Question

phân tich thành nhân tử

\(x\cdot\left(x+4\right)\cdot\left(x+6\right)\left(x+10\right)+128\)

Answer 1

\(A=x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

Đặt \(x^2+10x=a\) nên : 

 \(A=a\left(a+24\right)+128=a^2+24a+144-16=\left(a+12\right)^2-4^2=\left(a+16\right)\left(a+8\right)\)

\(=\left(x^2+10x+8\right)\left(x^2+10x+16\right)=\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)\)