Question

Question #ad6cf

Answer 1

`2x-1`

Explanation:

Let the third factor be `ax+b`, so that the polynomial is
`P(x) = (ax+b)(x+2)(x-4)`

`P(0) = 8 implies (a times 0+b)(0+2)(0-4) = 8 implies b=-1`

`P(1) = -9 implies (a times 1+b)(1+2)(1-4) = -9 implies a+b=1 implies a=2`

The third factor is `2x-1`

Answer 2

`"The answer is:" \qquad \qquad \qquad 2 x - 1.`

Explanation:

`"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."`