# Solve this problem that involves the factor theorem

• chwala

#### chwala

Gold Member
Homework Statement
If ##4x^3+kx^2+px+2## is divisible by ##x^2+λ^2##. Prove that ##kp=8##
Relevant Equations
Factor theorem
My attempt;
##4x^3+kx^2+px+2=(x^2+λ^2)(4x+b)##
##4x^3+kx^2+px+2=4x^3+bx^2+4λ^2x+bλ^2##
##⇒k=b, p=4λ^2 , bλ^2=2##

##\dfrac{4λ^2}{bλ^2}=\dfrac{p}{2}##
##\dfrac{4}{b}=\dfrac{p}{2}##
##⇒8=pb## but ##b=k##
##⇒8=kp##

Any other approach appreciated...

1. Same approach, but don't divide. Write kp = , and then plug in the formulas you derived for k and p. Otherwise you have to prove that you didn't divide by 0.
2. Different approach. Plug in iλ. Set the real and imaginary parts equal to 0.

chwala
chwala said:
Homework Statement:: If ##4x^3+kx^2+px+2## is divisible by ##x^2+λ^2##. Prove that ##kp=8##
Relevant Equations:: Factor theorem

My attempt;
##4x^3+kx^2+px+2=(x^2+λ^2)(4x+b)##
##4x^3+kx^2+px+2=4x^3+bx^2+4λ^2x+bλ^2##
##⇒k=b, p=4λ^2 , bλ^2=2##

##\dfrac{4λ^2}{bλ^2}=\dfrac{p}{2}##
##\dfrac{4}{b}=\dfrac{p}{2}##
##⇒8=pb## but ##b=k##
##⇒8=kp##

Any other approach appreciated...
Slightly different approach and maybe a bit cleaner: Since you've determined that ##k=b, p=4λ^2## , and ##bλ^2=2##, then ##b = \frac 2 {\lambda^2}##, so ##kp = b \cdot 4 \lambda^2 = \frac 2 {\lambda^2} \cdot 4\lambda^2 = 8##.

Prof B said:
1. Same approach, but don't divide.
@chwala didn't do any division. Since ##4x^3+kx^2+px+2## is divisible by ##x^2+λ^2##, then ##4x^3+kx^2+px+2## is equal to the product of ##x^2 + \lambda^2## and a linear polynomial whose x term has a coefficient of 4. That is, ##4x + b##.

chwala
You divided 4λ2 by bλ2

Prof B said:
1. Same approach, but don't divide. Write kp = , and then plug in the formulas you derived for k and p. Otherwise you have to prove that you didn't divide by 0.
2. Different approach. Plug in iλ. Set the real and imaginary parts equal to 0.
Different approach. Plug in iλ. Set the real and imaginary parts equal to 0. I do not seem to get this part where you're talking of complex numbers. Kindly be clear on this part.

chwala said:
Different approach. Plug in iλ. Set the real and imaginary parts equal to 0. I do not seem to get this part where you're talking of complex numbers. Kindly be clear on this part.
It goes like this.

Factor the sum of squares. ## ~~ x^2 + \lambda^2 = (x+i\lambda)(x-i\lambda)~.~~## Right?

So ##~(x-i\lambda) ~ ## is a factor of ##f(x)=4x^3+kx^2+px+2##.

What does the factor theorem say? - - - in part: If ##x-a## is a factor of polynomial ##P(x)##, then ##P(a)=0##.

So in your case, plug ##i\lambda## into ##f(x)## and set that equal to zero. Since you get a complex result, both the real part is zero and the imaginary part is zero, just as @Prof B said.

SammyS said:
It goes like this.

Factor the sum of squares. ## ~~ x^2 + \lambda^2 = (x+i\lambda)(x-i\lambda)~.~~## Right?

So ##~(x-i\lambda) ~ ## is a factor of ##f(x)=4x^3+kx^2+px+2##.

What does the factor theorem say? - - - in part: If ##x-a## is a factor of polynomial ##P(x)##, then ##P(a)=0##.

So in your case, plug ##i\lambda## into ##f(x)## and set that equal to zero. Since you get a complex result, both the real part is zero and the imaginary part is zero, just as @Prof B said.
Does it make any difference? Any justification for this step?
When you expand ## ~~ (x+i\lambda)(x-i\lambda)~~~= ~~ x^2 + \lambda^2## you will end up with the original factor ##x^2+λ^2##, which i simply used directly to work to solution. Is there a need or rather any advantage to factorize so as to help realize the envisaged solution?

Ok for the sake of the argument, let me go with you here;
If ##(x-iλ)## is a factor of the polynomial ##f(x)## then it follows that;
##0+0i=4λ^3i-kλ^2+pλi+2##
How will you solve the simultaneous equation;
##4λ^3+p=0##
##kλ^2+2=0##

Last edited:
chwala said:
Does it make any difference? Any justification for this step?
When you expand ## ~~ (x+i\lambda)(x-i\lambda)~~~= ~~ x^2 + \lambda^2## you will end up with the original factor ##x^2+λ^2##, which i simply used directly to work to solution. Is there a need or rather any advantage to factorize so as to help realize the envisaged solution?

Ok for the sake of the argument, let me go with you here;
If ##(x-iλ)## is a factor of the polynomial ##f(x)## then it follows that;
##0+0i=4λ^3i-kλ^2+pλi+2##
How will you solve the simultaneous equation;
##4λ^3+p=0##
##kλ^2+2=0##
To be fair, @Prof B merely said that this is a different approach. - So it might not be better. To see if you like it, you need to try it. I'm glad to see you do that. Now, on to your attempt.

You left a ##\lambda## out of one of those equations as well as a negative sign. By the way, in LaTeX that's \lambda .

##-4\lambda^3 +p\lambda=0~~## Factoring LHS gives ##~~(-\lambda)(4\lambda^2-p)=0 ##

Clearly, ##\lambda## can't be zero, since ##f(0)\ne 0##.

So ##4\lambda^2-p=0~~## and ##~~ -k\lambda^2+2=0## .

chwala
SammyS said:
To be fair, @Prof B merely said that this is a different approach. - So it might not be better. To see if you like it, you need to try it. I'm glad to see you do that. Now, on to your attempt.

You left a ##\lambda## out of one of those equations as well as a negative sign. By the way, in LaTeX that's \lambda .

##-4\lambda^3 +p\lambda=0~~## Factoring LHS gives ##~~(-\lambda)(4\lambda^2-p)=0 ##

Clearly, ##\lambda## can't be zero, since ##f(0)\ne 0##.

So ##4\lambda^2-p=0~~## and ##~~ -k\lambda^2+2=0## .
Ok, i can see that you end up with ##4\lambda^2=p##, therefore;
##-k\left[\dfrac {p}{4}\right] +2=0##
##-kp+8=0##
##⇒kp=8##

chwala said:
Ok, i can see that you end up with ##4\lambda^2=p##, therefore;
##-k\left[\dfrac {p}{4}\right] +2=0##
##-kp+8=0##
##⇒kp=8##
Or try elimination. (Same result, of course.)

Multiply ##4\lambda^2-p=0~## by ##k## .

Multiply ##~ -k\lambda^2+2=0~## by ##4## .

##~~~4k\lambda^2-pk=0~##
## -4k\lambda^2+8=0~##

chwala
SammyS said:
Or try elimination. (Same result, of course.)

Multiply ##4\lambda^2-p=0~## by ##k## .

Multiply ##~ -k\lambda^2+2=0~## by ##4## .