Mọi người ơi cíu tớ với
a, \(x^3-3x+2=\left(x^3+2x^2\right)-\left(2x^2+4x\right)+\left(x+2\right)\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+1\right)\)
\(=\left(x+2\right)\left(x-1\right)^2\)
b, \(2x^3+5x^2+4x+1=\left(2x^3+x^2\right)+\left(4x^2+2x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+4x+1\right)\)
c,
\(4x^4-3x^3-5x^2+3x+1\\ =\left(4x^4-4x^3\right)+\left(x^3-x^2\right)-\left(4x^2-4x\right)-\left(x-1\right)\)
\(=4x^3\left(x-1\right)+x^2\left(x-1\right)-4x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(4x^3+x^2-4x-1\right)\)
\(=\left(x-1\right)\left[\left(4x^3-4x^2\right)+\left(5x^2-5x\right)+\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[4x^2\left(x-1\right)+5x\left(x-1\right)+\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(4x^2+5x+1\right)\)
\(=\left(x-1\right)^2\left[\left(4x^2+x\right)+\left(4x+1\right)\right]\)
\(=\left(x-1\right)^2\left[x\left(4x+1\right)+\left(4x+1\right)\right]\)
\(=\left(x-1\right)^2\left(4x+1\right)\left(x+1\right)\)
