Section: 找到合理和真实的多元函数零 | 8.11 找到合理和真实的多元函数零

Section outline

A testing station collects data over a 10-month period about the number of mosquito eggs laid indoors. It is modeled by $\begin{aligned} y = 0.025 x (x^{3} - 42 x^{2} + 332 x + 840) \end{aligned}$ , where x is the month and y is the number of mosquito eggs ¹ . When does the model predict there will be no mosquito eggs? in this section, w e cover how to find zeros of a polynomial function .
::一个测试站在10个月期间收集关于室内投放的蚊子数量的数据,其模型为y=0.025x(x3-42x2+332x+840),而x是月份,y是蚊子蛋的数量。1. 模型预测何时不会有蚊子蛋?在本节中,我们涵盖了如何找到多元函数的零。

lesson content

Polynomials
::多元数

We considered in Chapter 6. Recall the following:
::我们在第六章中考虑了以下几点:

Polynomials and Important Terminology
::聚合和重要术语

A polynomial is an expression of the form
::多面性是表单的表达形式

\begin{aligned} a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + . . . + a_{2} x^{2} + a_{1} x + a_{0}, \end{aligned}

where the

\begin{aligned} a_{i} \end{aligned}

are real numbers and n is a non-negative integer .
::anxn+an-1-xn-1+an-2xn-2+...+a2x2+a1x+a0,其中Ai为实际数字,n为非负整数。

$\begin{aligned} a_{n} x^{n} \end{aligned}$ is the leading term .
::anxn是前一学期。

$\begin{aligned} a_{n} \end{aligned}$ is the leading coefficient .
::a 是主要系数。

$\begin{aligned} a_{0} \end{aligned}$ is the constant term.
::a0 是常数。

$\begin{aligned} n \end{aligned}$ is the degree of the polynomial. (If there is just a constant term, then the degree is 0 unless $\begin{aligned} a_{0} = 0 \end{aligned}$ ).
::n 是多元度。 (如果只有一个常数,则该度为0, 除非 a0=0 。)

We have talked about degree 0 and degree 1 polynomial functions, or lines, in Chapter 3, and degree 2 polynomial functions, or quadratic functions, in this chapter. In this section and the following sections, we will generalize to polynomial functions of any degree.
::我们在第三章中讨论了0和1级多元函数或线条,在本章中讨论了2级多元函数或二次函数。在本节和以下各节中,我们将概括到任何程度的多元函数。

Zeros of Polynomial Functions
::多元函数零

As we have seen in this chapter, it is important to know the roots or zeros of a quadratic function .
::正如我们在本章中所看到的那样,重要的是要知道二次函数的根或零。

Zeros of Polynomial Functions
::多元函数零

A zero of a function is a value $\begin{aligned} c \end{aligned}$ that satisfies $\begin{aligned} f (c) = 0 \end{aligned}$ .
::函数的零是满足 f(c)=0的 c 值。

If $\begin{aligned} f \end{aligned}$ is a polynomial function and $\begin{aligned} c \end{aligned}$ is a real number , then the following are equivalent:
::如果 f 是多式函数, c 是真实数字, 那么以下等值 :

$\begin{aligned} c \end{aligned}$ is a zero of $\begin{aligned} f \end{aligned}$ .
::c 为f的零。

$\begin{aligned} x = c \end{aligned}$ is a solution of $\begin{aligned} f (x) = 0 \end{aligned}$ .
::x=c 是 f( x) = 0 的溶液。

$\begin{aligned} x - c \end{aligned}$ is a factor of $\begin{aligned} f (x) \end{aligned}$ .
::x-c 是 f( x) 的系数。

$\begin{aligned} x = c \end{aligned}$ is an intercept on the graph of $\begin{aligned} f \end{aligned}$ .
::x=c 是 f 图形上的拦截。

The equivalency between statements 1 and 3 is also known as the Factor Theorem .
::报表1和报表3的等同性也称为 " 要素理论 " 。

Let's see how these relationships hold for a polynomial function we have already considered.
::让我们看看这些关系是如何维持我们已经考虑过的多元功能的

In Section 8, we graphed the quadratic function , $\begin{aligned} y = - \frac{1}{2} x^{2} - 2 x + 6 \end{aligned}$ . If we factor out the $\begin{aligned} - \frac{1}{2} \end{aligned}$ ,

\begin{aligned} y = - \frac{1}{2} x^{2} - 2 x + 6 = - \frac{1}{2} (x^{2} + 4 x - 12) = - \frac{1}{2} (x + 6) (x - 2) \end{aligned}

::在第8节中,我们绘制了四方函数的图,y= 12x2-2x+6. 如果我们将 -12,y= 12x2-2x+6= 12(x2+4x-12)= 12(x+6)(x-2)

