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\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=x²(x-3)-4x+3.4
=x²(x-3)-4(x+3)
=x²(x-3)+4(x-3)
=(x-3)(x²+4)
=(x-3)(x²+2²)
=(x-3)(x-2)(x+2)
\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
\(=x^2\left(x+y\right)-\left(x+y\right)=\left(x^2-1\right)\left(x+y\right)=\left(x-1\right)\left(x+1\right)\left(x+y\right)\)
\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Bài 2:
Sửa đề: \(x^3-3x^2-10x=0\)
\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)
\(x^3-4x^2+8x-8=x^2\left(x-2\right)-2x\left(x-2\right)+4\left(x-2\right)=\left(x-2\right)\left(x^2-2x+4\right)\)
\(x^3-4x^2+8x-8\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+4\right)\)
sửa lại
x^2+x-6
=x2+2x-3x-6
=x(x+2)-3(x+2)
=(x+2)(x-3)
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3.\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)