Câu hỏi của Alan Walker

Question

Phân tích đa thức thành nhân tử

x4+2008x2+2007x+2008

x5-x4+3x3+3x2-x+1

ngonhuminh · Accepted Answer

$A=x^5-x^4+3x^3+3x^2-x+1$

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

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

$A=\left(x+1\right)\left(x^4+2x^3+5x^2-2x+1\right)$Câu trả lời: $A=x^5-x^4+3x^3+3x^2-x+1$

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

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

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

Nhã Doanh · Answer

$x^4+2008x^2+2007x+2008$

$=x^4+2008x^2+2008x-x+2008$

$=x\left(x^3-1\right)+2008\left(x^2+x+1\right)$

$=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)$

$=\left(x^2+x+1\right)\left(x^2-x+2008\right)$Câu trả lời: $x^4+2008x^2+2007x+2008$

$=x^4+2008x^2+2008x-x+2008$

$=x\left(x^3-1\right)+2008\left(x^2+x+1\right)$

$=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)$

$=\left(x^2+x+1\right)\left(x^2-x+2008\right)$

Violympic toán 8