x3+5x2-4x-20=0
=>(x3-4x)+(5x2-20)=0
=>x(x2-4)+5(x2-4)=0
=>(x2-22)(x+5)=0
=>(x-2)(x+2)(x+5)=0
=>x=2 hoặc x=-2 hoặc x=-5
\(x^3+5x^2-4x-20=0\)
<=> \(x^3+2x^2+3x^2+6x-10x-20=0\)
<=> \(\left(x+2\right)\cdot\left(x^2+3x-10\right)=0\)=> x+2=0 hoặc
\(x^2+3x-10=0\)
<=> x=-2 hoặc x=-2 hặc x=-5
vậy tâp nghiệm : S={-2,-5,2}
\(x^3+5x^2-4x-20=0\)
\(\Leftrightarrow\left(x^3+5x^2\right)-\left(4x+20\right)=0\)
\(\Leftrightarrow x^2\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\x-2=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\\x=-2\end{array}\right.\)
