Find Real Solutions to x4-2x3+kx2+px+36 = 0

[DragonMaths]

One of the solutions to
x⁴-2x³+kx²+px+36 = 0 is x = 3i

Prove that this polynomial has no real solutions (roots) and find the real values of k and p.
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So far the only progress I seem to have made was possibly finding the value of p.

Since x - 3i is a solution , we can say that x + 3i is also a solution.

I then proceeded to replacing x with 3i on one side, and then setting the solution to itself but with x as -3i , as follows :

x⁴-2x³+kx²+px+36 = x⁴-2x³+kx²+px+36
(3i)⁴-2(3i)³+k(3i)²+p(3i)+36 = (-3i)⁴-2(-3i)³+k(-3i)²+p(-3i)+36
81i⁴-54i³+9ki+3pi+36 = 81i⁴+54i³+9ki-3pi+36
0 = 108i³-6pi
0=-108i-6pi
0=-(108-6p)i
0=108-6p
6p=108
p=18

This is as much progress as I have made , and I cannot find anything that can help me
Keep in mind this is asked pre-calculus on on 11th grade Advanced Math course with university grade math problems and so on.Any help would be grately appreciated as I have no idea how to continue at this moment.

 

[S.G. Janssens]

Are you sure about the sign of $p$?

Since $3i$ and $-3i$ have to satisfy the equation separately, substitution gives you a system of two independent linear equations in the two unknowns $p$ and $k$. As a starter, from this you can solve both $p$ and $k$.

 

[I like Serena]

One of the solutions to
x⁴-2x³+kx²+px+36 = 0 is x = 3i

Prove that this polynomial has no real solutions (roots) and find the values of k and p.
-------------------------------------------------------------------------------------------------------------------

So far the only progress I seem to have made was possibly finding the value of p.

Since x - 3i is a solution , we can say that x + 3i is also a solution.

I then proceeded to replacing x with 3i on one side, and then setting the solution to itself but with x as -3i , as follows :

x⁴-2x³+kx²+px+36 = x⁴-2x³+kx²+px+36
(3i)⁴-2(3i)³+k(3i)²+p(3i)+36 = (-3i)⁴-2(-3i)³+k(-3i)²+p(-3i)+36
81i⁴-54i³+9ki+3pi+36 = 81i⁴+54i³+9ki-3pi+36
0 = 108i³-6pi
0=-108i-6pi
0=-(108-6p)i
0=108-6p
6p=108
p=18

This is as much progress as I have made , and I cannot find anything that can help me
Keep in mind this is asked pre-calculus on on 11th grade Advanced Math course with university grade math problems and so on.Any help would be grately appreciated as I have no idea how to continue at this moment.

Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

 

[S.G. Janssens]

Note that the (as usual correct!) post #3 gives you a way to obtain the quotient of the polynomial $p(x) = x^4 - 2x^3 + kx^2 + px + 36$ upon division by the known factor $m(x) = x^2 + 9$. Namely, it sets you up for solving a system of equations for the coefficients of $q$.

Another way to obtain $q$ is by long division of $p$ by $m$. I don't know if this is part of your pre-calculus course, but here it usually is, and it is a useful technique to know.

 

[DragonMaths]

Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

Firstly,thanks for welcoming me :) .Secondly , the question states that we have to prove all the factors are complex,which answers your question that the remaining $x^2+ax+b=0$ would not have any real roots as it is a factor.Thanks you for your contribution.

 

[I like Serena]

Firstly,thanks for welcoming me :) .Secondly , the question states that we have to prove all the factors are complex,which answers your question that the remaining $x^2+ax+b=0$ would not have any real roots as it is a factor.Thanks you for your contribution.

Indeed, so we still need to find those values for a and b, and verify that the discriminant, which is $D=a^2-4b$ in this case, is negative, yielding complex solutions. (Nerd)

 

[DragonMaths]

Note that the (as usual correct!) post #3 gives you a way to obtain the quotient of the polynomial $p(x) = x^4 - 2x^3 + kx^2 + px + 36$ upon division by the known factor $m(x) = x^2 + 9$. Namely, it sets you up for solving a system of equations for the coefficients of $q$.

Another way to obtain $q$ is by long division of $p$ by $m$. I don't know if this is part of your pre-calculus course, but here it usually is, and it is a useful technique to know.

