Prove: if $ax^2-bx+c$ and $dx^3-bx+c$ have a common factor, then $a^3-abd+cd^2=0$.

Question

Prove: if $ax^2-bx+c$ and $dx^3-bx+c$ have a common factor, then $a^3-abd+cd^2=0$.

Not really sure how to proceed. I know that first I have to find the common factor. I guess it is $x-c$ because both expressions end with the constant $c$. Then plug in $x=c$ to get $ac^2-bc+c=0, c^3d-bc+c=0$. Hence:

$ac^2=c(b-1) $

$c^3d=c(b-1)$

$ac^2=c^3d \Rightarrow a =cd $

Plug into the original third expression:

$(cd)^3-(cd)bd+cd^2=0 \Rightarrow c^3d^3-bcd^2+cd^2=0 \Rightarrow c^3d^3=cd^2(b-1) \Rightarrow c^2d=b-1$

This is greatly simplified, but it doesn't prove it is equal to $0$. I think it would be helpful to know the proper way to first find the common factor instead of just guessing. Thanks!

Answer 1

Let $t$ be the common root. Then

$dt^3-bt+c=at^2-bt+c$

$dt^3-at^2=0$

$t^2(dt-a)=0$

$t=0$ or $t=\frac{a}{d}$

But $t=0$ is not a common root of the equations for $c\neq 0$. If $c=0$, $t$ is a common root but it does not satisfy the condition of the question.

For $t=\frac{a}{d}$

$a (\frac{a}{d})^2-b(\frac{a}{d})+c=0$

$a^3-abd+cd^2=0$.

Answer 2

The question is

Prove: if $ax^2-bx+c$ and $dx^3-bx+c$ have a common factor, then $a^3-abd+cd^2=0$.

The common factor can be of degree $0, 1,$ or $2$. If the common factor is of degree $0$, then it divides $a, b, c, d$ but otherwise there is no constraint on them. Thus, we assume it is of degree greater than zero. If $\,c=0\,$ then the polynomials have a common factor of $\,x\,$ and $\,a, b, d\,$ otherwise have no constraint. Thus, we assume further that $\,c\ne 0.$

Compute the linear combination of the two polynomials

$$a(dx^3-bx+c)-dx(ax^2-bx+c) = (dx-a)(bx-c). \tag1 $$

Any common factor of the two given polynomials must divide any linear combination of them and hence the product of the two linear factors. Assuming the common factor is quadratic, then from equation $(1)$ we must have (where $u\ne 0$)

$$ u(ax^2-bx+c) = uax^2 -ubx +uc = (dx-a)(bx-c) = bdx^2 -(ab+cd)x +ac. $$

Equating the constant term implies $\,u=a.\,$ Now equating the first degree term implies $\,d=0\,$ which implies $\,a=0\,$ and hence $\,u=0\,$ which contradicts $\,u\ne 0.\,$ Thus, the common factor must be linear.

Again, from equation $(1)$ the linear factor must divide $\,(dx-a)\,$ or $\,(bx-c).\,$ Assume the second case then (where $\,u\ne 0$)

$$ ax^2-bx+c = (bx-c)(ux-v) = bux^2 -(cu+bv)x + cv. $$

Again, this implies that $\,a=bu,\,$ and $\,v=1\,$ since $\,c\ne 0.\,$ But now, the degree $1$ term implies that $\,cu=0\,$ which contradicts both $\,c\ne 0\,$ and $\,u\ne 0.\,$

So, assume the first case then (where $\,du\ne 0$)

$$ ax^2-bx+c = (dx-a)(ux-v) = dux^2 -(au+dv)x + av. $$

Equating the degree $0$ and $2$ terms to zero implies that $\,a = du, c = av.\,$ Equating the degree $1$ term to zero implies that $\, du^2 + dv - b = 0.\,$ Now substitute the values of $\,a,c\,$ in terms of $\,d,u,v\,$ into the condition

$$ a^3 - abd + cd^2 = ud^2(du^2 + dv - b) $$

which now equates to zero.

The result just proven is this:

Theorem: Given two polynomials with real coefficients $\,ax^2-bx+c\,$ and $\,dx^3-bx+c.\,$ Assuming $\,c\ne 0,\,$ if they have have a common linear factor, then $a^3-abd+cd^2=0$.