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Question Number 145294 by imjagoll last updated on 04/Jul/21

Given a polynomial

p (x) = x^{4} + {4 x}^{3} + (2 p + 2) x^{2} + (2 p + 5 q + 2) x + 3 q + 2 r .

If p (x) = (x^{3} + {2 x}^{2} + 8 x + 6) Q (x)

then what the value of

(p + 2 q) r .

Answered by Rasheed.Sindhi last updated on 04/Jul/21

p (x) = x^{4} + {4 x}^{3} + (2 p + 2) x^{2} + (2 p + 5 q + 2) x + 3 q + 2 r .

If p (x) = (x^{3} + {2 x}^{2} + 8 x + 6) Q (x)

\underline{(p + 2 q) r = ?}

Let Q (x) = x + k

Given : D (x) = x^{3} + {2 x}^{2} + 8 x + 6

We can also consider :

D (x) = x + k & Q (x) = x^{3} + {2 x}^{2} + 8 x + 6

B y s y n t h e t i c d i v i s i o n :

\begin{array}{ccc} - k) & 1 & 4 & 2 p + 2 & 2 p + 5 q + 2 & 3 q + 2 r \\ - k & \dots & \dots & \dots \\ 1 & 2 & 8 & 6 & 0 \end{array}

4 - k = 2 \Rightarrow k = 2

\begin{array}{ccc} - 2) & 1 & 4 & 2 p + 2 & 2 p + 5 q + 2 & 3 q + 2 r \\ - 2 & - 4 & - 16 & - 12 \\ 1 & 2 & 8 & 6 & 0 \end{array}

2 p + 2 - 4 = 8 \Rightarrow p = 5

2 p + 5 q + 2 - 16 = 6

2 (5) + 5 q = 20 \Rightarrow q = 2

3 q + 2 r - 12 = 0

3 (2) + 2 r = 12 \Rightarrow r = 3

(p + 2 q) r = (5 + 2 (2)) (3) = 27

Commented by imjagoll last updated on 04/Jul/21

yes

Answered by liberty last updated on 04/Jul/21

b y {H o r n e r}^{'} s t h e o r e m

d (x) = x^{3} + 2 x^{2} + 8 x + 6 = 0

\Rightarrow x^{3} = - 2 x^{2} - 8 x - 6

\begin{array}{ccccc} * & 1 & 4 & 2 p + 2 & 2 p + 5 q + 2 & 3 q + 2 r \\ - 2 & * & - 2 & - 8 & - 6 & * \\ - 8 & * & * & - 4 & - 16 & - 12 \\ - 6 & * & * & * & * & * \\ * & 1 & 2 & 2 p - 10 & 2 p + 5 q - 20 & 3 q + 2 r - 12 \end{array}

w e g e t {\begin{cases} 2 p - 10 = 0 \to p = 5 \\ 10 + 5 q - 20 = 0 \to q = 2 \\ 6 + 2 r - 12 = 0 \to r = 3 \end{cases}

t h e n (p + 2 q) r = (5 + 4) \times 3 = 27

Commented by imjagoll last updated on 04/Jul/21

yes