"As" \ \ P(x) \ \ "has the two factors" \ x + 2 \quad "and" \quad x - 4, "we can write:"
\ P(x) \ = \ ( x + 2 ) ( x - 4) Q(x), \quad "for some polynomial" \ Q(x). \ (I)
"The polynomial," \ Q(x), "is the quantity requested in the"
"problem, and so is the solution to the problem."
"Now, taking the degree of both sides of this equation, we get:"
\qquad \qquad \qquad \qquad \qquad "deg" P(x) \ = \ "deg"( ( x + 2 ) ( x - 4) Q(x) ).
"Recalling that:" \quad "deg"( A(x) cdot B(x) ) \ = \ "deg"A(x) + "deg"B(x),
"we get:"
\qquad \qquad "deg" P(x) \ = \ "deg"( x + 2 ) + "deg" ( x - 4) + "deg" Q(x).
\qquad :. \qquad \qquad \qquad \quad "deg" P(x) \ = \ 1 + 1 + "deg" Q(x).
"And recalling, from the given, that:" \quad "deg" P(x) \ = \ 3, "we get:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 3 \ = \ 1 + 1 + "deg" Q(x).
\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad 3 \ = \ 2 + "deg" Q(x).
\qquad \qquad :. \qquad \qquad \qquad \qquad \qquad \quad "deg" Q(x) \ = \ 1.
\quad :. \quad \ Q(x) \ = \ a x + b; \quad \ "for some constants" \ a, b, \ a != 0. \quad \ \ (II)
"Substituting the previous into eqn. (I) above, we get:"
\qquad \qquad \qquad \qquad \qquad \quad P(x) \ = \ ( x + 2 ) ( x - 4) ( a x + b ). \qquad \qquad \qquad \qquad \quad \ (III)
"We are also given:" \ \ P(0) = 8.
"So substituting" \ \ x = 0 \ \ "into eqn. (III), we get:"
\qquad \qquad \qquad \qquad \qquad \quad \ P(0) \ = \ ( 0 + 2 ) ( 0 - 4 ) ( a cdot 0 + b )
\qquad \qquad \qquad \qquad \qquad \qquad \quad\quad \ 8 \ = \ ( 2 ) ( - 4 ) ( b )
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ 8 \ = \ -8 b.
\qquad \qquad :. \qquad \qquad \qquad \quad \ \ b \ = \ -1. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (IV)
"We are also given:" \ \ P(1) = -9.
"So substituting" \ \ x = 1 \ \ "into eqn. (III), we get:"
\qquad \qquad \qquad \qquad \qquad \quad \ P(1) \ = \ ( 1 + 2 ) ( 1 - 4 ) ( a cdot 1 + b )
\qquad \qquad :. \qquad \qquad \qquad \quad \ -9 \ = \ ( 3 ) ( - 3 ) ( a + b )
\qquad \qquad :. \qquad \qquad \qquad \quad \ -9 \ = \ -9 ( a + b ).
"Now, using the value we found for" \ \ b \ \ "in eqn. (IV), we get:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ -9 \ = \ -9 ( a - 1 )
\qquad \qquad :. \qquad \qquad \quad \ \ a - 1 \ = \ 1
\qquad \qquad :. \qquad \qquad \qquad \qquad \quad \ a \ = \ 2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (V)
"Now, using the values we found for" \ \ a, b \ \ "in eqns. (IV) and (V),"
"and substituting them into eqn. (II) for" \ \ Q(x), "we get:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ Q(x) \ = \ 2 x - 1.
"This is the answer to the question."