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).
评论
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.
评论
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
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这是几年级的题？
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