Solve and show work for #8 6. Use the factor theorem to determine if z -1 is a factor of f(z) = 2r* + 3r - 5z +11.f(1)-211)++3(1)-s1)+7-ス3-547= S-5+7f(1)=7not oSO(x-1) is not a factor of f(x )-2x443x²Sx+77. Use the remainder theorem to find the remainder when z10 + 78 is divided by a -1 without dividing.8. . Given f(x) = x4 – 7a3 + 14x? + 2x – 20, where a = 3 + i is a root, use synthetic division to find the otherroots of f. Write f(x) in its factored form.

Name: Solve and show work for #8 6. Use the factor theorem to determine if z
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Description: Solve and show work for #8 6. Use the factor theorem to determine if z -1 is a factor of f(z) = 2r* + 3r - 5z +11.f(1)-211)++3(1)-s1)+7-ス3-547= S-5+7f(1)=7not oSO(x-1) is not a factor of f(x )-2x443x²Sx+77. Use the remainder theorem to find the remai

Answered: 8. . Given f(x) = x* – 7x3 + 14r + 2x –…

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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Solve and show work for #8

6. Use the factor theorem to determine if z -1 is a factor of f(z) = 2r* + 3r - 5z +1
1.
f(1)-211)++3(1)-s1)+7
-ス3-547
= S-5+7
f(1)=7
not o
SO
(x-1) is not a factor of f(x )-2x443x²Sx+7
7. Use the remainder theorem to find the remainder when z10 + 78 is divided by a -1 without dividing.
8. . Given f(x) = x4 – 7a3 + 14x? + 2x – 20, where a = 3 + i is a root, use synthetic division to find the other
roots of f. Write f(x) in its factored form.

Expert Solution

Step 1: Given

8) $f (x) = x^{4} - 7 x^{3} + 14 x^{2} + 2 x - 20$ , where $x = 3 + i$ .

Step 2: To determine

To find the other roots of $f$ using the synthetic division and to write $f (x)$ in the factored form.

Let, $f (x) = x^{4} - 7 x^{3} + 14 x^{2} + 2 x - 20$ .

Using the synthetic division,

$\begin{matrix} 1 \end{matrix} \begin{matrix} - 7 \\ 3 + i \end{matrix} \begin{matrix} 14 \\ - 13 - i \end{matrix} \begin{matrix} 2 \\ 4 - 2 i \end{matrix} \begin{matrix} - 20 \\ 20 \end{matrix} \begin{matrix} 1 & - 4 + i & 1 - i & 6 - 2 i & 0 \end{matrix} 3 + i$

$\Rightarrow (x - 3 - i) (x^{3} + (- 4 + i) x^{2} + (1 - i) x + (6 - 2 i))$ .

If any complex number is a root of a polynomial then it's conjugate is also a root of that polynomial.

$\Rightarrow x = 3 - i$ is also a root of the given polynomial.

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