Question

Given that

Answer 1

\(f(x)=x^3+kx^2-2x+1\) has a remainder k when it is divided by (x-k),find the possible values of k.

Answer 2

My answer seems correct,but not sure whether my working is correct.

Answer 3

Here is the working:
If \(x=k\),\(f(x)=k\)
\(k^3+k(k^2)-2k+1=0\)
\(k^3+k^3-3k+1=0\)
\(2k^3-3k+1=0\)
Let \(k=x\)
\(2x^3-3x+1=0\)
The constant term for \(2x^3-3x+1=0\) is 1.
Hence,if \((x-k)\) were to be a factor of \(2x^3-3x+1=0\) ,then k must be a factor of 1, \(k=\pm1\)
If \(x=1\) , \(2(1)^3-3(1)+1=0\)
\(0=0\)
Hence, \((x-1)\) is a factor of \(2x^3-3x+1\)
If \(x=-1\) , \(2(-1)^3-3(-1)+1=0\)
\(2\neq0\)
Hence, \((x+1)\) is not a factor of \(2x^3-3x+1=0\)
Hence,by factor theorem, \((k-1)\) is a factor of \(2k3-3k+1=0\)
Using long division method,
\(2k^3-3k+1=0\)
\((k-1)(2k^2+2k+1=0\)
Factorise \(2k^2+2k-1=0\)
\(k=\frac{-2\pm\sqrt{2^2-4(2)(-1)}}{2(2)}\)
\(k=\frac{-1\pm\sqrt{3}}{2}\)
Therefore, \(k=1\) , \(\frac{-1\pm\sqrt{3}}{2}\)

Answer 4

@Zepdrix

Answer 5

We hope there is a rational zero...
the rational zero test is what you were trying to use
the possible rational zeros according to the rational zero test are
all the factors of the constant term which is 1 or -1
and all the factors of (constant term)/(coefficient of leading term)
so we would also include 1/2 or -1/2
but it just so happened there was one rational zero and it was the one you started with which was k=1
Also to find the other zeros you could use synthetic division instead of long division
your choice
your work looks fine other than that one statement
kudos
also you don't have to plug in -1 after realizing you have k=1 is a solution since the other factor will be a quadratic

see http://www.sosmath.com/algebra/factor/fac10/fac10.html

Answer 6

Uh oh!
I think you made a mistake right at the start.

If dividing out the factor (x-k) leaves a remainder of k,
then by the remainder theorem, f(k)=k.

You don't want to set your polynomial equal to zero.
You want to set it equal to k.

\(\large\rm k^3+k(k^2)-2k+1\ne0\)

\(\large\rm k^3+k(k^2)-2k+1=k\)

Answer 7

oops I didn't see the =0 part
he had the next line as if he did meant \[k^3+k(k^2)-2k+1=k\] though

he also stated when x=k he had f(x)=k as if he had seen that he was suppose to have f(k)=k

Answer 8

Ohhh I didn't even read the next line XD haha my bad

Answer 9

the next line being the
\[2k^3-3k+1=0\]

Answer 10

which would result from the equation you had @Zepdrix

Answer 11

ops,it was a typo error @Zepdrix xD

Answer 12

http://prntscr.com/dmw6dg
Can u show me how to use synthetic division in this question?
I will like to know. ^