Math
posted by stan .
Q.1)If one zero of the polynomial 3x2kx2 is 2 find the other zero.allso find the value of k.
Q.2)If sum of the zeroes of the polynomial x2xk(2x1) is 0,find the value of k
Q.3)If 2 and 3 are the zeroes of the polynomial 3x22kx+2m find the values of k and m
Q.4)Find the values of k so that the su of the zeros of the polynomial 3x2+(2k+1)xk5 is equal to the product of the zeros.
Q.5)Find the values of a and b so that x4+x3+8x2+ax+b is divisible by x2+1

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(2a1) = 0
f(a) = a^2  (a)  k(2a1) = 0
a^2  a  k(2a1) = a^2  (a)  k(2a1)
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(2x1)
= 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  k5 = 0
let the roots be m and n
m+n = (2k1)/3 and mn = (k5)/3
then (2k1)/3 = (k5)/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(a1), but there shouldn't have been a remainder, so
x(a1) = 0
so a = 1
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