Answer :
1)[tex]$\frac{3x^4-2x^2+x+5}{2x^2+1} = 3x^2 - 2 + \frac{x}{2x^2+1}.$[/tex]
2) [tex]$\frac{4x^4-2x^3-4x+2}{2x-1} = 2x^3+1.$[/tex]
3)(multiplicity 1)x = 2i (multiplicity 1)x = -2i (multiplicity 1)
4) the solutions to f(x)=0 are:[tex]x = -2x = $\frac{23 + 7\sqrt{131}}{-46}$x = $\frac{23 - 7\sqrt{131}}{-46}$x = 2ix = -2i[/tex]
1. To divide 3x4−2x2+x+5 by 2x2+1, use the following steps: [tex]$$\begin{array}{rrrrr}\multicolumn{1}{r|}{2x^2} & & -1 & & \\\cline{1-5}3x^4 & -2x^2 & +x & +5 & \\\multicolumn{1}{r|}{3x^4} & & +0x^2 & & \\\cline{1-3}& -2x^2 & +x & & \\& -2x^2 & & \\\cline{2-3}& & x & & \\& & x & & \\\cline{3-4}& & & 5 & \\\end{array}$$Therefore, $\frac{3x^4-2x^2+x+5}{2x^2+1} = 3x^2 - 2 + \frac{x}{2x^2+1}.$[/tex]
2. To divide 4x4−2x3−4x+2 by 2x−1 using synthetic division, use the following steps:
[tex]$$\begin{array}{r|rrrr}\multicolumn{1}{r|}{\frac{1}{2}} & 4 & -2 & 0 & 2 \\\cline{1-5}& & 2 & 0 & -1 \\\cline{2-5}& 4 & 0 & 0 & 1 \\\end{array}$$[/tex]
3. Given the function [tex]$f(x)=x^6−2x^5−13x^4+70x^3−160x^2+184x−80,$[/tex] the zeros (real and complex, nonreal) of f and their multiplicities can be found as follows:
Firstly, we can find the rational zeros using the Rational Root Theorem.
The rational roots of f can be written as $± \frac{a}{b}$ where a is a factor of -80 and b is a factor of 1.
That is, the possible rational roots of f are ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40.
To find the zeros of f using synthetic division, use the following steps:
[tex]$$\begin{array}{r|rrrrrrr}\multicolumn{1}{r|}{1} & -2 & -13 & 70 & -160 & 184 & -80 \\\cline{1-8}& & -2 & -15 & 55 & -105 & 79 & -1 \\\cline{2-8}& & -2 & -15 & 55 & -105 & 263 & -81 \\\end{array}$$[/tex]
Since the remainder is not 0, 1 is not a zero of f.
Then try using -1 as the test zero:
[tex]$$\begin{array}{r|rrrrrrr}\multicolumn{1}{r|}{-1} & -2 & -13 & 70 & -160 & 184 & -80 \\\cline{1-8}& & 2 & 11 & -81 & 241 & -57 & 137 \\\cline{2-8}& -2 & -11 & 81 & -241 & 425 & -137 & \color{red}{57} \\\end{array}$$[/tex]
Since the remainder is not 0, -1 is not a zero of f. Then try using -2 as the test zero:
[tex]$$\begin{array}{r|rrrrrrr}\multicolumn{1}{r|}{-2} & -2 & -13 & 70 & -160 & 184 & -80 \\\cline{1-8}& & 4 & -22 & 104 & -192 & 16 & \\\cline{2-7}& -2 & -9 & 48 & -56 & -8 & 16 & \\\end{array}$$[/tex]
Since the remainder is 0, -2 is a zero of f and (x+2) is a factor of f.
Divide f by (x+2) to find the remaining zeros:
[tex]$$\begin{array}{r|rrrrrr}\multicolumn{1}{r|}{-2} & 1 & -2 & -13 & 70 & -160 & 184 & -80 \\\cline{1-8}& & 2 & -10 & 47 & -46 & 246 & -276 \\\cline{2-8}& 1 & 0 & -23 & 117 & -206 & 430 & -356 \\\end{array}$$[/tex]
Use the quadratic formula to find the zeros of the remaining cubic polynomial:
[tex]$$\begin{aligned} x & = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ & = \frac{23 \pm \sqrt{23^2-4(-23)(-117)}}{2(-23)} \\ & = \frac{23 \pm 7\sqrt{131}}{-46} \end{aligned}$$[/tex]
Therefore, the zeros of f and their multiplicities are:x = -2 (multiplicity 1)x = [tex]$\frac{23 + 7\sqrt{131}}{-46}$[/tex]
(multiplicity 1)x =[tex]$\frac{23 - 7\sqrt{131}}{-46}$[/tex]
(multiplicity 1)x = 2i (multiplicity 1)x = -2i (multiplicity 1)
4. To write a complete linear factorization of f, use the following steps:
[tex]$$f(x)=(x+2)(x-\frac{23 + 7\sqrt{131}}{-46})(x-\frac{23 - 7\sqrt{131}}{-46})(x-2i)(x+2i)$$5.[/tex]
To solve f(x)=0, we can use the zeros of f found in question 3:
[tex]$$\begin{aligned} f(x) & = (x+2)(x-\frac{23 + 7\sqrt{131}}{-46})(x-\frac{23 - 7\sqrt{131}}{-46})(x-2i)(x+2i) \\ & = (x+2)(x-\frac{23 + 7\sqrt{131}}{-46})(x-\frac{23 - 7\sqrt{131}}{-46})(x^2+4) \\ & = 0 \end{aligned}$$[/tex]
Therefore, the solutions to f(x)=0 are:
[tex]x = -2x = $\frac{23 + 7\sqrt{131}}{-46}$x = $\frac{23 - 7\sqrt{131}}{-46}$x = 2ix = -2i[/tex]
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