# Thread: urgent.i have exam tmr.roots & remainder

1. ## urgent.i have exam tmr.roots & remainder

the cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k , k^2 . it is given that f(x) has a remainder of 7 when divided by x-2.

show that k^3 - 2k^2 - 2k - 3 =0.

thanks teach me how to solve. thank you very much

2. Originally Posted by helloying
the cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k , k^2 . it is given that f(x) has a remainder of 7 when divided by x-2.

show that k^3 - 2k^2 - 2k - 3 =0.

thanks teach me how to solve. thank you very much

$\displaystyle f(x)=ax^3+bx^2+cx+d$.. a=1 so

$\displaystyle f(x)=x^3+bx^2+cx+d$

$\displaystyle f(x)=(x-1)(x-k)(x-k^2)$ .. expand this part

$\displaystyle =x^3-(k^2+k+1)x^2+(k^3+k^2+k)x-k^3$

Here you can compare

$\displaystyle -k^2-k-1=b$--- 1

$\displaystyle k^3+k^2+k=c$ --- 2

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Now make use of the factor and remainder theorem

You know one of the factors is (x-1) so

f(1)=0

$\displaystyle 1+b+c+d=0\Rightarrow b+c+d=-1$ --- 3

Then it will give a remainder of 7 when divided by (x-2)

f(2)=7

$\displaystyle 8+4b+2c+d=7\Rightarrow 4b+2c+d=-1$--- 4

4-3 $\displaystyle 3b+c=0\Rightarrow c=-3b$

Now look back . From 2 above , $\displaystyle k^3+k^2+k=-3b$--6
(Replace the c with -3b)

Multiply 1 by 3 ... $\displaystyle -3(-k^2-k-1)=-3(b)$ -- 5

6-5 i will leave this for you .

3. Hi helloying
The factors are x = 1, x = k, and x = k^2, so :

(x-1)(x-k)(x-k^2) = 0

f(x) = (x-1)(x-k)(x-k^2)

Subs. x = 2