**Title: Solving Polynomial Division Problem****Problem Statement:**The polynomial $ x^4 + px^3 + qx^2 + rx + 6 $ is exactly divisible by each of $ (x - 1) $, $ (x - 2) $ and $ (x - 3) $.Find the values of $ p $, $ q $, and $ r $.**Explanation:**In this problem, we are given a polynomial $ x^4 + px^3 + qx^2 + rx + 6 $ which is divisible by $ (x - 1) $, $ (x - 2) $, and $ (x - 3) $. This implies that the given polynomial has roots at $ x = 1 $, $ x = 2 $, and $ x = 3 $.To find the values of $ p $, $ q $, and $ r $, we can set up the polynomial equation and apply the fact that when each of these values is substituted into the polynomial, it should satisfy the equation $ x^4 + px^3 + qx^2 + rx + 6 = 0 $.We can write our equations based on roots as follows:1. For $ x = 1 $: \[ 1^4 + p(1)^3 + q(1)^2 + r(1) + 6 = 0 \] Simplifies to: \[ 1 + p + q + r + 6 = 0 \] \[ p + q + r = -7 \]2. For $ x = 2 $: \[ 2^4 + p(2)^3 + q(2)^2 + r(2) + 6 = 0 \] Simplifies to: \[ 16 + 8p + 4q + 2r + 6 = 0 \] \[ 8p + 4q + 2r = -22 \] \[ 4p + 2q + r = -11 \]3. For $ x = 3 $: \[ 3^4 + p(3)^3 + q(3)^

Answered: The polynomial z + pr + qr + ra + 6 is…

The polynomial z + pr + qr + ra + 6 is exactly divisible by each of (z 1). (z - 2) and (z-3). Find the values of p. q andr.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Polynomials

Try substituting 1. You will get 6 and not 0. Try other real values also. What About -2? Let us check.

Question

**Title: Solving Polynomial Division Problem**

**Problem Statement:**

The polynomial $ x^4 + px^3 + qx^2 + rx + 6 $ is exactly divisible by each of $ (x - 1) $, $ (x - 2) $ and $ (x - 3) $.

Find the values of $ p $, $ q $, and $ r $.

**Explanation:**

In this problem, we are given a polynomial $ x^4 + px^3 + qx^2 + rx + 6 $ which is divisible by $ (x - 1) $, $ (x - 2) $, and $ (x - 3) $. This implies that the given polynomial has roots at $ x = 1 $, $ x = 2 $, and $ x = 3 $.

To find the values of $ p $, $ q $, and $ r $, we can set up the polynomial equation and apply the fact that when each of these values is substituted into the polynomial, it should satisfy the equation $ x^4 + px^3 + qx^2 + rx + 6 = 0 $.

We can write our equations based on roots as follows:

1. For $ x = 1 $:
\[
1^4 + p(1)^3 + q(1)^2 + r(1) + 6 = 0
\]
Simplifies to:
\[
1 + p + q + r + 6 = 0
\]
\[
p + q + r = -7
\]

2. For $ x = 2 $:
\[
2^4 + p(2)^3 + q(2)^2 + r(2) + 6 = 0
\]
Simplifies to:
\[
16 + 8p + 4q + 2r + 6 = 0
\]
\[
8p + 4q + 2r = -22
\]
\[
4p + 2q + r = -11
\]

3. For $ x = 3 $:
\[
3^4 + p(3)^3 + q(3)^

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