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,

$\dfrac{f(x)}{(x-2)(x-3)}=Q(x)+\dfrac{3x-2}{(x-2)(x-3)}$

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

$Q(x)=x-1$