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# 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

- TonyLv 47 years agoFavorite 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

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