The remainder that the polynomial f(x) leaves upon division by x−1 and x−2 is 2 and 3, respectively. If the remainder f(x) leaves upon division by x^2−3x+2 is g(x), what is the remainder g(x) leaves upon division by x−11?

To find the remainder that g(x) leaves upon division by x−11, we need to use the Remainder Theorem.

First, let's determine the polynomial f(x) based on the remainders given. We know that f(x) leaves a remainder of 2 when divided by x−1 and a remainder of 3 when divided by x−2.

Using the Remainder Theorem, we can write the following equations:

f(1) = 2
f(2) = 3

By substituting the values of x into the polynomial f(x), we can solve for the coefficients of f(x).

Let's express f(x) as follows:

f(x) = a(x−1) + 2 ---- (1)
f(x) = b(x−2) + 3 ---- (2)

where a and b are coefficients we want to find.

Substituting x=1 into equation (1):
f(1) = a(1−1) + 2
2 = a(0) + 2
a = 0

Substituting x=2 into equation (2):
f(2) = b(2−2) + 3
3 = b(0) + 3
b = 0

Therefore, the polynomial f(x) is f(x) = 0.

Now, we can find the remainder that g(x) leaves upon division by x−11.

Since the remainder that f(x) leaves upon division by x^2−3x+2 is g(x), we can write:

f(x) = (x^2−3x+2) p(x) + g(x)

where p(x) is the quotient.

We know that f(x) = 0, so the equation becomes:

0 = (x^2−3x+2) p(x) + g(x)

Since g(x) is the remainder, when f(x) = 0, is divided by x^2−3x+2, we can substitute 0 for f(x) in the equation above:

0 = (x^2−3x+2) p(x) + 0

Simplifying:

0 = p(x) * (x^2−3x+2)

Since the product of two polynomials is zero only if at least one of them is zero, we have:

p(x) = 0 --- (3)
(x^2−3x+2) = 0 --- (4)

From equation (3), we find that p(x) is zero.

From equation (4), we can solve for the values of x:

x^2−3x+2 = 0

Using the quadratic formula, we have:

x = (−(−3) ± √((−3)^2−4*1*2))/(2*1)

Simplifying:

x = (3 ± √(9−8))/2
x = (3 ± √1)/2
x = (3 ± 1)/2

Thus, the values of x are:

x₁ = (3+1)/2 = 2
x₂ = (3-1)/2 = 1

Therefore, the polynomial x^2−3x+2 has roots at x=2 and x=1.

Now, to find the remainder that g(x) leaves upon division by x−11, we substitute x=11 into g(x):

g(11) = 0 * (11−2)(11−1) [Substituting p(x)=0 from equation (3)]
g(11) = 0

Hence, the remainder that g(x) leaves upon division by x−11 is 0.