F.4 Maths

1.設a(x-3)(2x+1)+b(3-x)=4x^2-13x+3,求a,b的值
a(2x^2-5x-3)+b(3-x)-3a+3b =4x^2-13x+3
2ax^2 - (5a+b)x -3a+3b = 4x^2-13x+3
2a=4
a=2
Sub a=2 into -3a+3b = 3
-3(2) + 3b =3
3b = 9
b=3
2.設f(x)=x^3-3x^2+k,其中k為整數。若f(x)可被x-1整除,求f(x)除以x-1時的餘數
f(1)=1-3+k=0
k=2
3.設f(x)=2x^2+kx-2.已知2x+1為f(x)的因式,求k的值,並求f(x)的另一因式
己知2x+1為f(x)的因式
∴f(-1/2)=0
2(-1/2)^2+k(-1/2)-2=0
1/2k=3/2
k=3
2x^2+3x-2=0
(2x-1)(x+2)=0
另一因式是x+2
4.設f(x)=x^3-2x^2-5x+6.
(a)求f(3)
f(3)=3^3-2(3)^2-5(3)+6
=0
(b)因式分解f(x)
By4(a),
由於f(3)=0
∴x-3為f(x)的因式
x^3-2x^2-5x+6
=(x-3)(x^2+x-2)
=(x-1)(x+2)(x-3)
5.設f(x)=x^3-3x+k,其中k為常數。若f(-1)=0,因式分解f(x)
f(-1)=0
∴(-1)^3-3(-1)+k=0
k= -2
f(x)=x^3-3x-2
=(x+1)(x^2-x-2)
=(x+1)(x+1)(x-2)
=(x-2)(x+1)^2
6.設f(x)=5x^2-3x+k。已知f(x)除以x+2的餘數為3k.
(a)求k
f(-2)=3k
5(-2)^2-3(-2)+k=3k
26=2k
k=13
(b)因式分解f(x)
5x^2-3x+13
沒有任何因式
∴冇答案。
7.設f(x)=x^3+(k-1)x^2-(k+2)x-2k
(a)證明f(x)可被x-2整除
f(2) = (2)^3+(k-1)(2)^2-(k+2)(2)-2k
=0
∴f(x)可被x-2整除
(b)設f(x)亦可被x-1整除
(i)求k的值
f(1)=(1)^3+(k-1)(1)^2-(k+2)(1)-2k=0
k=-1
(ii)因式分解f(x)
x^3-2x^2-x+2
=(x+1)(x-1)(x-2)

1.a=2, b=3
2.0
3.2x^2+kx-2 （用十字相乘法）
2x 1
x -2
2x^2-3x-2
∴k=3
2x^2+kx-2=(2x+1)(x-2)
4a.f(3)=3^3-2(3)^2-5(3)+6
=27-18-15+6
=0
4b.由於f(3)=0，所以(x-3)是因式。
x^3-2x^2-5x+6
=(x-3)x^2+(x-3)x-2(x-3)
=(x-3)(x^2+x-2)
=(x+2)(x-1)(x-3)
5.f(-1)=(-1)^3-3(-1)+k
=-1+3+k
=2+k
=0(given in the question)
∴k=-2
由於f(-1)=0，所以(x+1)是因式。
x^3-3x-2
=(x+1)x^2-(x+1)x-2(x+1)
=(x+1)(x^2-x-2)
6a.f(-2)=5(-2)^2-3(-2)+k
=5*4+3*2+k
=20+6+k
=26+k
=3k(given in the question)
3k=26+k
2k=26
k=13
6b.f(x)不能被因式分解。
7a.x^3+(k-1)x^2-(k+2)x-2k
=x^3+kx^2-x^2-kx-2x-2k
=x^3-2x^2+x^2-2x+[kx^2-kx-2k]
=(x-2)x^2+(x-2)x+[(kx+k)(x-2)] [十字相乘法]
kx k
x -2
kx^2-kx-2k
=(x-2)(x^2+x+kx+k)
=(x-2)[(x+1)x+(x+1)k]
=(x+k)(x+1)(x-2)
7bi.f(x)=x^3+(k-1)x^2-(k+2)x-2k
=x^3+kx^2-x^2-kx-2x-2k
f(1)=1^3+k(1)^2-1^2-k(1)-2(1)-2k
=1+k-1-k-2-2k
=-2-2k
=0(given in the question)
-2-2k=0
2+2k=0
2k=-2
k=-1
7bii.f(x)=x^3+kx^2-x^2-kx-2x-2k
把7bi的答案代入k
f(x)=x^3+(-1)x^2-x^2-(-1)x-2x-2(-1)
=x^3-x^2-x^2+x-2x+2
=x^3-2x^2-x+2 （用十字相乘法）
x^2 -1
x -2
x^3-2x^2-x+2
=(x^2-1)(x-2)
=(x+1)(x-1)(x-2)