To find a monic quadratic polynomial with specific remainders, we utilized the Remainder Theorem and solved for constants to obtain f(x) = x² - x - 2.
To find a monic quadratic polynomial f(x) with specific remainders when divided by (x-1) and (x-3), we can use the Remainder Theorem.
We know the remainders when the polynomial is divided by these linear factors.
Therefore, we can express f(x) as:
f(x) = (x - 1)(x - 3)Q(x) + ax + b,
where Q(x) is a polynomial and a, b are constants that determine our remainders.
Since f(x) is monic, the coefficient of x² in Q(x) is 1.
We're given that f(1) = 2 and f(3) = 4, which gives us two equations:
f(1) = (1 - 1)(1 - 3)Q(1) + a*1 + b = 2
f(3) = (3 - 1)(3 - 3)Q(3) + a*3 + b = 4
Simplifying, we get:
a + b = 2
3a + b = 4
Subtracting the first equation from the second, we find 2a = 2, so a = 1. Substituting a into the first equation, b must be 1 as well.
Thus, our polynomial is f(x) = (x - 1)(x - 3) + x + 1 = x² - 3x + x - 3 + x + 1 = x² - x - 2.