Determine remainder when dividing polynomials if we know other remainders

Question

If 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?

I'm completely puzzled by this, I'm not sure where to start...

Any hint would be much appreciated.

Answer 1

Hint: $$P(x)=Q(x)(x-3)(x-4)+Ax+B$$ $$P(3)=3$$ $$P(4)=5$$

Can you solve for $A$ and $B$?

Answer 2

Let $$P(x)=(x-4)(x-3)Q(x)+ax+b$$

Now you have $$3a+b=3$$ and $$4a+b=5$$ So solve for $a$ and $b$.

Answer 3

If you carry out the division you should be able to show that $$P(x)=Q(x)(x-3)(x-4)+R(x)$$

Two questions: What is the maximum degree of the remainder $R(x)$? Can you see how to use the remainder theorem for the cases $x=3,4$ that you know already so that $Q(x)$ becomes irrelevant?

Answer 4

Notice $\ p\, = 3+(x\!-\!3)q\,\ $ by $\,\ p(3) = 3$

$5 = p(4) = 3+q(4)\,\Rightarrow\, {\color{#c00}2}\, =\, q(\color{#0a0}4) \, $

Hence $\, p = 3 + (x\!-\!3)(\underbrace{\color{#c00}2+(x\!-\!\color{#0a0}4)r}_{\Large q}\!)\, =\, 2x\!-\!3\, +\, (x\!-\!3)(x\!-\!4)r$

Answer 5

Euclidean division by $(x-3)(x-4)$ can be written as $$P(x)=(x-3)(x-4)Q(x)+R(x),\quad\deg R(x)\le1.$$ Instead of writing $R(x)$ with the standard basis $\{1,x\}$, use the basis $\{x-3,x-4\}$. Thus $$P(x)=(x-3)(x-4)Q(x)+A(x-3)+B(x-4)$$ Setting $x=3$, you get right away $B=-P(3)=-3$. Similarly $A=P(4)=5$, and finally $$R(x)=5(x-3)-3(x-4)=2x-3.$$

This formula can be generalised: the remainder when dividing a polynomial $P(x)$ by $(x-a)(x-b),\enspace a\ne b$, is $$R(x)=\frac1{b-a}\Bigl[P(b)(x--a)-P(a)(x-b)\Bigr]=\frac1{b-a}\begin{vmatrix}x-a&x-b\\[1ex]P(a)&P(b)\end{vmatrix}.$$