Question

 

Phân tích đa thức thành nhân tử

$x^5+x+1$

$x^{11}+x^4+1$

$6x^4-11x^2+3$

$xy(x-y)+yz(y-z)+zx(z-x)$

Answer 1

\(x^5+x+1\)

\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

\(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

\(x^{11}+x^4+1\)

\(=x^{11}+x^{10}+x^9-x^{10}-x^9-x^8+x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^4+x^2+1\)

\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2\right)+\left(x^4+x^2+1\right)\)

\(=\left(x^4+2x^2+1-x^2\right)+\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2\right)\)

\(=\left(x^2+1+x\right)\left(x^2+1-x\right)+\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2\right)\)

\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2+x^2-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x+1\right)\)

\(6x^4-11x^2+3\)

\(=6x^4-9x^2-2x^2+3\)

\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)