Anonymous
Anonymous asked in 科學及數學數學 · 7 years ago

F.4 MATH !!!!!! 急 20點=]

https://www.dropbox.com/s/c8aizoypx99ya7x/F.4%20BK...

Not homework asking but self practice.

Exam coming but I forgot a lot of the previous chapters' questions types.

Hope somebody can solve the polynomials questions

of course steps MUST be given coz i wanna know the methods to deal with the Qs.

thx for your kindness if you help me.

1 Answer

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  • Tony
    Lv 4
    7 years ago
    Favorite Answer

    Chp.7 PEQ 8

    as f(x) = (x^3 - 7x + 6) is divisible by (x^2 - 3x + k)

    let f(x) = (x + a) (x^2 - 3x + k) where a is integer

    (x + a) (x^2 - 3x + k) = (x^3 - 7x + 6)

    x^3 - 3x^2 + kx + ax^2 - 3ax + ak = x^3 - 7x + 6

    x^3 + (a - 3) x^2 + (k - 3a)x + ak = x^3 - 7x + 6

    compare the coefficient of similar terms

    a - 3 = 0

    a = 3

    k - 3a = -7

    k = -7 + 3(3) = -7 + 9 = 2

    ak = 6(it is for checking)

    so, k=2

    Chp.7 suppex.

    1. A (x + 1) (2x + 1) (x - 2) + C (x - 2)(x - 1) = x^2 + 2x - 1

    there is no B, probably there is an error in the question. please check again.

    2. A(x + 1) (x - 3) + B (4x - 1) + 5 = 5x^2 - 6x + C

    *method 1(comparsion)

    Ax^2 - 2Ax - 3A + 4Bx - B + 5 = 5x^2 - 6x + C

    Ax^2 + (4B - 2A)x + (-3A - B + 5) = 5x^2 - 6x + C

    compare the coefficient of similar terms

    A = 5

    4B - 2A = -6

    4B - 2(5) = -6

    4B - 10 = -6

    B = 1

    -3A - B + 5 = C

    C = -3(5) - 1 + 5 = -11

    so, A = 5, B = 1, C = -11

    method 2(substitution)

    (for your reference, for example, to eliminate A

    choose x = -1 or 3 so that x+1 = 0 or x - 3 = 0)

    when x = -1,

    -5B + 5 = 5 + 6 + C

    -5B = 6 + C

    5B + C = -6 (equation 1)

    when x = 3

    11B + 5 = 5(3)^2 - 6(3) + C

    11B - C = 22 (equation 2)

    when x = 1/4,

    A (1/4 + 1) (1/4 - 3) + 5 = 5(1/4)^2 - 6(1/4) + C

    (-55/16) A + 5 = 5/16 - 3/2 +C

    55A + 16C = 99 (equation 3)

    equation 1 + equation 2:

    16B = 16

    B = 1

    substitute B = 1 into equation 1,

    5(1) + C = -6

    C = -11

    substitute C = -11 into equation 3,

    55A + 16(-11) = 99

    A = 5

    so, A = 5, B = 1, C = -11

    *you can see that method 2 is so complicated, while method 1 is simple.

    method 2 is recommended.

    (sometimes, method 2 is better when you see Chp.7 suppex. Q3)

    2013-06-05 09:18:38 補充:

    3. x^3 = p (x+1) (x+2) (x+3) + q (x+2) (x+3) + r (x+3) + s

    method 2 in Q2

    when x = -3,

    s = -27

    when x = -2,

    (-2)^3 = r (-2+3) - 27

    r = 19

    when x = -1,

    (-1)^3 = q (-1+2) (-1+3) + r (-1+3) + s

    -1 = 2q + 2r + s

    2q = -1 -2(19) - (-27)

    q = -6

    when x = 0,

    0 = 6p +6q +3r + s

    p = 1

    Source(s): , self , but there is a limit on words.
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