Question

when the polynomial f(x) is divided by (x-2) the remainder is 4, and when it is divided by (x-3) the remainder is 7. Given that f(x) may be written in the form f(x)= (x-2)(x-3)Q(x)+ax+b, find the remainder when f(x) is divided by (x-2)(x-3). If also f(x) is a cubic function in which the coefficient of x^3 is unity and f(1)=1, determine Q(x)​

Answer 1

`f(x)=(x-2)(x-3)Q(x)+ax+b`

Recall the polynomial remainder theorem: the remainder upon dividing a polynomial
`p(x)` by
`x-c` is equal to
`p(c)`. This means that
`f(2)=4` and
`f(3)=7`, which tell us

`4=2a+b`

`7=3a+b`

From here we can solve for
`a,b`:

`4=2a+b\implies b=4-2a`

`7=3a+b=3a+(4-2a)\implies a=3\implies b=-2`

so that

`f(x)=(x-2)(x-3)Q(x)+3x-2`

Now,

`(f(x))/((x-2)(x-3))=Q(x)+(3x-2)/((x-2)(x-3))`

so the remainder upon dividing
`f(x)` by
`(x-2)(x-3)` is
`3x-2`.

Next, if
`f` is a cubic function, then
`Q(x)` is a linear polynomial that can be written as
`Q(x)=cx-d`. The coefficient of
`x^3` in
`f(x)` is 1 (unity), so that expanding
`f(x)` gives us

`f(x)=(x-2)(x-3)(cx-d)+3x-2`

`f(x)=(cx^3-(5c+d)x^2+(6c+5d)x-6d)+3x-2`

`f(x)=cx^3-(5c+d)x^2+(6c+5d+3)x-(6d+2)`

`\implies c=1`

and we also have that
`f(1)=1`, so that

`1=1-(5+d)+(6+5d+3)-(6d+2)`

`\implies2d=2\implies d=1`

so that

`Q(x)=x-1`