Q.1)If one zero of the polynomial 3x2-kx-2 is 2 find the other zero.allso find the value of k.

Q.2)If sum of the zeroes of the polynomial x2-x-k(2x-1) is 0,find the value of k
Q.3)If 2 and 3 are the zeroes of the polynomial 3x2-2kx+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)x-k-5 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(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

Q.1) To find the other zero of the polynomial 3x^2 - kx - 2 when one zero is 2, we can use the fact that the sum of the zeros of a polynomial is equal to the negative coefficient of the linear term (kx) divided by the leading coefficient (3).

Let's assume the other zero is denoted by "a". We have the equation:

2 + a = -(-k) / 3

Simplifying this equation, we get:

2 + a = k/3

To find the value of k, we can substitute one of the zeros (2) into the original polynomial equation:

3(2)^2 - k(2) - 2 = 0

Simplifying this equation, we get:

12 - 2k - 2 = 0
10 - 2k = 0
2k = 10
k = 5

So the other zero is a = -1 and the value of k is 5.

Q.2) To find the value of k when the sum of the zeros of the polynomial x^2 - x - k(2x - 1) is 0, we use the fact that the sum of the zeros is equal to the negative coefficient of the linear term (-x) divided by the leading coefficient (1).

Let's assume the other zero is denoted by "b". We have the equation:

b + 1 = -(k(2b - 1)) / 1

Simplifying this equation, we get:

b + 1 = -k(2b - 1)

Expanding and rearranging, we get:

b + 1 = -2bk + k

Rearranging again, we get:

b + 1 = k - 2bk

Simplifying further, we get:

b + 1 = k(1 - 2b)

Since the sum of the zeros is 0, we have:

0 = b + 1

From this equation, we get b = -1.

Substituting this value into the previous equation, we get:

-1 + 1 = k(1 - 2(-1))
0 = 3k

Therefore, the value of k is 0.

Q.3) To find the values of k and m when the zeros of the polynomial 3x^2 - 2kx + 2m are 2 and 3, we can use the fact that the sum of the zeros is equal to the negative coefficient of the linear term (-2kx) divided by the leading coefficient (3) and the product of the zeros is equal to the constant term (2m) divided by the leading coefficient (3).

The sum of the zeros is:

2 + 3 = -(-2k) / 3

5 = 2k / 3

Cross-multiplying, we get:

15 = 2k

k = 15 / 2

The product of the zeros is:

(2)(3) = 2m / 3

6 = 2m / 3

Cross-multiplying, we get:

18 = 2m

m = 18 / 2

m = 9

Therefore, the values of k and m are k = 15/2 and m = 9.

Q.4) To find the values of k so that the sum of the zeros of the polynomial 3x^2 + (2k + 1)x - k - 5 is equal to the product of the zeros, we can use the fact that the sum of the zeros is equal to the negative coefficient of the linear term ((2k + 1)x) divided by the leading coefficient (3) and the product of the zeros is equal to the constant term (-(k + 5)) divided by the leading coefficient (3).

The sum of the zeros is:

(1st zero) + (2nd zero) = -((2k + 1)) / 3

Sum of zeros = -2k - 1 / 3

The product of the zeros is:

(1st zero)(2nd zero) = -(k + 5) / 3

Product of zeros = -k - 5 / 3

Since the sum of the zeros is equal to the product of the zeros, we have:

Sum of zeros = Product of zeros

-2k - 1 / 3 = -k - 5 / 3

Cross-multiplying and simplifying, we get:

-2k - 1 = -k - 5

k = -4

Therefore, the value of k is -4.

Q.5) To find the values of a and b so that x^4 + x^3 + 8x^2 + ax + b is divisible by x^2 + 1, we can use the fact that if a polynomial is divisible by a given polynomial, its remainder when divided by that polynomial will be zero.

Since x^2 + 1 is a polynomial of degree 2, the remainder term will be of degree less than 2. Therefore, the term containing x will be zero.

Dividing x^4 + x^3 + 8x^2 + ax + b by x^2 + 1, we get:

______________________
x^2 + 1 | x^4 + x^3 + 8x^2 + ax + b

-(x^4 + x^2)
_______________
(x^3 + 7x^2)

-(x^3 + x)
_______________
(6x^2 + ax)

-(6x^2 + 6)
_______________
(ax + a + b + 6)

Since the term containing x in the remainder is zero, we have:

6x^2 + ax + a + b + 6 = 0

This equation must hold for all values of x. Therefore, the coefficients of x^2, x, and the constant term must all be zero.

Equating the coefficients, we get:

6 = 0 (coefficient of x^2)
a = 0 (coefficient of x)
a + b + 6 = 0 (constant term)

From the second equation, we get a = 0. Substituting this into the third equation, we get:

0 + b + 6 = 0
b = -6

Therefore, the values of a and b are a = 0 and b = -6.