Are you sure this is the problem? Unless it seems to me impossible
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.
Hello, ohm!
if is divisible by ,
then find the relation between and
This is a Factoring problem.
The final result should be true for all values of
. . We are not allowed to insert constant values for
Otherwise, this example would be correct.
. . Is divisible by ?
. . Answer: .Yes, let
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I agree with Dinkydoe . . . The problem is faulty.
Since is divisible by ,
. . we have: .
. . . . for some binimial,
Hence: .
Equate coefficients: .
From [4], we have: .
In [2] we have: .
In [1] we have: . .??
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
then
giving
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
or
Equating coefficients works if we are dealing with x as the set of reals or complex numbers.
If we perform polynomial division, we get
If we say that the numerator of the rightmost term must be zero, we impose the following restriction on x...
is one way to satisfy that.
The restriction is the above restriction.
Also, in saying the rightmost numerator must be zero, we get "in general" 7=0 and by equating coefficients.