Question

Find the condition that $x^3-3px+2q$ may be divisible by a factor of the form $x^2+2ax+a^2$.

Answer 1

As $x^3-3px+2q$ is divisible by $x^2+2ax+a^2$ that is ${(x+a)}^2$ which means the cubic equation must have three roots .

Let the three roots be $\alpha ,\beta ,\gamma$ .

Thus $\alpha =-a$

$\beta =-a$

Thus sum of three roots of cubic equation will be $\alpha +\beta +\gamma =(-a)+(-a)+\gamma =-\frac{x^2\left(coefficient\right)}{x^3\left(coefficient\right)}$

$-2a+\gamma =0$

$\gamma =2a$

Thus , $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{x\left(coefficient\right)}{x^3\left(coefficient\right)}$

Therefore,

$a^2-a\gamma -a\gamma =-3p$

$a^2-2a\gamma =-3p$

$a^2-2a\cdot 2a=-3p$

$a^2-4a^2=-3p$

$-3a^2=-3p$

$p=a^2$

$p^3=a^6$

Product of three roots of cubic equation will be $\alpha \cdot \beta \cdot \gamma =(-a)\cdot (-a)\cdot \gamma =\frac{constant\left(coefficient\right)}{x^2\left(coefficient\right)}$

Thus , $a^2\gamma =-2q$

$a^2\cdot 2a=-2q$

$2a^3=-2q$

$q=-a^3$

squaring both side ,

$q^2=a^6$

Thus , if $p^3=q^2$ then $x^3-3px+2q$ may be divisible by the factor of $x^2+2ax+a^2$