请教数学好的同学啊，来帮忙做个题

missdrafter

The polynomial p(x) has a remainder of -2 when divided by x-2 and has a reminder of 4 when devided by x+1, find the reminder when p(x) is divided by (x-2)(x+1).

zw_323

Let's use the remainder theorem to solve this problem. According to the theorem, when a polynomial P(x) is divided by x-a, the remainder is P(a).

Given that the polynomial p(x) has a remainder of -2 when divided by x-2, we have:

p(2) = -2

Similarly, when p(x) is divided by x+1, the remainder is 4. Thus:

p(-1) = 4

Now, let's use these two equations to find the remainder when p(x) is divided by (x-2)(x+1). We can use the Chinese remainder theorem to combine the remainders for the two divisors.

Since (x-2) and (x+1) are coprime, we can find two constants a and b such that:

a(x+1) + b(x-2) = 1

Solving for a and b, we get a = 1/3 and b = -1/3.

Now, the remainder when p(x) is divided by (x-2)(x+1) can be expressed as:

p(x) = Q(x)(x-2)(x+1) + r(x)

where Q(x) is the quotient, and r(x) is the remainder we want to find.

We can use the remainders we already know to set up two equations:

r(2) = p(2) = -2

r(-1) = p(-1) = 4

We can solve for the remainder r(x) by combining these two equations:

r(x) = p(x) - Q(x)(x-2)(x+1)

r(2) = p(2) - Q(2)(2-2)(2+1) = -2 - 0 = -2

r(-1) = p(-1) - Q(-1)(-1-2)(-1+1) = 4 - 0 = 4

Thus, the remainder when p(x) is divided by (x-2)(x+1) is -2x + 2.

不是小胖子

let p(x) = (q(x))(x-2)(x+1) + ax + b

then：
(ax+b) ÷ (x-2) = a ... -2
(ax+b) ÷ (x+1) = a ... 4

∴ b = -2a - 2 = a + 4

∴ a = -2, b = 2; remainder is -2x + 2

iope23

这是几年级的题？

act_nan

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