Phân tích đa thức thành nhân tử:
x4 + 8x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
\(8x^2-2x-3=8x^2+4x-6x-3=4x\left(2x+1\right)-3\left(2x+1\right)=\left(4x-3\right)\left(2x+1\right)\)
\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)
\(=4x^2-8x+4-3\)
\(=\left(2x-2-\sqrt{3}\right)\left(2x-2+\sqrt{3}\right)\)
\(14x^2-14xy-8x+8y=14x\left(x-y\right)-8\left(x-y\right)=\left(x-y\right)\left(14x-8\right)\)
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
\(x^4+8x=x\left(x+2\right)\left(x^2-2x+4\right)\)