ok, this might be a bit torturous, but if anyone is patient enough and help me with my working, i'd REALLY APPRECIATE IT!!!!
Question:
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, and 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 __________________________________________________________________________
this is what i did, it might be a stupid or completely wrong way of doing..... please pardon my stupidity
x^3 + ax^2 + bx^2 + c = (x-1)(x-k)(x-k^2)
f(2)=7
8+4a+2b+c=7
4a+2b+c=-1
(x-1)(x-k)(x-k^2)= x^3 + (-k-1-k^2)x^2 + (-k^3+k^2+k)x -k^3
so... (-k-1-k^2)= a
(-k^3 + k^2 + k)=b
-k^3 =c
4(-k^2-k-1)+2(-k^3+k^2+k)-k^3= -1
-3k^3-2k^2-2k-3= 0
!!??? which is different from what i;m supposed to show!?!?!?
I honestly don't know what im doing.........
ok... i know how to do this question already! :D
For the previous part, this was how i got the answer:
f(x) = (x-1)(x-k)(x-k^2)
= Q(x)(2-x)+7
Let x=2, so (2-k)(2-k^2)=7, and i get the answer,
But there is another part to this question which i don't understand...
please help:D
it says: Hence, find a value of k and show that there are no other real values of k which satisfy this equation.
How do i do this??
k^3 - 2k^2 - 2k - 3 = 0
after a few quick trial & error guesses
I found k=3 to work
so by division I got
k^3 - 2k^2 - 2k - 3 = 0
(k-3)(k^2 + k + 1) = 0
so any other solutions for k must come from
k^2 + k + 1=0
Using the formula it can be quickly shown that there is no "real" solution to this
No worries, let's go through the steps together to find where the mistake might be.
Given that the cubic polynomial f(x) has the roots 1, k, and k^2, we can write f(x) in factored form as (x-1)(x-k)(x-k^2).
To find the remainder when f(x) is divided by (x-2), we need to perform polynomial long division. Here's how you can do it step by step:
1. Write the dividend and divisor:
Dividend: x^3 + ax^2 + bx^2 + c
Divisor: x - 2
2. Divide the first term of the dividend (x^3) by the divisor (x), which gives the quotient as (x^2). Multiply the divisor (x - 2) by the quotient (x^2), which gives (x^3 - 2x^2). Write this below the dividend.
x^2
____________
x - 2 | x^3 + ax^2 + bx^2 + c
x^3 - 2x^2
_______________
ax^2 + bx^2
3. Subtract (x^3 - 2x^2) from (ax^2 + bx^2), which gives (ax^2 + bx^2 - x^3 + 2x^2). Write this as the new dividend.
ax^2 + bx^2 - x^3 + 2x^2
________________________________
x - 2 | x^3 + ax^2 + bx^2 + c
x^3 - 2x^2
_______________
ax^2 + bx^2
4. Repeat steps 2 and 3 with the new dividend until you have no more terms to divide.
Continuing the division:
ax^2 + bx^2 - x^3 + 2x^2
________________________________
x - 2 | x^3 + ax^2 + bx^2 + c
x^3 - 2x^2
_______________
ax^2 + bx^2
ax^2 - 2ax
___________________
bx^2 + 2ax
bx^2 - 2bx
______________________
2ax + 2bx + c
2ax - 4
_______________________________
c + 4
5. The remainder is c + 4.
Now, we know that the remainder is 7 when divided by (x - 2). So, we have c + 4 = 7.
c + 4 = 7
c = 3
Now, let's go back to the factored form of f(x) = (x-1)(x-k)(x-k^2).
We know that c = -k^3, so -k^3 = 3.
This implies k^3 = -3.
Now, substitute k^3 = -3 into the equation you need to show: k^3 - 2k^2 - 2k - 3 = 0.
(-3) - 2k^2 - 2k - 3 = 0
-2k^2 - 2k - 6 = 0
Divide both sides by -2 to simplify:
k^2 + k + 3 = 0
So, the correct equation we have is k^2 + k + 3 = 0, not k^3 - 2k^2 - 2k - 3 = 0.
I hope this helps clarify the process and point out the mistake. Let me know if you have any further questions!