[tex]x^3=x^2\cdot x[/tex], and [tex]x^2(x-1)=x^3-x^2[/tex]. This gives a remainder of
[tex](x^3+4x^2+x-6)-(x^3-x^2)=5x^2+x-6[/tex]
[tex]5x^2=5x\cdot x[/tex], and [tex]5x(x-1)=5x^2-5x[/tex]. This gives a new remainder of
[tex](5x^2+x-6)-(5x^2-5x)=6x-6[/tex]
[tex]6x=6\cdot x[/tex], and [tex]6(x-1)=6x-6[/tex]. This gives a new remainder of
[tex](6x-6)-(6x-6)=0[/tex]
and so there is no remainder.
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Quicker method: Use the polynomial remainder theorem, which says the remainder upon dividing a polynomial [tex]p(x)[/tex] by [tex]x-c[/tex] is [tex]p(c)[/tex]. Here we have
[tex]p(x)=x^3+4x^2+x-6[/tex]
[tex]c=1\implies p(c)=1+4+1-6=0[/tex]
