Be 發問於 科學及數學數學 · 1 十年前

# S4 maths polynomial!!!~~~~

1) When a polynomial f(x) is divided by x-5, the remainder is 9; when it is divided by x+2, the remainder is -5. Find the remainder when f(x) is divided by (x-5)(x+2).

2) Let f(x) = 3x^3 + mx^2 - nx - 7. When f(x) is divided by (x+1)(x-3), the remainder is 2x-4.

(a) Find the values of f(-1) and f(3).

(b) Set up two equations connecting m and n.

(c) Find the values of m and n.

### 2 個解答

• 1 十年前
最愛解答

1)Let f(x) be Q(x) * (x-5)(x+2) + k(x-5) + 9

f(-2) = 0 + k(-2-5) + 9 = - 5

-7k = - 5 - 9

k = 2

So f(x) = Q(x) * (x-5)(x+2) + 2(x-5) + 9

= Q(x) * (x-5)(x+2) + 2x - 1

The remainder = 2x - 1

2a)

The remainder when f(x) is divided by (x+1) or (x-3) = the remainder when 2x-4 is

divided by (x+1) or (x-3)

By remainder theoreom :

f(-1) = 2(-1) - 4 = - 6

f(3) = 2(3) - 4 = 2

b)

f(-1) = 3(-1)^3 + m(-1)^2 - n(-1) - 7 = - 3 + m + n - 7

= m+n-10 = - 6....(1)

f(3) = 3(3)^3 + m(3)^2 - n(3) - 7 = 81 + 9m - 3n - 7

= 9m - 3n + 74 = 2....(2)

c)

(2) + (1)*3 :

9m + 74 + 3m - 30 = 2 - 18

m = - 5

- 5 + n - 10 = - 6 , n = 9

2010-03-16 20:16:05 補充：

係let出來 , (x-5)孖住(x+2)時餘數是一次式k(x-5) + 9,

(x-5)不孖住(x+2)時k(x-5) + 9可繼續被(x-5)除,所以咁設。

2010-03-16 20:47:12 補充：

9係remainder , 係f(x)被(x-5)除的remainder.

成個[k(x-5) + 9] 係(x)被(x-5)(x+2)除的remainder.

f(x)除以2次式(x-5)(x+2)的remainder是 x 的 一 次式 ax + b 的樣子,

f(x)除以1次式(x-5)的remainder是一個數,這裡是 9.

即是說 ax + b 除以 (x-5) 餘數是 9 , 所以 ax+b 可設成 k(x-5)+9,

k是 ax+b 除以(x-5)的商。

• 1 十年前

請問一下第一題既k係let出來?

我唔明白..

2010-03-16 20:35:40 補充：

sorry...我仲係唔明...

其實成個k(x-5) + 9我都唔係好明..

題目到話9係remainder...

咁所設既 k(x-5) 係唔係remanider呢?

而k既用處又係d咩呢..

2010-03-16 21:39:56 補充：

THX啊 好details..

開始有少少明白..

我會慢慢咁揣摩一下嫁喇...