Posted by stan on Saturday, April 24, 2010 at 5:01am.
Please type 3x^2 ... to indicate powers in this format.
1. f(x) = 3x^2 - kx - 2
if -2 is a "zero", then
f(-2) = 3(4) + 2k - 2 - 0
2k = -10
k - -5
so f(x) = 3x^2 + 5x - 2
then 3x^2 + 5x - 2 = (x+2)(.......)
by inspection
3x^2 + 5x - 2 = (x+2)(3x - 1)
making the other root, or zero, equal to 1/3
2. let the zeroes be a and b
a+b = 0 , so b = -a
f(a) = a^2 - a - k(2a-1) = 0
f(-a) = a^2 - (-a) - k(-2a-1) = 0
a^2 - a - k(2a-1) = a^2 - (-a) - k(-2a-1)
-a - 2ak + k = a + 2ak + k
-2a -4ak = 0
a( -2 - 4k)=0
k =-1/2
or (easier way)
If the roots add up to zero, then they must be opposite, (see above)
and the function would have to be a difference of squares.
x^2 - x - k(2x-1)
= x^2 - x -2kx + k
to be a difference of squares, no x term should show up, so
-x - 2kx = 0
1 + 2k = 0
k = -1/2
3. in f(x) = 3x^2 - 2kx + 2m find
f(2) and f(3), set those equal to 0
You will have 2 equations in k and m, solve them.
4. recall that for ax^2 + bx + c = 0,
the sum of the roots is -b/a and the product of the roots is c/as
so for 3x^2 + (2k+1)x - k-5 = 0
let the roots be m and n
m+n = (-2k-1)/3 and mn = (-k-5)/3
then (-2k-1)/3 = (-k-5)/3
-2k - 1 = -k - 5
k=4
5. see next post
5. if x^2 + 1 is a factor then x = ± i are roots,
remember that i^1 - -1 and i^4 = +1
f(i) = 1 - i - 8 - ai + b = 0
f(-i) = 1 + i -8 + ai + b = 0
add them
2 - 16 + 2b = 0
b = 7
so the function is
f(x) = x^4 + x^3 + 8x^2 + ax + 7
I then did a long division of that function by x^2 + 1.
This left me with a remainder of x(a-1), but there shouldn't have been a remainder, so
x(a-1) = 0
so a = 1