Question

find a polynomial of least degree which leaves remainders 1, 2, -2 when divided by x-1 , x+2 , x-2​

Answer 1

Answer:

We can write p(x)=x100g(x)+1=(x−2)3h(x)+2 for some polynomial g and h.

Then, p′(x)=x99(xg′(x)+100g(x))=(x−2)2((x−2)h′(x)+3h(x)),

whence p′ is divisible by x99 and (x−2)2. So p′ is also divisible by x99(x−2)2.

Let p′(x)=x99(x−2)2f(x) for some polynomial f.

Then, p(x)=f(x)∫x99(x−2)2dx=f(x)(x102102−4x101101+4x100100)+c

Now, p(0)=1⇒c=1 and p(2)=2⇒f(2)(2102102−4×2101101+4×2100100)+1

ANSWER=2,

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

(x+3)

Let the quadratic polynomial be denoted as P(x).

The polynomial when divided by x+2 gives a remainder of 1. So, from remainder theorom, P(−2)=1.

Similarly, the polynomial when divided by x−1 gives a remainder of 4. So, from remainder theorom, P(1)=4.

Now, if P(x) is divided by the product (x+2)(x−1), the remainder can be at most be a linear function.

We can write P(x)=C(x+2)(x−1)+(Ax+B), where A, B, and C are constants.

Use P(1)=4 and P(−2)=1.

We get two equations: A+B=1 and −2A+B=1.

Solving, we get A=1 and B=3. Hence, the remainder is

Ax+B=x+3