So I have tried using long division and synthetic division,but since we have to either divide using a complex number (which is not too much of a problem) or we use the known factor of $x^2 + 9$ ,but the unknown coefficients are still both present in the new polynomial created, namely being $x^2 -2x + k - 9$ with a remainder of $(p + 18)x - 9k +117$ which should be equal to 0 , as there should be no remainder when dividing by a factor.But as both $p$ and $k$ are present in the remainder , I'm stumped.

 

[I like Serena]

So I have tried using long division and synthetic division,but since we have to either divide using a complex number (which is not too much of a problem) or we use the known factor of $x^2 + 9$ ,but the unknown coefficients are still both present in the new polynomial created, namely being $x^2 -2x + k - 9$ with a remainder of $(p + 18)x - 9k +117$ which should be equal to 0 , as there should be no remainder when dividing by a factor.But as both $p$ and $k$ are present in the remainder , I'm stumped.

Indeed. The remainder must be zero.
So that puts restrictions on p and k.
It means for starters that $p=-18$ (with the correct sign now, as Krylov pointed out).
Otherwise $3i$ and $-3i$ wouldn't be solutions.

Btw, is it given that k and p are real numbers?
As it is, that is not given in the problem statement.
And otherwise we cannot assume that $x=-3i$ is a solution.

 

[DragonMaths]

Hi DragonMaths, welcome to MHB! (Wave)

Indeed, if $x=3i$ is a solution then $x=-3i$ is also a solution assuming that all coefficients are real numbers.
It means that we can factorize the polynomial.
And we can pick as a known factor $(x-3i)(x+3i)=(x^2+9)$.

So we can write the polynomial as:
$$(x^2+9)(x^2+ax+b) = x^4+ax^3+(b+9)x^2+9ax+9b$$

It follows that $a=-2,\ b+9=k,\ 9a=p,\ 9b=36$.
Can we find the constants from that?
Does the remaining $x^2+ax+b=0$ have any real roots? (Wondering)

Also , this is a great way to solve it and I did not even think that would so simple.It seems like a valid way to get the coefficients,and yet I've never seen someone do it in that way?

- - - Updated - - -

Indeed. The remainder must be zero.
So that puts restrictions on p and k.
It means for starters that $p=-18$ (with the correct sign now, as Krylov pointed out).
Otherwise $3i$ and $-3i$ wouldn't be solutions.

Btw, is it given that k and p are real numbers?
As it is, that is not given in the problem statement.
And otherwise we cannot assume that $x=-3i$ is a solution.

Just updated the original post thank you,the question does state that it is indeed a real number.

 

[I like Serena]

Also , this is a great way to solve it and I did not even think that would so simple.It seems like a valid way to get the coefficients, and yet I've never seen someone do it in that way?

It's just one of the many tricks of the trade.
A textbook may focus on explaining long division, and leave this approach out.
It's certainly easier to explain, and I've always preferred the simplest, most intuitive approach. ;)

 

[DragonMaths]

So after calculating and factorising a little it seems that $x^4 - 2x^3 + 13x^2 - 18x - 36 = 0$ is the solution to the problem , with $k = 13$ and $p = - 18$ for the coefficients , and the roots of the polynomial being equal to $(x + 3i)(x - 3i)(x - 1 + \sqrt{3}i)(x - 1 - \sqrt{3}i)$

 

[I like Serena]

Yep. That's what I found as well. (Nod)

 

[Greg]

Alternatively (and assuming $p$ and $k$ are real),

Substituting $3i$ and $-3i$ for $x$ in $f(x)=x^4-2x^3+kx^2+px+36$ we arrive at

$$81+54i-9k+3ip+36=0$$

and

$$81-54i-9k-3ip+36=0$$

Adding these two equations gives

$$162-18k+72=0\implies k=13$$

It can then be found that $p=-18.$

Now, if the other two roots were purely imaginary, $f(x)$ would have no odd powers of $x$ so we have

$$f(x)=(x^2+9)(x-(a+bi))(x-(a-bi))=(x^2+9)(x^2-(a+bi)x-(a-bi)x+(a+bi)(a-bi))=(x^2+9)(x^2-2ax+a^2+b^2)$$

Now, equating coefficients with $x^4-2x^3+13x^2-18x+36$, we have

$$9(a^2+b^2)=36\implies a^2+b^2=4$$

$$-2a=-2\implies a=1\Rightarrow b=\pm\sqrt3$$

The roots of $x^2-2x+4=0$ are $1\pm\sqrt3i$, as required.