Question

Phân tích đa thức sau thành nhân tử

a) \(x^8+98x^4+1\)

b)\(x^7+x^5+1\)

Answer 1

a) \(x^8+98x^4+1=\left(x^8+2x^4+1\right)+96x^4\) \(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)

\(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)\)

\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^2-1\right)^2\)

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

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

Answer 2

b))

Answer 3

b) \(x^7+x^5+1\)

\(=\left(x^7-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

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

\(=\left(x^2+x+1\right)\left(x-1\right)\left(x^4+x\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[\left(x^5-x^4+x^2-x\right)+\left(x^3-x^2\right)+1\right]\)

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