Maths plz show it step by step

method 1:

you can do this step again , there are some tech.

as you know f(x) can be factorize as this form :

f(x)=(x-a)(x-b)(x-c) , where a , b and c are the roots of f(x)=0

then f(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-(abc)

by comparing the coef , you can know that the product of roots

equal to the constant term x -1

use x^3-2x^2-7x-4 as an example

product of roots = -4 x-1= 4

and you know one of the root is -1 , then product of another two roots is -4

you can guess some of the combination : 1,-4 ; -2 ,2 and -1 ,4

f(4)=4^3-2*4^2-7*4-4=0 then another roots are -1 and 4

so f(x)=(x+1)(x+1)(x-4)=(x+1)^2(x-1)

method 2 :

let f(x)=(x+1)(x^2+bx+c)=x^3+(1+b)x^2+(c+b)x+c

by comparing coef , we have

1+b=-2 and c+b=-7

=>b=-3 and c=-4

then f(x)=(x+1)(x^2-3x-4)=(x+1)(x-4)(x+1)=(x+1)^2(x-4)//

it may be useful for you =]

2010-02-28 22:55:22 補充：
x(x-1)(x-2)-2(x^2+2x-12)

2010-02-28 22:57:24 補充：
you can use the same method :

x(x-1)(x-2)-2(x^2+2x-12) = (x-4)(x^2+ax+b)

=>x^3-5x^2-2x+24=x^3+(a-4)x^2+(b-4a)x-4b

=>a-4=-5 and -4b=24

=>a=-1 and b=-6

=>x(x-1)(x-2)-2(x^2+2x-12) =(x-4)(x^2-x-6)=(x-4)(x-3)(x+2)//

1. Then do this :

(x^3-2x^2-7x-4) / (x+1)

2(a) Let f(x) = x(x-1)(x-2)-2(x^2+2x-12)

Remainder = f(4)

x - 4 = 0

x = 4

Subs. f(4) into f(x)

= 24 - 2(12)

= 0

Because f(x)=0 , then x-4 is a factor of x(x-1)(x-2)-2(x^2+2x-12)

(b)

1.Since f(-1)=0
So, 0=(x+1)(x^2-3x-4)
0=(x+1)(x+1)(x-4)
0=(x+1)^2(x-4)

2(a).f(4)=4(4-1)(4-2)-2(4^2+2(4)-12)
=4(3)(2)-2(16+8-12)
=24-24
=0

(b) Since (x-4)=0
So 0=x(x-1)(x-2)-2(x^2+2x-12)
0=(x^3-5x^2-2x+24)
0=(x-4)(x^2-x-6)
0=(x-4)(x+2)(x-3)