# Thread: Remainder theorem

1. ## Remainder theorem

given that f(x)=$\displaystyle x^3+kx^2-2x+1$ and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

i have $\displaystyle k^3+k(k)^2-kx+1= k$
$\displaystyle 2k^3 - 2k+1 = k$

am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
may i get some help with this?

2. Originally Posted by sigma1
given that f(x)=$\displaystyle x^3+kx^2-2x+1$ and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

i have $\displaystyle k^3+k(k)^2-kx+1= k$
$\displaystyle 2k^3 - 2k+1 = k$

am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
may i get some help with this?
Try this:
$\displaystyle 2k^3 - 2k+1 - k = 0$
$\displaystyle 2k(k^2-1)-(k-1)=0$
$\displaystyle 2k(k-1)(k+1) - (k-1)=0$
$\displaystyle (k-1)(2k(k+1)-1)=0$
etc

3. Originally Posted by sigma1
given that f(x)=$\displaystyle x^3+kx^2-2x+1$ and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

i have $\displaystyle k^3+k(k)^2-kx+1= k$
$\displaystyle 2k^3 - 2k+1 = k$

am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
may i get some help with this?
Perhaps you tried to divide $\displaystyle 2k^3-2k+1$ by $\displaystyle k-1$? - That doesn't work out as expected to be sure. But what about dividing $\displaystyle 2k^3-3k+1$ by $\displaystyle k-1$?
In this case I get $\displaystyle (2k^3-3k+1)k-1)=2k^2+2k-1$.
Moral: you have to move everything to one side of the equation before trying the division by $\displaystyle k-1$.

4. Originally Posted by Failure
Perhaps you tried to divide $\displaystyle 2k^3-2k+1$ by $\displaystyle k-1$? - That doesn't work out as expected to be sure. But what about dividing $\displaystyle 2k^3-3k+1$ by $\displaystyle k-1$?
In this case I get $\displaystyle (2k^3-3k+1)k-1)=2k^2+2k-1$.
Moral: you have to move everything to one side of the equation before trying the division by $\displaystyle k-1$.

ok thanks for that but how do you factorise that to get the other values for k?

5. Originally Posted by Debsta
Try this:
$\displaystyle 2k^3 - 2k+1 - k = 0$
$\displaystyle 2k(k^2-1)-(k-1)=0$
$\displaystyle 2k(k-1)(k+1) - (k-1)=0$
$\displaystyle (k-1)(2k(k+1)-1)=0$
etc
thanks . i follow your steps but how do i solve for the values of k

6. Originally Posted by sigma1
ok thanks for that but how do you factorise that to get the other values for k?
The other two values of k that are solutions to the original equation are irrational, so you can't find them by "educated guessing": you have to solve the quadratic equation $\displaystyle 2k^2+2k-1=0$ by whatever other means available (such as applying a memorized formula for that, or simply "completing the square").

7. Originally Posted by Failure
The other two values of k that are solutions to the original equation are irrational, so you can't find them by "educated guessing": you have to solve the quadratic equation $\displaystyle 2k^2+2k-1=0$ by whatever other means available (such as applying a memorized formula for that, or simply "completing the square").
i have tried completing the square but was unable to over several attempts.

8. Originally Posted by sigma1
i have tried completing the square but was unable to over several attempts.
$\displaystyle 2k^2+2k-1=2(k^2+k)-1=2\left[\left(k+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right]-1=2\left(k+\frac{1}{2}\right)^2-\frac{3}{2}$.
So to solve that quadratic equation, you solve

$\displaystyle 2\left(k+\frac{1}{2}\right)^2-\frac{3}{2}=0$ for k, which gives $\displaystyle k_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$.

9. Originally Posted by sigma1
thanks . i follow your steps but how do i solve for the values of k
$\displaystyle (k-1)(2k^2+2k-1)=0$
Use the null factor law to get:
$\displaystyle k-1=0 {or} 2k^2+2k-1=0$
The first will lead to the solution k=1
The second can be solved using the quadratic formula, so you will end up with 3 solutions overall.

10. Originally Posted by sigma1
given that f(x)=$\displaystyle x^3+kx^2-2x+1$ and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

i have $\displaystyle k^3+k(k)^2-kx+1= k$
$\displaystyle 2k^3 - 2k+1 = k$
$\displaystyle 2k^3- 3k+ 1= 0$
By the rational root theorem, the only possible rational roots are 1, -1, 1/2, and -1/2. Putting those into the equation, $\displaystyle 2(1)^3- 3(1)+ 1= 0$, $\displaystyle 2(-1)^3- 3(-1)+ 1= 2$, $\displaystyle 2(1/2)^3- 3(1/2)+ 1= -1/4$, and $\displaystyle 2(-1/2)^3- 3(-1/2)+ 1= 2$.

Since 1 is a root, $\displaystyle 2k^3- 3k+ 1$ has k- 1 as a factor: $\displaystyle 2k^3- 3k+ 1= (k- 1)(2k^2+ 2k+ 1)$. You can use the quadratic formula to find the other two roots.

am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
may i get some help with this?
Then youare dividing wrong!