Alan

As follows

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

f(x) = p(x)*(x-1)(x-2)(x-3) + r(x)    where r(x) is the remainder (which must be a quadratic at most).  say r(x) = a*x^2 +b*x + c

2 = a(1²) + b(1) + c          or   a + b + c = 2             (1)

3 = a(2²) + 2b + c            or   4a + 2b + c = 3          (2)

-1 = a(3²) + 3b + c           or  9a + 3b + c = -1          (3)

(2) - (1):    3a + b = 1           (4)

(3 - (2):     5a + b = -4          (5)

(5) - (4)    2a = - 5     or    a = -5/2 

Use this in (4) to find       b = 1 - 3(-5/2)  or   b = 17/2

Use these two in (1) to find c.  I'll leave the rest to you.

I get the following:

Guest #1 has correctly identified the roots (approximately), but I think he/she has made a mistake in the final multiplication. 

Let's see you have a go at this, making use of the answer I gave here: https://web2.0calc.com/questions/please-help-thank-you_42#r1

Note that

A^-1 = [ 0  1]

         [-1  0]

T = [-1]

      [-1]

U = [2]

      [3]

V = [3]

      [1]

W = [ 2]

       [-1]

"For what values of a does the equation (a^2 + 2a)x^2 + (3a - 2)x+1 = 0 yield no real solutions x?".

"How many ordered pairs of positive integers (x,y) satisfy the equation x/y = 100/(xy) + y/x?"

Multiply both sides by xy

x² = 100 + y²  

Rearrange

x² - y² = 100

Rewrite as 

(x + y)(x - y) = 100

100 can be expressed as  

50x2   For this we set

x+y = 50

x-y = 2

Add to get 2x = 52 so x = 26, hence y = 24

100 can also be expressed as 

25x4

20x5

10x10

I'll leave you to find the x and y values corresponding to these products.

As follows (I think!):

I don't know if the following is what is wanted or not (I'm guessing, based on what the vectors looked like to me!).  a, b and k are constants (not necessarily the same for each matrix).

By definition:    arc length = angle*radius

 You can crunch the numbers.

"Find all pairs (x,y) such that x^2 + y^2 - 2x + 4y + 5 = 0"