Question

Phân tích đa thức thành nhân tử: \(M=x^4+6x^3+7x^2-6x+1\)

Answer 1

xét \(x\ne0\)ta có :

\(M=\)\(^{x^2\cdot\left(x^2+6x+7-\frac{6}{x}+\frac{1}{x^2}\right)}\)

Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2-2+\frac{1}{x^2}\Leftrightarrow t^2+2=x^2+\frac{1}{x^2}\)

Do đó \(M=x^2\cdot\left(t^2+2+6t+7\right)\Leftrightarrow x^2\cdot\left(t^2+6t+9\right)\)

\(\Leftrightarrow M=x^2\cdot\left(t+3\right)^2\)

Answer 2

M=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)

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

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