# "Given:" #
# "1)" \ \ "Unknown polynomial, call it" \ \ p(x). #
# "2)" \ \ p(x), \ \ "divided by" \ \ ( x - 2 ), \ \ "leaves remainder" \ \ 1. #
# "3)" \ \ p(x), \ \ "divided by" \ \ ( 2 x - 1/2 ), \ \ "leaves remainder" \ \ 2. #
# "Determine:" #
# "The remainder, when" \ \ p(x) \ \ "is divided by" \ \ ( x - 2 )( 4 x - 1 ). #
# \ #
# "Analysis." #
# "1)" \ \ p(x), \ \ "divided by" \ \ ( x - 2 ), \ \ "leaves remainder" \ \ 1 rArr #
# \qquad \quad \ p(x) \ = \
( x - 2 ) \cdot u(x) + 1; \qquad \ "for some polynomial" \ \ u(x). #
# "2)" \ \ \ p(x), \ \ "divided by" \ \ ( 2 x - 1/2 ), \ \ "leaves remainder" \ \ 2 rArr #
# \ \ p(x) \ = \
( 2 x - 1/2 ) \cdot v(x) + 2; \qquad \ "for some polynomial" \ \ v(x). #
# "3) Consider the division of" \ \ p(x) \ \ "by" \ \ ( x - 2 )( 4 x - 1 ); \ \ "resulting in a" \ "quotient," \ q(x), \ "and a remainder," \ r(x). "We write:" #
# \ \ \ p(x) \ = \ q(x) ( x - 2 )( 4 x - 1 ) + r(x); \quad "for some polynomials" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad q(x), r(x), #
# \quad \ \ \ "where:" \ \ "deg" \ r(x) <= "deg" \ [ ( x - 2 )( 4 x - 1 ) ] \quad "or" \quad r(x) = 0. #
# \qquad \qquad \quad \ \ :. \qquad \qquad \qquad \ \ "deg" \ r(x) <= 2 \qquad "or" \qquad r(x) = 0. #
# \qquad \quad :. \qquad \qquad \ \ r(x) \ = \ a x + b ; \qquad \qquad \quad "for some" \ \ a, b \in RR. #
# \qquad \ "We seek to find the remainder:" \qquad \quad \r(x). #
# "4) Substitute" \ \ r(x) \ \ "into the first equation of part (3):" #
# \quad \quad \ \ p(x) \ = \ q(x) ( x - 2 )( 4 x - 1 ) + a x + b; \qquad \quad a, b \in RR. \qquad \ \ (1) #
# "5) a) Substitute" \ \ x = 2 \ \ "into eqn. (1) above:" #
# \qquad \qquad \qquad \quad p(2) \ = \ color{red}cancel{ q(2) ( 2 - 2 )( 4 \cdot 2 - 1) } + a ( 2 ) + b #
# :. \qquad \qquad \qquad \qquad p(2) \ = \ 2 a + b. #
# \qquad "b) Now substitute" \ \ x = 1/4 \ \ "into eqn. (1) above:" #
# \quad \quad \ p(1/4) \ = \ color{red}cancel{ q(1/4) ( 1/4 - 2 )( 4 \cdot 1/4 - 1 ) } + a ( 1/4 ) + b #
# :. \qquad \qquad \quad \ p(1/4) \ = 1/4 a + b. #
# \qquad "c) Collecting the two results from parts (5a) and (5b):" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ p(2) \ = \ 2 a + b. #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (2)#
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ p(1/4) \ = 1/4 a + b. #
# "6) a) Substituting:" \qquad x = 2 \qquad "into part (1) above:" #
# \qquad \qquad \qquad p(2) = \color{red}cancel{ ( 2 - 2 ) \cdot u(2) } + 1 = 1. #
# \qquad \qquad :. \qquad \qquad p(2) = 1 #
# \qquad "b) Substituting:" \qquad x = 1/4 \qquad "into part (2) above:" #
# \qquad \qquad \qquad \quad p(1/4) = \color{red}cancel{ ( 2 \cdot 1/4 - 1/2 ) \cdot v(1/4) } + 2 = 2. #
# \qquad \qquad :. \qquad \qquad p(1/4) = 2. #
# \qquad"c) Collecting the two results from parts (6a) and (6b):" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad p(2) = 1. #
# \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad p(1/4) = 2. #
# \qquad"d) Substituting the two results from part (6c) into the two" #
# "results in part (5c), we find:" #
# \qquad \qquad \qquad \qquad \quad \ \ 1 \ = \ 2 a + b \quad rArr \quad \ \ 1 \ = \ 2 a + b . #
# \qquad \qquad \qquad \qquad \quad \ 2 \ = \ 1/4 a + b \quad rArr \quad \ 8 \ = \ a + 4 b. #
# \qquad"e) Solving the" \ \ 2 xx 2 \ \ "system in the RHS of part (6d) above," #
# "we find easily:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad a \ = \ - 4/7, \qquad \quad b \ = \ 15/7. #
# \qquad"f) Substituting the results from part (6e) into the expression" #
# "for" \ \ r(x) \ \ "at the bottom of part (3) above," # # \"we obtain:" #
# \qquad \ \ r(x) \ = \ a x + b \ = \ ( - 4/7 ) x + 15/7 \ = \ - 4/7 x + 15/7. #
# \qquad \qquad "Thus:" \qquad \qquad \qquad \qquad \quad r(x) \ = \ - 4/7 x + 15/7. #
# "This is the result we sought: the remainder, when" \ \ p(x) \ \ "is divided by" \ \ ( x - 2 )( 4 x - 1 ). #
# \ #
# "Hence, summarizing, we have:" #
# "The remainder, when" \ \ p(x) \ \ "is divided by" \ \ ( x - 2 )( 4 x - 1 ) , \ \ "is:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad - 4/7 x + 15/7. #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ square #
