Question

tìm đa thức dư trong phép chia đa thức f(x)=x^312-x^313+x^314-x^315+x^446 cho đa thức g(x)=x^2-1

 

 

Answer 1

\(f\left(x\right)=x^{312}-x^{313}+x^{314}-x^{315}+x^{446}\)

\(=\left(x^{312}-1\right)-\left(x^{313}-x\right)+\left(x^{314}-1\right)-\left(x^{315}-x\right)+\left(x^{446}-1\right)-2x+3\)

\(=\left[\left(x^2\right)^{156}-1\right]-x\left[\left(x^2\right)^{156}-1\right]+\left[\left(x^2\right)^{157}-1\right]-x\left[\left(x^2\right)^{157}-1\right]+\left[\left(x^2\right)^{223}-1\right]-2x+3\)

\(=\left(x^2-1\right)A_{\left(x\right)}-x\left(x^2-1\right)B_{\left(x\right)}+\left(x^2-1\right)C_{\left(x\right)}-x\left(x^2-1\right)D_{\left(x\right)}+\left(x^2-1\right)E_{\left(x\right)}-2x+3\)

Vậy số dư là -2x+3