3.3E: Exercises

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Jan 10, 2023
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- 3.3: Real Zeros of Polynomials
- 3.4: Complex Zeros and the Fundamental Theorem of Algebra

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Exercises

In Exercises 1 - 10, for the given polynomial:

Use Cauchy’s Bound to find an interval containing all of the real zeros.
Use the Rational Zeros Theorem to make a list of possible rational zeros.

$f(x)=x^{3}-2 x^{2}-5 x+6$
$f(x)=x^{4}+2 x^{3}-12 x^{2}-40 x-32$
$f(x)=x^{4}-9 x^{2}-4 x+12$
$f(x)=x^{3}+4 x^{2}-11 x+6$
$f(x)=x^{3}-7 x^{2}+x-7$
$f(x)=-2 x^{3}+19 x^{2}-49 x+20$
$f(x)=-17 x^{3}+5 x^{2}+34 x-10$
$f(x)=36 x^{4}-12 x^{3}-11 x^{2}+2 x+1$
$f(x)=3 x^{3}+3 x^{2}-11 x-10$
$f(x)=2 x^{4}+x^{3}-7 x^{2}-3 x+3$

In Exercises 11 - 30, find the real zeros of the polynomial using the techniques specified by your instructor. State the multiplicity of each real zero.

$f(x)=x^{3}-2 x^{2}-5 x+6$
$f(x)=x^{4}+2 x^{3}-12 x^{2}-40 x-32$
$f(x)=x^{4}-9 x^{2}-4 x+12$
$f(x)=x^{3}+4 x^{2}-11 x+6$
$f(x)=x^{3}-7 x^{2}+x-7$
$f(x)=-2 x^{3}+19 x^{2}-49 x+20$
$f(x)=-17 x^{3}+5 x^{2}+34 x-10$
$f(x)=36 x^{4}-12 x^{3}-11 x^{2}+2 x+1$
$f(x)=3 x^{3}+3 x^{2}-11 x-10$
$f(x)=2 x^{4}+x^{3}-7 x^{2}-3 x+3$
$f(x)=9 x^{3}-5 x^{2}-x$
$f(x)=6 x^{4}-5 x^{3}-9 x^{2}$
$f(x)=x^{4}+2 x^{2}-15$
$f(x)=x^{4}-9 x^{2}+14$
$f(x)=3 x^{4}-14 x^{2}-5$
$f(x)=2 x^{4}-7 x^{2}+6$
$f(x)=x^{6}-3 x^{3}-10$
$f(x)=2 x^{6}-9 x^{3}+10$
$f(x)=x^{5}-2 x^{4}-4 x+8$
$f(x)=2 x^{5}+3 x^{4}-18 x-27$

In Exercises 31 - 33, use graphing technology to help you find the real zeros of the polynomial. State the multiplicity of each real zero.

$f(x)=x^{5}-60 x^{3}-80 x^{2}+960 x+2304$
$f(x)=25 x^{5}-105 x^{4}+174 x^{3}-142 x^{2}+57 x-9$
$f(x)=90 x^{4}-399 x^{3}+622 x^{2}-399 x+90$
Find the real zeros of $f(x)=x^{3}-\frac{1}{12} x^{2}-\frac{7}{72} x+\frac{1}{72}$ by first finding a polynomial $q(x)$ with integer coefficients such that $q(x)=N \cdot f(x)$ for some integer $N$ . (Recall that the Rational Zeros Theorem required the polynomial in question to have integer coefficients.) Show that $f$ and $q$ have the same real zeros.

In Exercises 35 - 44, find the real solutions of the polynomial equation.

$9 x^{3}=5 x^{2}+x$
$9 x^{2}+5 x^{3}=6 x^{4}$
$x^{3}+6=2 x^{2}+5 x$
$x^{4}+2 x^{3}=12 x^{2}+40 x+32$
$x^{3}-7 x^{2}=7-x$
$2 x^{3}=19 x^{2}-49 x+20$
$x^{3}+x^{2}=\frac{11 x+10}{3}$
$x^{4}+2 x^{2}=15$
$14 x^{2}+5=3 x^{4}$
$2 x^{5}+3 x^{4}=18 x+27$

In Exercises 45 - 54, solve the polynomial inequality and state your answer using interval notation.

