Câu hỏi của Hân Điền

Question

Phân tích đa thức thành nhân tử:x4 + 2x3 + 2x3 + 2x + 1

Dũng · Accepted Answer

x4−2x3+2x−1x4−2x3+2x−1=x4−x3−x3+x2−x2+x+x−1=x4−x3−x3+x2−x2+x+x−1=x3(x−1)−x2(x−1)−x(x−1)+(x−1)=x3(x−1)−x2(x−1)−x(x−1)+(x−1)=(x−1)(x3−x2−x+1)=(x−1)(x3−x2−x+1)=(x−1)[Câu trả lời: x4−2x3+2x−1x4−2x3+2x−1=x4−x3−x3+x2−x2+x+x−1=x4−x3−x3+x2−x2+x+x−1=x3(x−1)−x2(x−1)−x(x−1)+(x−1)=x3(x−1)−x2(x−1)−x(x−1)+(x−1)=(x−1)(x3−x2−x+1)=(x−1)(x3−x2−x+1)=(x−1)[

ILoveMath · Answer

$x^4+2x^3+2x^2+2x+1\
=\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\
=x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\
=\left(x^3+x^2+x+1\right)\left(x+1\right)\
=\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\
=\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\
=\left(x^2+1\right)\left(x+1\right)^2$Câu trả lời: $x^4+2x^3+2x^2+2x+1\
=\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\
=x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\
=\left(x^3+x^2+x+1\right)\left(x+1\right)\
=\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\
=\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\
=\left(x^2+1\right)\left(x+1\right)^2$

Nguyễn Lê Phước Thịnh · Answer

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

Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

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