)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{5}{2}\end{matrix}\right.\)