Question

What is the remainder to x^3+4x^2+x-6 divided by (x-1)

Answer 1

$x^3=x^2\cdot x$, and $x^2(x-1)=x^3-x^2$. This gives a remainder of

$(x^3+4x^2+x-6)-(x^3-x^2)=5x^2+x-6$

$5x^2=5x\cdot x$, and $5x(x-1)=5x^2-5x$. This gives a new remainder of

$(5x^2+x-6)-(5x^2-5x)=6x-6$

$6x=6\cdot x$, and $6(x-1)=6x-6$. This gives a new remainder of

$(6x-6)-(6x-6)=0$

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 $p(x)$ by $x-c$ is $p(c)$. Here we have

$p(x)=x^3+4x^2+x-6$

$c=1\implies p(c)=1+4+1-6=0$