$-2 x^{3}+19 x^{2}-49 x+20>0$
$x^{4}-9 x^{2} \leq 4 x-12$
$(x-1)^{2} \geq 4$
$4 x^{3} \geq 3 x+1$
$x^{4} \leq 16+4 x-x^{3}$
$3 x^{2}+2 x<x^{4}$
$\frac{x^{3}+2 x^{2}}{2}<x+2$
$\frac{x^{3}+20 x}{8} \geq x^{2}+2$
$2 x^{4}>5 x^{2}+3$
$x^{6}+x^{3} \geq 6$
In Section 3.1, a box with no top is constructed from a 10 inch × 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. We determined the volume of that box (in cubic inches) is given by $V(x)=4 x^{3}-44 x^{2}+120 x$ , where $x$ denotes the length of the side of the square which is removed from each corner (in inches), $0<x<5$ . Solve the inequality $V(x) \geq 80$ analytically and interpret your answer in the context of that example.
From Exercise 32 in Section 3.1, $C(x)=.03 x^{3}-4.5 x^{2}+225 x+250$ , for $x \geq 0$ models the cost, in dollars, to produce $x$ PortaBoy game systems. If the production budget is $5000, find the number of game systems which can be produced and still remain under budget.
Let $f(x)=5 x^{7}-33 x^{6}+3 x^{5}-71 x^{4}-597 x^{3}+2097 x^{2}-1971 x+567$ . With the help of your classmates, find the x- and y- intercepts of the graph of $f$ . Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Sketch the graph of $f$ , using more than one picture if necessary to show all of the important features of the graph.
With the help of your classmates, create a list of five polynomials with different degrees whose real zeros cannot be found using any of the techniques in this section.

