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Question Number 156059 by cortano last updated on 07/Oct/21

Commented by john_santu last updated on 07/Oct/21

$g (x) = 3 x^{3} + {(h (x) (x^{4} - 3 x + 1) + 2 x^{3} - 7 x)}^{2}$ $\frac{g (x)}{x^{4} - 3 x + 1} = u (x) (x^{4} - 3 x + 1) + {a x}^{3} + {b x}^{2} + c x + d$ $w h e r e {a x}^{3} + {b x}^{2} + c x + d i s r e m a i n d e r$ $w h e n 3 x^{3} + 4 x^{6} - 28 x^{4} + 49 x^{2} d i v i d e d$ $b y x^{4} - 3 x + 1$
	Answered by ajfour last updated on 07/Oct/21
	$f(x)=(2x^3 −3x)+[h(x)](x^4 −3x+1) 3x^3 +{(2x^3 −7x)+[h(x)](x^4 −3x+1)}^2 =(x^4 −3x+1)[s(x)] +ax^3 +bx^2 +cx+d now let x_1 ^4 +1=3x_1 ⇒ 3x_1 ^3 +(2x_1 ^3 −7x_1 )^2 =ax_1 ^3 +bx_1 ^2 +cx_1 +d ⇒ (3−a)(3x_1 −1)+4(3x_1 ^4 −x_1 ^3 ) −28(3x_1 ^2 −x_1 )+49x_1 ^3 =bx_1 ^3 +cx_1 ^2 +dx_1 ⇒ 9x_1 −3−3ax_1 +a+12(3x_1 −1) −4x_1 ^3 −84x_1 ^2 +28x_1 +(49−b)x_1 ^3 −cx_1 ^2 −dx_1 =0 ⇒ (45−b)x_1 ^3 −(84+c)x_1 ^2 +(73−3a−d)x_1 −15+a=0 ........(I) now again ×x_1 gives (45−b)(3x_1 −1)−(84+c)x_1 ^3 +(73−3a−d)x_1 ^2 −(15−a)x_1 =0 ⇒ (84+c)x_1 ^3 −(73−3a−d)x_1 ^2 −(120+3b−a)x_1 +(45−b)=0 .......(II) let me assume (I)≡(II) ⇒ ((45−b)/(84+c))=((84+c)/(73−3a−d))=((−73+3a+d)/(120+3b−a)) = ((15−a)/(b−45)) let a=15, c=−84, b=45 d=73−3(15)=28 ⇒ (a+b)−(c+d) =15+45+84−28 =60+56 = 116 .$
	$f (x) = (2 x^{3} - 3 x) + [h (x)] (x^{4} - 3 x + 1)$ $3 x^{3} + {(2 x^{3} - 7 x) + [h (x)] (x^{4} - 3 x + 1)}^{2}$ $= (x^{4} - 3 x + 1) [s (x)]$ $+ {a x}^{3} + {b x}^{2} + c x + d$ $n o w l e t x_{1}^{4} + 1 = 3 x_{1} \Rightarrow$ $3 x_{1}^{3} + {(2 x_{1}^{3} - 7 x_{1})}^{2}$ $= {a x}_{1}^{3} + {b x}_{1}^{2} + {c x}_{1} + d$ $\Rightarrow$ $(3 - a) (3 x_{1} - 1) + 4 (3 x_{1}^{4} - x_{1}^{3})$ $- 28 (3 x_{1}^{2} - x_{1}) + 49 x_{1}^{3}$ $= {b x}_{1}^{3} + {c x}_{1}^{2} + {d x}_{1}$ $\Rightarrow$ $9 x_{1} - 3 - 3 {a x}_{1} + a + 12 (3 x_{1} - 1)$ $- 4 x_{1}^{3} - 84 x_{1}^{2} + 28 x_{1} + (49 - b) x_{1}^{3}$ $- {c x}_{1}^{2} - {d x}_{1} = 0$ $\Rightarrow (45 - b) x_{1}^{3} - (84 + c) x_{1}^{2}$ $+ (73 - 3 a - d) x_{1} - 15 + a = 0$ $. . . . . . . . (I)$ $n o w a g a i n \times x_{1} g i v e s$ $(45 - b) (3 x_{1} - 1) - (84 + c) x_{1}^{3}$ $+ (73 - 3 a - d) x_{1}^{2} - (15 - a) x_{1} = 0$ $\Rightarrow$ $(84 + c) x_{1}^{3} - (73 - 3 a - d) x_{1}^{2}$ $- (120 + 3 b - a) x_{1} + (45 - b) = 0$ $. . . . . . . (I I)$ $l e t m e a s s u m e (I) \equiv (I I) \Rightarrow$ $\frac{45 - b}{84 + c} = \frac{84 + c}{73 - 3 a - d} = \frac{- 73 + 3 a + d}{120 + 3 b - a}$ $= \frac{15 - a}{b - 45}$ $l e t a = 15, c = - 84, b = 45$ $d = 73 - 3 (15) = 28$ $\Rightarrow (a + b) - (c + d)$ $= 15 + 45 + 84 - 28$ $= 60 + 56 = 116 .$
	Commented by Tawa11 last updated on 08/Oct/21

	$Great sir$
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