Question Number 186292 by ajfour last updated on 03/Feb/23

(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....