If the HCF of the polynomials $x^3+px+q $ and $x^3+rx^2+lx+x$ is $x^2+ax+b$, then their LCM is? (provided that $r≠0$) I tried multiplying the two polynomials together and then dividing them by the HCF (as product=HCF*LCM), but reached nowhere.
Then, I used the factor theorem but also got stuck.
Can somebody help?
 A: Your idea works. First, comparing coef's of $\,x^{\large 2}\,$ below yields $\,\color{#0a0}{f_1/g}\, =\, \color{#c00}{x-a}$ 
$\quad\ \ \smash[t]{\overbrace{x^{\large 3}+px+q}^{\Large\color{#0a0}{ f_1}}\, =\, (\color{#c00}{x+c})(\overbrace{x^{\large 2}+ax+b}^{\Large\color{#0a0} g})\,\Rightarrow\, \overbrace{0 = c+a}^{\Large {\rm coef\ of\ } x^2}\,\Rightarrow\, \color{#c00}{c = -a}} \phantom{|^{|^{|^{|^{|^I}}}}}$
Therefore $\ {\rm lcm}(f_1,f_2) =  \dfrac{f_1 f_2}{\gcd(f_1,f_2)} = \dfrac{f_1 f_2}{g} = \color{#0a0}{\dfrac{f_1}{g}} f_2 = (\color{#c00}{x-a})f_2$

Remark $ $ Similarly: $ $ lcm $= (x\!+\!r\!-\!a)f_1,\ $ so $\ (x\!+\!r\!-\!a)f_1 = (x\!-\!a) f_2.\ $  Thus $\ r = 0\,\Rightarrow\, f_1 = f_2.\,$ Conversely $\, f_1 = f_2\,\Rightarrow\, r = 0\,$ by comparing coef's of $\,x^{\large 2}$. Thus the hypothesis $\, r\neq 0\,$ is redundant, since if $\,r = 0\,$ then $\,f_1 = f_2\,$ so their gcd $= f_1\,$ has degree $3$, contra hypothesis.
A: If $b \neq 0$,
$$x^2+ax+b \mid x(x^2+rx+l+1) $$
This would imply that $x^2+ax+b = x^2+rx+l+1 $.
And the LCM would be,
$$\frac{x(x^2+rx+l+1)(x^3+px+q)}{x^2+ax+b}=x^4+px^2+q$$
If $b=0$,
$$x+a|x^2+rx+l+1$$
if you divide you'll find the quotient to be $x+(r-a)$ and remainder to be $l+1-a(r-a)=0$
LCM is,
$$(x^3+px+q)(x+r-a)$$
