F4 Math

f(x)=(x^2+3x-1)(x+k)
=(x^3+3x^2-x+kx^2+3kx-k)
=x^3+(k+3)x^2+(3k-1)x-k

(i) Therefore, coefficient of x is 3k-1

(ii) Coefficient of x^2 = k+3

(b) 3k-1 = k+3
2k=4
k=2

(c)x^3+(k+3)x^2+(3k-1)x-k
x^3+5x^2+5x-5

By Remainder Theorem
f(1) = 1+5+5-5
=6

Therefore,the reminder is 6

(x^3+5x^2+5x-5) = (x-1)Q(x)+6 Q(x) is the quotient
x^3+5x^2+5x-11 = (x-1)(x^2+Ax+11) A is a real number
Since (-1)(11)= -11, x(x^2) = x^3

By comparing the coefficient of x^2
A-1 = 5
A = 6

Therefore Q(x) = x^2+6x+11
The quotient is x^2+6x+11

2006-12-14 20:00:30 補充：
Sorry, part (c)should be like this(c)x^3+5x^2+5x-2Remainderf(1) = 1+5+5-2= 9(x^3+5x^2+5x-2) = (x-1)Q(x)+9 Q(x) is the quotientx^3+5x^2+5x-11 = (x-1)(x^2+Ax+11)By comparing the coefficient of x^2A-1 = 5A = 6The quotient is x^2+6x+11

f(x)=(x^2+3x-1)(x+k)=(x^3+3x^2-x)+(kx^2+3kx-k)=x^3+(k+3)x^2+(3k-1)x-k
coeff. of x = 3k-1
coeff. of x^2 = k+3

coeff. of x and x^2 in f(x) are equal,
3k-1 = k+3
k=2

When k=2, f(x)=x^3+5x^2+5x-2
reminder=f(1)=1+5+5-2=9
f(x)-9=x^3+5x^2+5x-11=(x-1)(x^2+6x+11)
quotient=(x^2+6x+11)