\(B=x\left(x-3\right)\left(x+1\right)\left(x+4\right)\)
\(B=\left[x\left(x+1\right)\right]\left[\left(x-3\right)\left(x+4\right)\right]\)
\(B=\left(x^2+x\right)\left(x^2+x-12\right)\)
Đặt \(x^2+x=a\)ta được;
\(B=a\left(a-12\right)=a^2-12a=\left(a^2-2.a.6+36\right)-36\)\(=\left(a-6\right)^2-36\)
Vì \(\left(a-6\right)^2\ge0\)\(\Rightarrow\left(a-6\right)^2-36\ge-36\)
Dấu ''='' xảy ra khi \(a-6=0\Rightarrow a=6\Rightarrow x^2+x-6=0\)\(\Rightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\)\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy GTNN của B là B=-36 khi x=-3 hoặc x=2
