+0

# Super Polynomial Challenge!

0
423
2
+115

Suppose f(x) is a polynomial of degree 4 or greater such that f(1)=2, f(2)=3, and f(3)=5. Find the remainder when f(x) is divided by (x-1)(x-2)(x-3).

Dec 16, 2018

#1
+770
-1

Note that f(1) = 2 means that the sum of the coefficients of f(x) = 2. Just try x = 2 for now and experiment to get your answer.

Dec 16, 2018
#2
+6196
+2

$$p(x)=(x-1)(x-2)(x-3)\\ r(x) = f(x) - p(x)q(x),~\text{ where }q(x) \text{ is the quotient of }f(x) \text{ and }p(x)\\ \text{note that }p(1)=p(2)=p(3) = 0\\ r(1) = f(1) - p(1)q(1) = f(1) = 2\\ r(2) = f(2) = 3\\ r(3) = f(3) = 5$$

$$\text{we have 3 equations so we are looking for a polynomial with 3 degrees of freedom}\\ \text{i.e. a 2nd degree polynomial }a x^2 + b x+ c\\ \text{plugging things in we get 3 equations}\\ a+b+c = 2\\ 4a+2b+c=3\\ 9a+3b+c=5\\ \text{This can be solved to obtain }\\ a=\frac{1}{2},b= -\frac{1}{2},c=2, \text{ and thus the remainder polynomial is}\\ r(x) = \dfrac 1 2(x^2 - x + 4)$$

Dec 16, 2018
edited by Rom  Dec 16, 2018
edited by Rom  Dec 16, 2018