Question 68111
Find the polynomial f(x) of degree three that has zeroes at 1, 2 and 4 such that f(0)=-16.
f(x)=a(x-1)(x-2)(x-4)
f(0)=a(-1)(-2)(-4)=-16
-8a=-16
a=2
Therefore: f(x)=2(x-1)(x-2)(x_4)
--------------------------------------------- 
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. 
If the coefficients are Real and 3+i is a root, 3-1 must be a root.
f(x)=(x-4)(x-(3+i))(x-(3-i))
f(x)=(x-4)((x-3)-i)((x-3)+i)
f(x)=(x-4)((x-3)^2-i^2)
f(x)=(x-4)(x^2-6x+9+1)
f(x)=(x-4)(x^2-6x+10)
----------------------------------
Given that (3x-a)(x-2)(x-7)=3x^2-32x^2+81x-70, determine the value of a.
The product of the constants on the left is (-a)(-2)(-7)=-14a
The constant on the right is -70a
Therefore: -14a=-70a
a=5
----------------------------- 
Find all roots of the polynomial x^3-x^2+16x-16.
The coefficients add up to zero, so x=1 is a root.
Using synthetic division to divide by x-1 you get:
.....-1)1....-1....16....-16
.........1.....0.....16.|.0
So the quotient is x^2+16
This factors as (x+4i)(x-4i)
By the factor theorem this shows 4i and -4i are roots.
Conclusion: 1, 4i, -4i are the roots
-----------------------------
Find the vertical asymptote of the rational function f(x)=3x-12/4x-2.
You have vertical asymptote candidates when the denominator is zero.
If 4x-2=0, then x=1/2
The numerator is not zero when x=1/2 so you have a vertical asymptote
at x=1/2.
---------------------------- 
Find the horizontal asymptote of the rational function f(x)=8x-12/4x-2.
The highest power of x in both numerator and denominator is x^1
The coefficients in the numerator and denominator of x^1 give 8/4=2
You have a horizontal asymptote at x=2
----------------
Cheers,
Stan H.