Question

已知多项式x^4-x^3+kx^2+x-6,能被x-1整除,求k的值和商式

Answer 1

设相除的商是A

则x^4-x^3+kx^2-x-6=A(x-1)

x=1,右边=0,所以左边等于0

所以1-1+k-1-6=0

k=7

x^4-x^3+7x^2-x-6

=x^3(x-1)+(7x+6)(x-1)

=(x-1)(x^3+7x+6)

所以商式是x^3+7x+6