Math

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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

  • Math -

    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

  • Math -

    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

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