Find the polynomial f(x) of degree three that has zeroes 
at 1, 2 and 4 such that f(0)=-16.

Two ways to do it:

First way:

Rule 1: If a polynomial f(x) has roots r1, r2, r3, ···, rn
and leading coefficient a, then

f(x) = a(x-r1)(x-r2)···(x-rn)

So for this problem:

f(x) = a(x - 1)(x - 2)(x - 4)

 -16 = a(0 - 1)(0 - 2)(0 - 4)
 -16 = a(-1)(-2)(-4)
 -16 = -8a
   2 = a

f(x) = 2(x - 1)(x - 2)(x - 4)
f(x) = 2(x - 1)(x² - 6x + 8)
f(x) = 2(x³ - 6x² + 8x - x² + 6x - 8)
f(x) = 2(x³ - 7x² + 14x - 8)
f(x) = 2x³ - 14x² + 28x - 16

Here's the second method: 
 
Let the polynomial be 

f(x) = ax³ + bx² + cx + d

Then plugging in the values for x, and
setting it to -16 when 0 is plugged in,
and 0 when the others are plugged in:

f(0) = a(0)³ + b(0)² + c(0) + d = -16
f(1) = a(1)³ + b(1)² + c(1) + d = 0
f(2) = a(2)³ + b(2)² + c(2) + d = 0
f(4) = a(4)³ + b(4)² + c(4) + d = 0

The above simplifies to this system

                              d = -16
                  a + b + c + d = 0
               8a + 4b + 2c + d = 0 
             64a + 16b + 4c + d = 0

Substitute =16 for d in the bottom three:

                  a + b + c - 16 = 0
               8a + 4b + 2c - 16 = 0 
             64a + 16b + 4c - 16 = 0

                  a + b + c = 16 
               8a + 4b + 2c = 16  
             64a + 16b + 4c = 16

Solve that system of equations as you get

a = 2, b = -14, c = 28

So the polynomial

f(x) = ax³ + bx² + cx + d becomes

f(x) = 2x³ - 14x + 28x - 16               

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The degree three polynomial f(x) with real coefficients and 
leading coefficient 1, has 4 and 3 + i among its roots. 
Express f(x) as a product of linear and quadratic polynomials 
with real coefficients.

Rule 2:
If a polynomial has real coefficients, then if a complex
number a + bi is a root, then so is its conjugate a - bi

So the roots are 4, 3+i, 3-i

Using rule 1

f(x) = 1(x - 4)[x - (3+i)][x - (3-i)]
f(x) = (x - 4)[x² - (3-i)x - (3+i)x + (3+i)(3-i)]
f(x) = (x - 4)[x² - 3x + ix - 3x - ix + 9 - i²]
f(x) = (x - 4)[x² - 6x + 9 - i²]
Since i² = -1, substitute -1 for i²
f(x) = (x - 4)[x² - 6x + 9 - (-1)]
f(x) = (x - 4)(x² - 6x + 9 + 1]
f(x) = (x - 4)(x² - 6x + 10)
f(x) = x³ - 6x² + 10x - 4x² + 24x - 40 
f(x) = x³ - 10x² + 34x - 40

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Given that (3x-a)(x-2)(x-7)=3x³-32x²+81x-70, determine the 
value of a.

Since this is an identity, we may substitute any number for x
and it will be true.  It wouldn't do any good to substitute a 
root 2 or 7 since that would just give 0 = 0. So let's 
substitute x = 0, since that is the easiest number to 
substitute, and 0 is not a root:

          (3x-a)(x-2)(x-7) = 3x³-32x²+81x-70
        [3(0)-a](0-2)(0-7) = 3(0)³-32(0)²+81(0)-70
              (-a)(-2)(-7) = -70
                      -14a = -70 
                         a = 5       

Find all roots of the polynomial x^3-x^2+16x-16.

Set it = 0

   x³ - x² + 16x - 16 = 0

Factor by grouping:
Factor x² out of the first two terms
and 16 out of the last two terms:

x²(x - 1) + 16(x - 1) = 0

Factor out common factor (x - 1)

(x - 1)(x² + 16) = 0

Set each factor = 0

x - 1 = 0 
    x = 1

x² + 16 = 0
     x² = -16  
      x = ±Ö-16
      x = ±4i

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Find the vertical asymptote of the rational 
function f(x)=(3x-12)/(4x-2).

Set denominator = 0

4x - 2 = 0
    4x = 2
     x = 1/2
  
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Find the horizontal asymptote of the rational function 

f(x) = .

Rule: Divide every term on top and bottom by the greatest
power of x

f(x) =  

Simplify

f(x) = 

As x grows very large in absolute value, the fractions 
get extremely small, and become negligible, so f(x)
approaches the fraction 8/2 or 4, so the horizontal
asymptote is y = 4.

Or you can learn the rule that if the numerator and
denominator have the same degree, the horizontal 
asymptote's equation is the quotient of the leading
coefficients, y = 8/4  or y = 2 

Edwin