Under normal conditions, a reply would not be so complete as is the one that follows. However, Purplemath does not, to my knowledge, currently offer tutorials on these particular topics. Therefore, I will provide a more extensive response.

cruxxfay wrote:"Find the polynomial f such that f(2x-1) = 16x^4 - 32x^3 +12x^2"

The result polynomial is the composition of the unknown f(x) with g(x) = 2x - 1. By nature of

composition of functions, we know that (f o g o g

^{-1})(x) = f(x). Since g(x) = 2x - 1, then g

^{-1}(x) = (x + 1)/2. We are given that (f o g)(x) = 16x

^{4} - 32x

^{3} + 12x

^{2}. Combining these results, we find:

. . . . .. . . . . . .Simplify to obtain the required polynomial answer.

cruxxfay wrote:"When polynomial f is divided by (x-3) the remainder is 2, and when divided by (x+1) the remainder is -4. Find the remainder when polynomial f is divided by (x^2 -2x - 3)"

The

Factor Theorem states that any polynomial f(x) may be restated as p(x)q(x) + r(x), where p(x) is the (polynomial) divisor, q(x) is the result (that is, the quotient), and r(x) is the remainder. The

Remainder Theorem, related to the Factor Theorem, states that the value of a polynomial f(x) at x = a is the same as the value of the remainder when f(x) is divided by x - a. We use these as follows:

We are given that f(x) must be divisible by the given quadratic. Therefore:

. . . . .Immediately, we note that the specified linear divisors are factors of the quadratic divisor. Continuing directly and applying the Remainder Theorem, we find:

. . . . .So r(3) = 2.

. . . . .So r(-1) = -4. Since we are dividing by a quadratic, then logically the remainder can have a degree of 0 or 1. Since r(x) is clearly not constant, then r(x) must be linear. Then r(x) = ax + b for some values a and b. One may then find the equation of the remainder as being r(x) = (3/2)x - (5/2).

Since f(x) = (x

^{2} - 2x - 2)p(x) + r(x), then the remainder, upon division by (x

^{2} - 2x - 2), must be r(x).

Kindly please reply with queries, should any of the above leave questions in your mind. Thank you.