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Question

Phân tích đa thức thành nhâ tử:a, x4+2002x2+2001x+2002b. x4+2007x2-2006x+2007

Phạm Thị Thu Ngân · Accepted Answer

$x^4+2002x^2+2001x+2002$

$=x^4+x^2+1+2001x^2+2001x+2001$

$=\left(x^4+2x^2+1\right)-x^2+2001\left(x^2+x+1\right)$

$=\left(x^2+1-x\right)\left(x^2+1+x\right)+2001\left(x^2+x+1\right)$

$=\left(x^2+x+1\right)\left(x^2+1-x+2001\right)$

$=\left(x^2+x+1\right)\left(x^2-x+2002\right)$Câu trả lời: $x^4+2002x^2+2001x+2002$

$=x^4+x^2+1+2001x^2+2001x+2001$

$=\left(x^4+2x^2+1\right)-x^2+2001\left(x^2+x+1\right)$

$=\left(x^2+1-x\right)\left(x^2+1+x\right)+2001\left(x^2+x+1\right)$

$=\left(x^2+x+1\right)\left(x^2+1-x+2001\right)$

$=\left(x^2+x+1\right)\left(x^2-x+2002\right)$

Phạm Thị Thu Ngân · Answer

$x^4+2007x^2-2006x+2007$

$=x^4+2x^2+1-x^2+2006\left(x^2-x+1\right)$

$=\left(x^2+1\right)^2-x^2+2006\left(x^2-x+1\right)$

$=\left(x^2+1+x\right)\left(x^2+1-x\right)+2006\left(x^2-x+1\right)$

$=\left(x^2-x+1\right)\left(x^2+x+1+2006\right)$

$=\left(x^2-x+1\right)\left(x^2+x+2007\right)$Câu trả lời: $x^4+2007x^2-2006x+2007$

$=x^4+2x^2+1-x^2+2006\left(x^2-x+1\right)$

$=\left(x^2+1\right)^2-x^2+2006\left(x^2-x+1\right)$

$=\left(x^2+1+x\right)\left(x^2+1-x\right)+2006\left(x^2-x+1\right)$

$=\left(x^2-x+1\right)\left(x^2+x+1+2006\right)$

$=\left(x^2-x+1\right)\left(x^2+x+2007\right)$

Đại số lớp 8