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August 21, 2014

August 21, 2014

Posted by **stan** on Saturday, April 24, 2010 at 5:01am.

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

- Math -
**Reiny**, Saturday, April 24, 2010 at 9:56amPlease 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

- Math -
**Reiny**, Saturday, April 24, 2010 at 10:16am5. 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

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