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Question Number 138872 by peter frank last updated on 19/Apr/21

	Answered by bramlexs22 last updated on 19/Apr/21
	$p(x)=x^3 +(A−7)x^2 +Ax−8 p(x)=0⇒x^3 +(A−7)x^2 +Ax−8=0 let { ((x_1 =(a/r))),((x_2 =a)),((x_3 =ar)) :} by Vieta′s rule ⇒x_1 .x_2 .x_3 = 8 ; a^3 =8 ; a=2 ⇒x_2 =2⇒p(2)=0 ⇒8+(A−7).4+2A−8=0 ⇒6A= 28 ; A=((14)/3) p(x)=x^3 −(7/3)x^2 +((14)/3)x−8$
	$p (x) = x^{3} + (A - 7) x^{2} + Ax - 8$ $p (x) = 0 \Rightarrow x^{3} + (A - 7) x^{2} + Ax - 8 = 0$ $let {\begin{cases} x_{1} = \frac{a}{r} \\ x_{2} = a \\ x_{3} = ar \end{cases} by {Vieta}^{'} s rule$ $\Rightarrow x_{1} . x_{2} . x_{3} = 8; a^{3} = 8; a = 2$ $\Rightarrow x_{2} = 2 \Rightarrow p (2) = 0$ $\Rightarrow 8 + (A - 7) . 4 + 2 A - 8 = 0$ $\Rightarrow 6 A = 28; A = \frac{14}{3}$ $p (x) = x^{3} - \frac{7}{3} x^{2} + \frac{14}{3} x - 8$
	Commented by peter frank last updated on 19/Apr/21

	$t h a n k y o u$
	Answered by Rasheed.Sindhi last updated on 21/Apr/21

	$A n O t h e r W a y$ $P (x) = x^{3} + (A - 7) x^{2} + A x - 8 . . . (i)$ $L e t a, a r, {a r}^{2} a r e r o o t s .$ $P (x) = (x - a) (x - a r) (x - {a r}^{2})$ $= x^{3} - a (r^{2} + r + 1) x^{2} + a^{2} (r^{2} + r + 1) x - a^{3} . . . (i i)$ $C o m p a r i n g (i) & (i i)$ $^{∙} - a (r^{2} + r + 1) = A - 7 . . . (i i i)$ $^{∙} a^{2} (r^{2} + r + 1) = A . . . (i v)$ $^{∙} a^{3} = 8 \Rightarrow a = 2 . . . . . . . . . . . . . . . (v)$ $(i v) / (i i i) : - a = \frac{A}{A - 7} = - 2$ $A + 2 A = 14$ $A = \frac{14}{3}$ $(i i i) : a^{2} (r^{2} + r + 1) = A$ $\Rightarrow 2^{2} (r^{2} + r + 1) = \frac{14}{3}$ $6 r^{2} + 6 r - 1 = 0$ $r = \frac{- 6 \pm \sqrt{36 + 24}}{12} = \frac{- 3 \pm \sqrt{15}}{6}$ $P (x) = x^{3} + (\frac{14}{3} - 7) x^{2} + \frac{14}{3} x - 8$ $∙ P (x) = x^{3} - \frac{7}{3} x^{2} + \frac{14}{3} x - 8$ $∙ R o o t s a r e : a, a r, {a r}^{2}$ $2, 2 (\frac{- 3 \pm \sqrt{15}}{6}), 2 {(\frac{- 3 \pm \sqrt{15}}{6})}^{2}$
	Commented by peter frank last updated on 12/May/21

	$t h a n k y o u$
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