p-x-x-1-2-Q-x-3x-8-p-x-x-2-Q-x-R-R-

Question Number 144060 by Khalmohmmad last updated on 21/Jun/21

p (x) = {(x - 1)}^{2} Q (x) + 3 x + 8

p (x) = (x - 2) Q (x) + R

R = ?

Commented by Rasheed.Sindhi last updated on 21/Jun/21

Q (x) i s s a m e i n b o t h c a s e s!!! ?

Answered by Rasheed.Sindhi last updated on 21/Jun/21

p (x) = {(x - 1)}^{2} Q_{1} (x) + 3 x + 8 \dots . . (i)

p (x) = (x - 2) Q_{2} (x) + R \dots . . (i i); R = ?

^{∙} L e t Q_{1} (x) = q (c o n s t a n t

p (x) = {(x - 1)}^{2} q + 3 x + 8

= {q x}^{2} - 2 x q + q + 3 x + 8

= {q x}^{2} + (3 - 2 q) x + q + 8

W h e n d i v i d e d b y x - 2

R = q {(2)}^{2} + (3 - 2 q) (2) + q + 8

R = 4 q + 6 - 4 q + q + 8

= q + 14

^{∙} Q_{1} (x) = a x + b

p (x) = {(x - 1)}^{2} Q_{1} (x) + 3 x + 8

p (x) = {(x - 1)}^{2} (a x + b) + 3 x + 8

= (x^{2} - 2 x + 1) (a x + b) + 3 x + 8

= {a x}^{3} + {b x}^{2} - 2 {a x}^{2} - 2 b x + a x + b + 3 x + 8

= {a x}^{3} + (b - 2 a) x^{2} + (a + 3 - 2 b) x + b + 8

W h e n d i v i d e d b y x - 2

R = p (2) = a {(2)}^{3} + (b - 2 a) {(2)}^{2} + (a + 3 - 2 b) (2) + b + 8

= \cancel 8 a + \cancel 4 b - \cancel 8 a + 2 a + 6 - \cancel 4 b + b + 8

= 2 a + b + 14

N o u n i q u e a n s w e r .

R d e p e n d s u p o n Q_{1} (x) .

Answered by Rasheed.Sindhi last updated on 21/Jun/21

p (x) = {(x - 1)}^{2} Q_{1} (x) + 3 x + 8

p (x) = (x - 2) Q_{2} (x) + R; R = ?

p (x) w h e n i s d i v i d e d b y x - 2 g i v i n g

r e m a i n d e r R

{(x - 1)}^{2} Q_{1} (x) + 3 x + 8 i s d i v i d e d b y

x - 2 g i v i n g r e m a i n d e r R

R = p (2) = {(2 - 1)}^{2} Q_{1} (2) + 3 (2) + 8

R = p (2) = Q_{1} (2) + 14

^{∙} L e t Q_{1} (x) = q (c o n s t a n t)

R = q + 14

^{∙} L e t Q_{1} (x) = a x + b

R = Q_{1} (2) + 14 = a (2) + b + 14

R = 2 a + b + 14

^{∙} L e t Q_{1} (x) = {a x}^{2} + b x + c

R = a {(2)}^{2} + b (2) + c + 14

R = 4 a + 2 b + c + 14

R d e p e n d s o n Q_{1} (x) (o r Q_{2} (x)) .

{I t}^{'} s n o t u n i q u e .