Câu hỏi của Trần Thu Phương

Question

Phân tích thành nhân tử :$x^4+2004x^2+2003x+2004$

Pham Van Hung · Accepted Answer

$x^4+2004x^2+2003x+2004$$=x^4-x+2004x^2+2004x+2004$$=x\left(x^3-1\right)+2004\left(x^2+x+1\right)$$=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left(x^2-x+2004\right)$Câu trả lời:        $x^4+2004x^2+2003x+2004$$=x^4-x+2004x^2+2004x+2004$$=x\left(x^3-1\right)+2004\left(x^2+x+1\right)$$=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left(x^2-x+2004\right)$

Trần Thùy Dương · Answer

$x^4+2004x^2+2003x+2004$$=x^4+2004x^2+2004x-x+2004$$=\left(x^4-x\right)+2004\left(x^2+x+1\right)$$=x\left(x^3-1\right)+2004\left(x^2+x+1\right)$$=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left(x^2-x+2004\right)$Câu trả lời: $x^4+2004x^2+2003x+2004$$=x^4+2004x^2+2004x-x+2004$$=\left(x^4-x\right)+2004\left(x^2+x+1\right)$$=x\left(x^3-1\right)+2004\left(x^2+x+1\right)$$=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left(x^2-x+2004\right)$

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