Question

phân tích đa thức sau thành nhân tử: x\(^4\)-4x\(^3\)+8x+3

Mấy chế giúp e ạ!!!!!!!!!

Answer 1

Đặt

\(A=x^4-4x^3+8x+3\)

Giả sử 

\(A=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

\(=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)

\(=x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd\)

\(\left[\begin{array}{nghiempt}a+c=-4\\b+ac+d=0\\ad+bc=8\\bd=3\end{array}\right.\)

\(\left[\begin{array}{nghiempt}a=-2\\b=-3\\c=-2\\d=-1\end{array}\right.\)

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

 

Answer 2

dài dòng

\(x^4-4x^3+8x+3=x^4-2x^3-2x^3-x^2+4x^2-3x^2+2x+6x+3\)
\(=\left(x^4-2x^3-x^2\right)-\left(2x^3-4x^2-2x\right)-\left(3x^2-6x-3\right)\)

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

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