F.5 績多項式 1題~

(p+1)x^2+qx+2p =(x-2) (ax^2+bx+c)-16
When x=2,
(p+1)2^2+2q+2p=0-16
(p+1)4+2q+2p=-16
4p+4+2q+2p=-16
6p+2q+20=0
3p+q+10=0........(1)

(p+1)x^2+qx+2p =(x-3) (Ax^2+Bx+C)-13
When x =3,
(p+1)3^2+3q+2p=0-13
(p+1)9+3q+2p=-13
9p+9+3q+2p=-13
11p+3q+22=0.......(2)

(1)*3-(2)
9p+3q+30-11p-3q-22=0
-2p+8=0
p=-8/-2
p=4//

sub p=4 into (1)
3*4+q+10=0
12+q+10=0
22+q=0
q=-22//

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問：

當多項式 (p+1)x^2+qx+2p 除以 x-2 及 x-3 時,所得的餘數分別為 -16 及 -13 ,求p 及 q 的值.

答：

較簡單的計法：

(p+1)x^2+qx+2p 除以 x-2 ... -16 = -16 + 3(x-2) = 3x-22

(p+1)x^2+qx+2p 除以 x-3 ... -13 = -13 + 3(x-3) = 3x-22

即 (p+1)x^2+qx+2p+3x-22 能除盡 x-2 和 x-3 ,

(p+1)x^2+qx+2p+3x-22 ≡ k(x-2)(x-3)
(p+1)x^2+(3+q)x+(2p-22) ≡ kx^2 - 5kx + 6k

得出

p + 1 = k --- (1)
3 + q = -5k --- (2)
2p - 22 = 6k --- (3)

(1) × 2 - (3)

24 = -4k
k = -6 --- (4)

Sub (4) into (1)

p = -7

Sub (4) into (2)

3 + q = 30
q = 27

Check :

(p+1)x^2+qx+2p
= -6x^2 + 27x - 14
= -3(x-3)(2x-3) - 13

對

-6x^2 + 27x - 14
= -3(x-2)(2x-5) - 16

對！

答案： p = -7 , q = 27。