polynomials 習題

1a)f(1)=-6
(1)2+a(1)+b=6
a+b=5_______________________(1)

f(-2)=0
(-2)2+a(-2)+b=6
2a-b=-2______________________(2)

Solve (1) and (2),we have
a=1,b=4

b)the remainder =f(-3)
=(-3)2 +(-3)+4
=10

2a)f(-2k)= 2(-2k)3+ k(-2k)2 - 5k2(-2k)+2k3
=-16k3+4k3+10k3+2k3
=0
∴x+ 2k is a factor of f(x)

b) using the result of (a) ,x+1 is a factor of2x3- x2-5x-2.
2x3- x2-5x-2=(x-1)(x-2)(2x+1)

3a)f(1) =6
a(1)3-b(1)2+2(1)+(b+3)=6
a-b+2+b+3=6
a=1

g(1)=12
(1)3-6a(1)2+b(1)+12=12
1-6a+b+12=12
6a-b=1_________________________(1)
Sub a=1 into (1),we have
6-b=1
b=5

b)f(x)=g(x)
x3-5x2+2x+8=x3-6x2+5x+12
x2-3x-4=0
(x-4)(x+1)=0
x=4 or x=-1

f(x)=x^2+ax+b
f(1)=-6 and f(-2)=0. So 1+a+b=-6 and 4-2a+b=0
So 1+a+2a-4=-6,a=-1 and then b=2a-4=-6
(b) f(-3)=9-3a+b=9+3-6=6. The remainder is 6
2 f(-2k)=2(-2k)^3+k(-2k)^2-5k^2(-2k)+2k^3=(-16+4+10+2)k^3=0
So x+2k is a factor of f(x)
(b) Sub. k=-1,2x^3-x^2-5x-2=(x-2)(2x^2+3x+1)=(x-2)(x+1)(2x+1)
3(a) f(1)=6 and g(1)=12
a+2+3=6, 1-6a+b+12=12
So a=1,b=5
(b) f(x)=g(x)
x^3-5x^2+2x+8=x^3-6x^2+5x+12
x^2-3x-4=0
(x-4)(x+1)=0
x=-1 or 4