Question ad6cf

Feb 24, 2018

$2 x - 1$

Explanation:

Let the third factor be $a x + b$, so that the polynomial is
$P \left(x\right) = \left(a x + b\right) \left(x + 2\right) \left(x - 4\right)$

$P \left(0\right) = 8 \implies \left(a \times 0 + b\right) \left(0 + 2\right) \left(0 - 4\right) = 8 \implies b = - 1$

$P \left(1\right) = - 9 \implies \left(a \times 1 + b\right) \left(1 + 2\right) \left(1 - 4\right) = - 9 \implies a + b = 1 \implies a = 2$

The third factor is $2 x - 1$

Feb 24, 2018

$\text{The answer is:} \setminus q \quad \setminus q \quad \setminus q \quad 2 x - 1.$

Explanation:

$\text{As" \ \ P(x) \ \ "has the two factors" \ x + 2 \quad "and" \quad x - 4, "we can write:}$

$\setminus P \left(x\right) \setminus = \setminus \left(x + 2\right) \left(x - 4\right) Q \left(x\right) , \setminus \quad \text{for some polynomial} \setminus Q \left(x\right) . \setminus \left(I\right)$

$\text{The polynomial," \ Q(x), "is the quantity requested in the}$
$\text{problem, and so is the solution to the problem.}$

$\text{Now, taking the degree of both sides of this equation, we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{deg" P(x) \ = \ "deg} \left(\left(x + 2\right) \left(x - 4\right) Q \left(x\right)\right) .$

$\text{Recalling that:" \quad "deg"( A(x) cdot B(x) ) \ = \ "deg"A(x) + "deg} B \left(x\right) ,$
$\text{we get:}$

$\setminus q \quad \setminus q \quad \text{deg" P(x) \ = \ "deg"( x + 2 ) + "deg" ( x - 4) + "deg} Q \left(x\right) .$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{deg" P(x) \ = \ 1 + 1 + "deg} Q \left(x\right) .$

$\text{And recalling, from the given, that:" \quad "deg" P(x) \ = \ 3, "we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad 3 \setminus = \setminus 1 + 1 + \text{deg} Q \left(x\right) .$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 3 \setminus = \setminus 2 + \text{deg} Q \left(x\right) .$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{deg} Q \left(x\right) \setminus = \setminus 1.$

 \quad :. \quad \ Q(x) \ = \ a x + b; \quad \ "for some constants" \ a, b, \ a != 0. \quad \ \ (II) #

$\text{Substituting the previous into eqn. (I) above, we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad P \left(x\right) \setminus = \setminus \left(x + 2\right) \left(x - 4\right) \left(a x + b\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(I I I\right)$

$\text{We are also given:} \setminus \setminus P \left(0\right) = 8.$

$\text{So substituting" \ \ x = 0 \ \ "into eqn. (III), we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus P \left(0\right) \setminus = \setminus \left(0 + 2\right) \left(0 - 4\right) \left(a \cdot 0 + b\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus 8 \setminus = \setminus \left(2\right) \left(- 4\right) \left(b\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \setminus 8 \setminus = \setminus - 8 b .$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus b \setminus = \setminus - 1. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(I V\right)$

$\text{We are also given:} \setminus \setminus P \left(1\right) = - 9.$

$\text{So substituting" \ \ x = 1 \ \ "into eqn. (III), we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus P \left(1\right) \setminus = \setminus \left(1 + 2\right) \left(1 - 4\right) \left(a \cdot 1 + b\right)$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus - 9 \setminus = \setminus \left(3\right) \left(- 3\right) \left(a + b\right)$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus - 9 \setminus = \setminus - 9 \left(a + b\right) .$

$\text{Now, using the value we found for" \ \ b \ \ "in eqn. (IV), we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus - 9 \setminus = \setminus - 9 \left(a - 1\right)$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus a - 1 \setminus = \setminus 1$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus a \setminus = \setminus 2. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(V\right)$

$\text{Now, using the values we found for" \ \ a, b \ \ "in eqns. (IV) and (V),}$
$\text{and substituting them into eqn. (II) for" \ \ Q(x), "we get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus Q \left(x\right) \setminus = \setminus 2 x - 1.$

$\text{This is the answer to the question.}$