Question

Let P(x) be a polynomial such that when P(x) is divided by x-19, the remainder is 99, and when P(x) is divided by x-99, the remainder is 19. Find the remainder when P(x) is divided by (x-19)(x-99)

Answer 1

$$
\cfrac{P(x)}{x-19}+99=\cfrac{P(x)}{x-99}+19\\
\implies \\
\cfrac{P(x)}{x-19}+\cfrac{P(x)}{x-99}+99-19=0\\
\cfrac{P(x)(x-99)+P(x)(x-19)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)(x-99-19)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)(x-118)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)}{(x-19)(x-99)}+\cfrac{80}{(x-118)}=0\\
\implies\\
\cfrac{P(x)}{(x-19)(x-99)}=-\cfrac{80}{(x-118)}\\
$$
Therefore, \(\cfrac{P(x)}{(x-19)(x-99)}=-\cfrac{80}{(x-118)}\).

Answer 2

\[
P(x)(x-99)+P(x)(x-19)=P(x)(x-99+x-19)=P(x)(2x-118)
\]

Answer 3

O.O two separate answers... i received a hint that one might use Little Bezout's Theorem to solve it?

Answer 4

I forgot the negative sign in step 2 above:
$$
\cfrac{P(x)}{x-19}+99=\cfrac{P(x)}{x-99}+19\\
\implies \\
\cfrac{P(x)}{x-19}-\cfrac{P(x)}{x-99}+99-19=0\\
\cfrac{P(x)(x-99)-P(x)(x-19)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)(-99+19)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)(-80)}{(x-19)(x-99)}+80=0\\
\cfrac{P(x)}{(x-19)(x-99)}+\cfrac{80}{(-80)}=0\\
\implies\\
\cfrac{P(x)}{(x-19)(x-99)}=1\\
$$

Answer 5

retriceumed when P(x) divided by (x-19)(x-99) has remainder ax+b.
according information above, use the remainder theeorem :
P(19) = 99 , it means -----> a(19) + b = 99 or 19a + b = 99 ... (1)
P(99) = 19 , it means -----> a(99) + b = 99 or 99a + b = 19 ... (2)

solve for a and b. then distribute into ax+b

Answer 6

look elimination and subtitution method is good

Answer 7

thank you both I think I understand now :D