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 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.
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 $\displaystyle q= 0$
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 $\displaystyle x = 4.$
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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: .$\displaystyle aq = 1,\;q = 0 \quad\Rightarrow\quad a\!\cdot\!0 = 1$ .??
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.
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.