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r,q are constants. I need to factor this equation such that there is a double root.

[tex]-\frac{r}{q}u^3+ru^2-\left(\frac{r}{q}+1\right)u+r=0[/tex]

Are there any tricks for this because this just a nasty equation.

I don't know if that is a wise approach but:

[tex](au+b)(cu+d)^2 = ac^2u^3+(2acd+c^2b)u^2+(ad^2+2bcd)u+bd^2[/tex]

Then

[tex]ac^2 = -\frac{r}{q}[/tex]

[tex]2acd+c^2b=bd^2=r[/tex]

[tex]ad^2+2bcd = \frac{r}{q}+1[/tex]

[tex]-\frac{r}{q}u^3+ru^2-\left(\frac{r}{q}+1\right)u+r=0[/tex]

Are there any tricks for this because this just a nasty equation.

I don't know if that is a wise approach but:

[tex](au+b)(cu+d)^2 = ac^2u^3+(2acd+c^2b)u^2+(ad^2+2bcd)u+bd^2[/tex]

Then

[tex]ac^2 = -\frac{r}{q}[/tex]

[tex]2acd+c^2b=bd^2=r[/tex]

[tex]ad^2+2bcd = \frac{r}{q}+1[/tex]

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