When a polynomial is divided by (x+2) the remainder is 19. When the same polynomial is divided by (x-1) the remainder is 2. Determine the remainder when the polynomial is divided by (x-1)(x+2).
It doesn't matter the degree of the polynomial, as long as it degree 3 or more it can divide by two factors. When you divide a polynomial by (x-a), the result is another polynomial with degree one less than the previous one and a possible remainder term, unless it divides evenly.
I've been thinking about this problem and I can't figure it out 100%. I am almost sure you either add them or multiply, but I can't see which one.
HI
$\displaystyle f(1)=2$ , $\displaystyle f(-2)=19$
Lets assume that the remainder of f(x) is Ax+B . We can make this assumption because :
When f(x) is divided by $\displaystyle (x-1)(x+2)$ , there is a quotient q(x) and a remainder r(x) .
$\displaystyle f(x)=(x-1)(x+2)q(x)+r(x)$
where $\displaystyle r(x)=0$ or $\displaystyle \deg r(x)<\deg (x-1)(x+2)=2 $
so now $\displaystyle f(x)=(x-1)(x+2)q(x)+Ax+B$
$\displaystyle f(1)=A+B\Rightarrow 2=A+B$ --- 1
$\displaystyle f(-2)=-2A+B\Rightarrow 19=-2A+B$ ---- 2
Solve the simultaneous equations .