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`