Question

phân tích đa thức thành nhân tử

(x-1)^5+(x+3)^5-242(x+1)

Answer 1

Đặt y=x+1

\(\left(x-1\right)^5=\left(y-2\right)^5\)

\(=y^5-5y^4+40y^3-80y^2+80y-32\)

\(\left(y+2\right)^5=y^5+5y^4+40y^3+80y^2+80y+32\)

\(A=\left(y+2\right)^5+\left(y-2\right)^5-242y\)

\(=2y^4+80y^3+160y-242y\)

\(=2y^4+80y^3-82y\)

\(=2y\left(y^2-1\right)\left(y^2+41\right)\)

\(=2y\left(y+1\right)\left(y-1\right)\left(y^2+41\right)\)

\(=2\left(x+1\right)\left(x+2\right)\cdot x\cdot\left[\left(x+1\right)^2+41\right]\)