a.\(3x^3+x^2-10x-8=\left(3x^3+3x^2\right)-\left(2x^2+2x\right)-\left(8x+8\right)\)
\(=3x^2\left(x+1\right)-2x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x^2-2x-8\right)\)
Ta lại có : \(3x^2-2x-8=\left(3x^2+4x\right)-\left(6x+8\right)=3x\left(x+\dfrac{4}{3}\right)-6\left(x+\dfrac{4}{3}\right)\)
\(=\left(x+\dfrac{4}{3}\right)\left(3x-6\right)\)
\(\Rightarrow3x^3+x^2-10x-8=\left(x+1\right)\left(x+\dfrac{4}{3}\right)\left(3x-6\right)=3\left(x+1\right)\left(x+\dfrac{4}{3}\right)\left(x-2\right)\)
a) \(3x^3+x^2-10x-8\)
\(=3x^3+3x^2-2x^2-2x-8x-8\)
\(=3x^2\left(x+1\right)-2x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x^2-2x-8\right)\)
b) \(-3x^3+4x^2+x+6\)
\(=-3x^3+6x-2x+4x-3x+6\)
\(=-\left(3x^3-6x^2+2x^2-4x+3x-6\right)\)
\(=-\left[3x^2\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(3x^2+2x+3\right)\)
c) \(x^3-4x^2+7x-4\)
\(=x^3-x^2-3x^2+3x+4x-4\)
\(=x^2\left(x-1\right)-3x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-3x+4\right)\)