Answers

For f(x)=x3−2x2−5x+6
- All of the real zeros lie in the interval [−7, 7]
- Possible rational zeros are ±1, ±2, ±3, ±6
For f(x)=x4+2x3−12x2−40x−32
- All of the real zeros lie in the interval [−41, 41]
- Possible rational zeros are ±1, ±2, ±4, ±8, ±16, ±32
For f(x)=x4−9x2−4x+12
- All of the real zeros lie in the interval [−13, 13]
- Possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12
For f(x)=x3+4x2−11x+6
- All of the real zeros lie in the interval [−12, 12]
- Possible rational zeros are ±1, ±2, ±3, ±6
For f(x)=x3−7x2+x−7
- All of the real zeros lie in the interval [−8, 8]
- Possible rational zeros are ±1, ±7
For f(x)=−2x3+19x2−49x+20
- All of the real zeros lie in the interval $\left[-\frac{51}{2}, \frac{51}{2}\right]$
- Possible rational zeros are ± 1 2 , ±1, ±2, ± 5 2 , ±4, ±5, ±10, ±20
For f(x)=−17x3+5x2+34x−10
- All of the real zeros lie in the interval [−3, 3]
- Possible rational zeros are $\pm \frac{1}{17}, \pm \frac{2}{17}, \pm \frac{5}{17}, \pm \frac{10}{17}, \pm 1, \pm 2, \pm 5, \pm 10$
For f(x)=36x4−12x3−11x2+2x+1
- All of the real zeros lie in the interval $\left[-\frac{4}{3}, \frac{4}{3}\right]$
- Possible rational zeros are $\pm \frac{1}{36}, \pm \frac{1}{18}, \pm \frac{1}{12}, \pm \frac{1}{9}, \pm \frac{1}{6}, \pm \frac{1}{4}, \pm \frac{1}{3}, \pm \frac{1}{2}, \pm 1$
For f(x)=3x3+3x2−11x−10
- All of the real zeros lie in the interval $\left[-\frac{14}{3}, \frac{14}{3}\right]$
- Possible rational zeros are $\pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3}, \pm 1, \pm 2, \pm 5, \pm 10$
For f(x)=2x4+x3−7x2−3x+3
- All of the real zeros lie in the interval $\left[-\frac{9}{2}, \frac{9}{2}\right]$
- Possible rational zeros are $\pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 3$
$\begin{aligned} &f(x)=x^{3}-2 x^{2}-5 x+6 \\ &x=-2, x=1, x=3(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{4}+2 x^{3}-12 x^{2}-40 x-32 \\ &x=-2(\text { mult. } 3), x=4(\text { mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{4}-9 x^{2}-4 x+12 \\ &x=-2 \text { (mult. } 2), x=1(\text { mult. } 1), x=3(\text { mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{3}+4 x^{2}-11 x+6 \\ &x=-6(\text { mult. } 1), x=1(\text { mult. } 2) \end{aligned}$
$\begin{aligned} &f(x)=x^{3}-7 x^{2}+x-7 \\ &x=7 \text { (mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=-2 x^{3}+19 x^{2}-49 x+20 \\ &x=\frac{1}{2}, x=4, x=5(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=-17 x^{3}+5 x^{2}+34 x-10 \\ &x=\frac{5}{17}, x=\pm \sqrt{2}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=36 x^{4}-12 x^{3}-11 x^{2}+2 x+1 \\ &x=\frac{1}{2}(\text { mult. } 2), x=-\frac{1}{3}(\text { mult. } 2) \end{aligned}$
$\begin{aligned} &f(x)=3 x^{3}+3 x^{2}-11 x-10 \\ &x=-2, x=\frac{3 \pm \sqrt{69}}{6}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=2 x^{4}+x^{3}-7 x^{2}-3 x+3 \\ &x=-1, x=\frac{1}{2}, x=\pm \sqrt{3}(\text { each mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=9 x^{3}-5 x^{2}-x \\ &x=0, x=\frac{5 \pm \sqrt{61}}{18}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=6 x^{4}-5 x^{3}-9 x^{2} \\ &x=0 \text { (mult. } 2), x=\frac{5 \pm \sqrt{241}}{12} \text { (each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{4}+2 x^{2}-15 \\ &x=\pm \sqrt{3} \text { (each has mult. } 1 \text { ) } \end{aligned}$
$\begin{aligned} &f(x)=x^{4}-9 x^{2}+14 \\ &x=\pm \sqrt{2}, x=\pm \sqrt{7}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=3 x^{4}-14 x^{2}-5 \\ &x=\pm \sqrt{5} \text { (each has mult. } 1 \text { ) } \end{aligned}$
$\begin{aligned} &f(x)=2 x^{4}-7 x^{2}+6 \\ &x=\pm \frac{\sqrt{6}}{2}, x=\pm \sqrt{2}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{6}-3 x^{3}-10 \\ &x=\sqrt[3]{-2}=-\sqrt[3]{2}, x=\sqrt[3]{5}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=2 x^{6}-9 x^{3}+10 \\ &x=\frac{\sqrt[3]{20}}{2}, x=\sqrt[3]{2}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{5}-2 x^{4}-4 x+8 \\ &x=2, x=\pm \sqrt{2}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=2 x^{5}+3 x^{4}-18 x-27 \\ &x=-\frac{3}{2}, x=\pm \sqrt{3}(\text { each has mult. } 1) \end{aligned}$
$\begin{aligned} &f(x)=x^{5}-60 x^{3}-80 x^{2}+960 x+2304 \\ &x=-4(\text { mult. } 3), x=6 \text { (mult. } 2) \end{aligned}$
$\begin{aligned} &f(x)=25 x^{5}-105 x^{4}+174 x^{3}-142 x^{2}+57 x-9 \\ &x=\frac{3}{5} \text { (mult. } 2), x=1 \text { (mult. } 3) \end{aligned}$
$\begin{aligned} &f(x)=90 x^{4}-399 x^{3}+622 x^{2}-399 x+90 \\ &x=\frac{2}{3}, x=\frac{3}{2}, x=\frac{5}{3}, x=\frac{3}{5}(\text { each has mult. } 1) \end{aligned}$
We choose $q(x)=72 x^{3}-6 x^{2}-7 x+1=72 \cdot f(x)$ . Clearly $f(x) = 0$ if and only if $q(x) = 0$ so they have the same real zeros. In this case, $x=-\frac{1}{3}$ , $x=\frac{1}{6}$ and $x=\frac{1}{4}$ are the real zeros of both $f$ and $q$ .
$x=0, \frac{5 \pm \sqrt{61}}{18}$
$x=0, \frac{5 \pm \sqrt{241}}{12}$
$x$ = −2, 1, 3
$x$ = −2, 4
$x$ = 7
$x=\frac{1}{2}, 4,5$
$x=-2, \frac{3 \pm \sqrt{69}}{6}$
$x=\pm \sqrt{3}$
$x=\pm \sqrt{5}$
$x=-\frac{3}{2}, \pm \sqrt{3}$
$\left(-\infty, \frac{1}{2}\right) \cup(4,5)$
$\{-2\} \cup[1,3]$
$(-\infty,-1] \cup[3, \infty)$
$\left\{-\frac{1}{2}\right\} \cup[1, \infty)$
[−2, 2]
$(-\infty,-1) \cup(-1,0) \cup(2, \infty)$
$(-\infty,-2) \cup(-\sqrt{2}, \sqrt{2})$
$\{2\} \cup[4, \infty)$
$(-\infty,-\sqrt{3}) \cup(\sqrt{3}, \infty)$
$(-\infty,-\sqrt[3]{3}) \cup(\sqrt[3]{2}, \infty)$
$V(x) \geq 80 \text { on }[1,5-\sqrt{5}] \cup[5+\sqrt{5}, \infty)$ . Only the portion $[1,5-\sqrt{5}]$ lies in the applied domain, however. In the context of the problem, this says for the volume of the box to be at least 80 cubic inches, the square removed from each corner needs to have a side length of at least 1 inch, but no more than $5-\sqrt{5} \approx 2.76$ inches.
$C(x) \leq 5000$ on (approximately) $(-\infty, 82.18]$ . The portion of this which lies in the applied domain is (0, 82.18]. Since x represents the number of game systems, we check $C$ (82) = 4983.04 and C(83) = 5078.11, so to remain within the production budget, anywhere between 1 and 82 game systems can be produced.

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