\(2x^4+3x^3-9x^2-3x+2\)
\(=2x^4+5x^3-2x^2-2x^3-5x^2+2x-2x^2-5x+2\)
\(=x^2\left(2x^2+5x-2\right)-x\left(2x^2+5x-2\right)-\left(2x^2+5x-2\right)\)
\(=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
b/
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4-4x^3-x^2+x^3-4x^2-x-x^2+4x+1\)
\(=x^2\left(x^2-4x-1\right)+x\left(x^2-4x-1\right)-\left(x^2-4x-1\right)\)
\(=\left(x^2+x-1\right)\left(x^2-4x-1\right)\)
c/
\(x^4-6x^3+12x^2-14x+3\)
\(=x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3\)
\(=x^2\left(x^2-4x+1\right)-2x\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)\)
\(=\left(x^2-2x+3\right)\left(x^2-4x+1\right)\)
e/
Đề sai, sao có 2 hạng tử chứa \(x^4\) thế kia?
\(x^4-8x+63=x^4+16x^2+64-16x^2-8x-1=\left(x^2+8\right)^2-\left(4x+1\right)^2=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
\(c,=\left(x^2-3x+2\right)^2-\left(x+1\right)^2=\left(x^2-4x+1\right)\left(x^2+2x+3\right)\)
