1. Polynomial divison.

if x^3+px+7 is completely divisible by qx^2+r, then find the relation between p,q and r.
I have tried to solve it by putting the value of x^2=-r/q but unable to reach any solution.

2. Are you sure this is the problem? Unless $\displaystyle q=0$ it seems to me impossible

3. What makes you say that, Dinky? There's limitless solutions.
Example: x=2,p=18,q=3,r=5
x^3 + px + 7 = 51
qx^2 + r = 17
51/17 = 3

4. That's true however, I'm not so sure how to interpret the problem.

If we just see it as polynomial division (and don't give x a specific value)

Then it's just like I said.

$\displaystyle x^3+px+7 = (qx^2+r)(ax+b) = aqx^3+bqx^2+arx+br \Rightarrow$

$\displaystyle bq = 0$ but $\displaystyle b= 7/r\neq 0$

We get $\displaystyle q= 0$

5. Originally Posted by Wilmer
What makes you say that, Dinky? There's limitless solutions.
Example: x=2,p=18,q=3,r=5
x^3 + px + 7 = 51
qx^2 + r = 17
51/17 = 3
Sorrư! How have you reached this solution? And what is the relation between p,q and r?

6. Hello, ohm!

if $\displaystyle x^3+px+7$ is divisible by $\displaystyle qx^2+r$,
then find the relation between $\displaystyle p,q$ and $\displaystyle r.$

This is a Factoring problem.
The final result should be true for all values of $\displaystyle x.$
. . We are not allowed to insert constant values for $\displaystyle x.$

Otherwise, this example would be correct.

. . Is $\displaystyle x^2 + 9$ divisible by $\displaystyle x+1$ ?

. . Answer: .Yes, let $\displaystyle x = 4.$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I agree with Dinkydoe . . . The problem is faulty.

Since $\displaystyle x^3 + px + 7$ is divisible by $\displaystyle qx^2 + r$,

. . we have: .$\displaystyle (ax+b)(qx^2+r) \:=\:x^2 + px + 7$

. . . . for some binimial, $\displaystyle ax+b.$

Hence: .$\displaystyle aqx^3 + bqx^2 + arx + br \;=\;x^3 + px + 7$

Equate coefficients: .$\displaystyle \begin{Bmatrix}aq &=& 1 & [1] \\ bq &=& 0 & [2] \\ ar &=& p & [3] \\ br &=& 7 & [4] \end{Bmatrix}$

From [4], we have: .$\displaystyle b \ne 0,\;r \ne 0$

In [2] we have: .$\displaystyle bq = 0,\;b \ne 0 \quad\Rightarrow\quad q = 0$

In [1] we have: .$\displaystyle aq = 1,\;q = 0 \quad\Rightarrow\quad a\!\cdot\!0 = 1$ .??

7. I think you shuld solve it thus;
Let A(x)=x^3+px+7.
By Polynomial long division we know that A(x)=(qx^2+r)Q(x)+R(x) and To be A(x) divisible by qx^2+r, we must have R(x)=0.
If we divide A(x) by qx^2+r, we earn A(x)= [(x/q)(qx^2+r)]+((pq-r)/q)x+7.
Therefore by my last assertion we have ((pq-r)/q)x+7=0 => x=-7/(p+(r/q)).
This means that if x=-7/(p+(r/q)), then R(x)=0,
So subsituate this value for x in the main division expression:
A(-7/(p+(r/q)))=q [(-7/(p+(r/q)))^2+r]Q(x).
Now you can find the relation between p,q and r.

8. Originally Posted by ohm
if x^3+px+7 is completely divisible by qx^2+r, then find the relation between p,q and r.
I have tried to solve it by putting the value of x^2=-r/q but unable to reach any solution.
If we write $\displaystyle x^3+px+7=\left(\frac{x}{q}+\frac{7}{r}\right)\left (qx^2+r\right)$

then $\displaystyle x^3+px+7=x^3+\frac{7q}{r}x^2+\frac{r}{q}x+7$

giving $\displaystyle px=\frac{7q}{r}x^2+\frac{r}{q}x$

which imposes a restriction on x rather than allowing x to be the set of reals.

For example, q=1, r=1... then choosing x=2 and p=15 is a solution
but not for x in general.

For all x, we'd require $\displaystyle px=\frac{7q}{r}x^2+\frac{r}{q}x$

or $\displaystyle q=0\ if\ \frac{r}{q}=p$

Equating coefficients works if we are dealing with x as the set of reals or complex numbers.

If we perform polynomial division, we get

$\displaystyle \frac{x^3+px+7}{qx^2+r}=\frac{x}{q}+\frac{\left(\f rac{pq-r}{q}\right)x+7}{qx^2+r}$

If we say that the numerator of the rightmost term must be zero, we impose the following restriction on x...

$\displaystyle x^3+px+7=x^3+\frac{r}{q}x\ \Rightarrow\ px+7=\frac{r}{q}x$

$\displaystyle x=1,\ p=-6,\ q=r=2$ is one way to satisfy that.

The restriction $\displaystyle \frac{r-pq}{q}x=7$ is the above restriction.

Also, in saying the rightmost numerator must be zero, we get "in general" 7=0 and $\displaystyle p=\frac{r}{q}$ by equating coefficients.