Find the polynomial with roots that exceed the roots of f(x)=3x^3 −14x^2 +x+62=0 by 3. Hence determine the value of (1/(a+3))+(1/(b+3))+(1/(c+3)), where a,b and c are roots.


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Question Number 198968 by necx122 last updated on 26/Oct/23

$F i n d t h e p o l y n o m i a l w i t h r o o t s t h a t$ $e x c e e d t h e r o o t s o f$ $f (x) = 3 x^{3} - 14 x^{2} + x + 62 = 0 b y 3 . H e n c e$ $d e t e r m i n e t h e v a l u e o f \frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3},$ $w h e r e a, b a n d c a r e r o o t s .$
	Answered by Rasheed.Sindhi last updated on 26/Oct/23

	$L e t a, b, c a r e r o o t s o f 3 x^{3} - 14 x^{2} + x + 62 = 0$ $a + b + c = \frac{14}{3}, a b + b c + c a = \frac{1}{3}, a b c = - \frac{62}{3}$ $T h e r e q u i r e d e q u a t i o n h a v e r o o t s a + 3,$ $b + 3, c + 3$ $∙ (a + 3) + (b + 3) + (c + 3) = a + b + c + 9 = \frac{14}{3} + 9 = \frac{41}{3}$ $∙ (a + 3) (b + 3) + (b + 3) (c + 3) + (c + 3) (a + 3)$ $= a b + 3 (a + b) + 9 + b c + 3 (b + c) + 9 + c a + 3 (c + a) + 9$ $= a b + b c + c a + 6 (a + b + c) + 27$ $= \frac{1}{3} + 6 (\frac{14}{3}) + 27 = \frac{1 + 84 + 81}{3} = \frac{166}{3}$ $∙ (a + 3) (b + 3) (c + 3)$ $= a b c + 3 (a b + b c + c a) + 9 (a + b + c) + 27$ $= - \frac{62}{3} + 3 (\frac{1}{3}) + 9 (\frac{14}{3}) = \frac{- 62 + 3 + 126}{3} = \frac{67}{3}$ $H e n c e t h e r e q u i r e d e q u a t i o n w i l l b e :$ $x^{3} + \frac{41}{3} x^{2} + \frac{166}{3} x + \frac{67}{3} = 0$ $\Rightarrow 3 x^{3} + 41 x^{2} + 166 x + 67 = 0$ $\frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3}$ $= \frac{(a + 3) (b + 3) + (b + 3) (c + 3) + (a + 3) (c + 3)}{(a + 3) (b + 3) (c + 3)}$ $= \frac{\frac{166}{3}}{\frac{67}{3}} = \frac{166}{67}$
	Commented by necx122 last updated on 26/Oct/23
	This is great and I'm grateful. well understood. Thank you sir.
	Commented by mr W last updated on 26/Oct/23

	$i t h i n k t h e q u e s t i o n s u g g e s t s o n e t o$ $s h i f t t h e g i v e n p o l y n o m i a l f (x) t o t h e$ $r i g h t d i r e c t i o n b y 3 t o g e t t h e n e w$ $p o l y n o m i a l a t f i r s t . t h i s i s t h e e a s i e s t$ $w a y t o d e t e r m i n e t h e n e w p o l y n o m i a l .$ $p l e a s e r e c h e c k y o u r c a l c u l a t i o n s i r!$ $i t h i n k s o m e t h i n g i s w r o n g i n i t,$ $s e e f o l l o w i n g d i a g r a m w i t h$ $(1) = o r i g i n a l p o l y n o m i a l f (x)$ $(2) = w h a t y o u g o t$ $(3) = w h a t i g o t (s e e b e l o w)$
	Commented by mr W last updated on 26/Oct/23

	Commented by deleteduser1 last updated on 26/Oct/23

	$T h e c o r r e c t e q u a t i o n s h o u l d b e :$ $3 x^{2} - 41 x^{2} + 166 x - 67 = 0$
	Commented by mr W last updated on 26/Oct/23

	$i t h i n k t h e c o r r e c t e q u a t i o n i s$ $k (3 x^{2} - 41 x^{2} + 166 x - 148) = 0 w i t h k \in R, k \neq 0$
	Commented by mr W last updated on 26/Oct/23

	Commented by deleteduser1 last updated on 26/Oct/23
	$I′m referring to the solution above. The solution would have been correct if a,b and c were all real numbers. (a+3)(b+3)(c+3) =abc+3(ab+bc+ca)+9(a+b+c)+27 is only true when a,b,c∈R When two(in the case of a cubic equation) come from C\R,then we are no longer dealing with variables(a,b,c) but (a,b^− ,c^− ) where b^− =x+yi. ⇒b^− +3=(x+3)+yi;so we are only increasing the real part by 3$
	$I^{'} m r e f e r r i n g t o t h e s o l u t i o n a b o v e .$ $T h e s o l u t i o n w o u l d h a v e b e e n c o r r e c t i f a, b a n d$ $c w e r e a l l r e a l n u m b e r s .$ $(a + 3) (b + 3) (c + 3)$ $= a b c + 3 (a b + b c + c a) + 9 (a + b + c) + 27 i s o n l y t r u e$ $w h e n a, b, c \in R$ $W h e n t w o (i n t h e c a s e o f a c u b i c e q u a t i o n) c o m e$ $f r o m C ∖ R, t h e n w e a r e n o l o n g e r d e a l i n g w i t h$ $v a r i a b l e s (a, b, c) b u t (a, \bar{b}, \bar{c}) w h e r e \bar{b} = x + y i .$ $\Rightarrow \bar{b} + 3 = (x + 3) + y i; s o w e a r e o n l y i n c r e a s i n g$ $t h e r e a l p a r t b y 3$
	Answered by mr W last updated on 26/Oct/23

	$s a y t h e n e w p o l y n o m i a l i s p (x) .$ $r o o t s o f p (x) e x c e e d t h e r o o t s o f f (x)$ $b y 3, t h a t m e a n s p (x) = f (x - 3) .$ $p (x) = 3 {(x - 3)}^{3} - 14 {(x - 3)}^{2} + (x - 3) + 62$ $p (x) = 3 x^{3} - 41 x^{2} + 166 x - 148$ $s a y r o o t s o f p (x) = 0 a r e A, B, C,$ $t h e n A = a + 3, B = b + 3, C = c + 3$ $\frac{1}{a + 3} + \frac{1}{b + 3} + \frac{1}{c + 3} = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}$ $= \frac{A B + B C + C A}{A B C} = \frac{\frac{166}{3}}{\frac{148}{3}} = \frac{166}{148} = \frac{83}{74} ✓$ $◼$
	Commented by necx122 last updated on 26/Oct/23
	Wow!! Having to exclaim is all i can do at this point. Funnily, both answers appear in the options.
	Commented by Rasheed.Sindhi last updated on 27/Oct/23

	$Most efficient way!$
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