The factors $\begin{aligned} x + 6 \end{aligned}$ and $\begin{aligned} x - 2 \end{aligned}$ give us solutions to the quadratic equation , showing that statements 2 and 3 above are equivalent.
::系数x+6和系数x-2为我们提供了四方方程式的解决方案,表明以上表2和表3等值。

\begin{aligned} 0 & = - \frac{1}{2} x^{2} - 2 x + 6 \\ = - \frac{1}{2} (x + 6) (x - 2) \\ x + 6 = 0 & x - 2 = 0 \\ x = - 6 & x = 2 \end{aligned}

::0=12x2--2x6=12(x+6)(x-2)x6=0x-2=0x=0x=6x=2

When we input -6 and 2 into the function, we get 0, showing they are zeros of the function and that statement 1 is equivalent to statements 2 and 3.
::当我们输入 -6 和 2 到此函数时, 我们得到 0, 显示它们为函数的零, 而报表 1 相当于报表 2 和 3 。

\begin{aligned} y & = - \frac{1}{2} (- 6)^{2} - 2 (- 6) + 6 = - \frac{1}{2} (36) + 12 + 6 = - 18 + 12 + 6 = 0 \\ y & = - \frac{1}{2} (2)^{2} - 2 (2) + 6 = - \frac{1}{2} (4) - 4 + 6 = - 2 - 4 + 6 = 0 \end{aligned}

::y=-12(-6)2-2(-6)2-2(-6)+6=-12(36)+12+6=18+12+6=6=0y=12(2)-2-2(2)+6=12(4)-4+6=-2-4-6=-2-4+6=0

Lastly, as we can see in the graph below, there are two x- intercepts at (-6,0) and (2,0), showing statement 4 is equivalent to the other three statements.
::最后,如下图所示,(6,0)和(2,0)有两个X界面,显示报表4相当于其他三项报表。

lesson content

In this example, we were able to factor the quadratic polynomial. However, for polynomial functions of degree 3 and higher, it is not always clear how to factor the polynomial. We can use the equivalency above, specifically the Factor Theorem, to help us determine the factors of a polynomial function. If we input values into the function and get 0 as a result, then we know that zero came from a factor of the form $\begin{aligned} x - c \end{aligned}$ .
::在此例子中,我们得以将四边多面体函数乘以四边体数。但是,对于3级及以上多面体函数,并不总是很清楚如何将多面体函数乘以多面体函数。我们可以使用上面的等值, 特别是因子理论, 帮助我们确定多面体函数的因子。如果我们将值输入函数, 结果得到0, 那么我们就会知道零来自窗体 x- c 的一个因子。

But what values do we input? We can start our search with the help of the Rational Root Theorem.
::但是,我们输入什么价值?我们可以在理性根理论的帮助下开始搜索。

Rational Root Theorem
::有理根定理

For a polynomial function, $\begin{aligned} f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} \end{aligned}$ , where $\begin{aligned} a_{n}, a_{n - 1}, \dots a_{0} \end{aligned}$ are integers, the rational roots can be determined from the factors of $\begin{aligned} a_{n} \end{aligned}$ and $\begin{aligned} a_{0} \end{aligned}$ . Any rational roots will have the form $\begin{aligned} \pm \frac{p}{q} \end{aligned}$ , where $\begin{aligned} p \end{aligned}$ is a factor of $\begin{aligned} a_{0} \end{aligned}$ and $\begin{aligned} q \end{aligned}$ is a factor of $\begin{aligned} a_{n} \end{aligned}$ . In other words, the factors of the constant divided by the factors of the leading coefficient will yield all the possible rational solutions to $\begin{aligned} f (x) \end{aligned}$ .
::对于多数值函数 f(x) = anxn+an- 1xn-1ña1x+10, 其中, a- 1, @a1x+10 是整数, 理性根可以从和 a0 的因数中确定。任何理性根将具有 \ pq 的形式, 其中 p 是 a0 的因数, q 是 a 的因数。换句话说, 恒定因数除以主要系数的因数, 会给 f( x) 带来所有可能的理性解决办法。

It is important to note that the Rational Root Theorem does not guarantee that the polynomial has any rational roots.
::必须指出,有理的根理论不能保证多元性具有任何合理的根基。

Example 1
::例1

Find all the possible rational solutions to $\begin{aligned} f (x) = 6 x^{4} - 43 x^{3} + 66 x^{2} - 3 x - 10 \end{aligned}$ .
::查找 f( x) = 6x4 - 43x3+66x2 - 3x- 10 的所有可能的理性解决方案。

Solution: All the possible factors of 10 are 1, 2, 5, and 10. All the possible factors of 6 are 1, 2, 3, and 6.
::解决办法:所有10种可能因素为1、2、5和10。所有6种可能因素为1、2、3和6。

The possible combinations are

\begin{aligned} \frac{\pm 1. \pm 2, \pm 5, \pm 10}{\pm 1, \pm 2, \pm 3, \pm 6} = \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm 10, \pm \frac{10}{3} \end{aligned}

::可能的组合是 1,2,5,10,1,2,3,6,1,12,13,16,2,23,5,52,53,56,10,103。

Therefore , there are 24 possibilities.
::因此,有24种可能性。

by Mathispower4u provides an example of how to list the possible rational roots or zeros of a polynomial function.
::由 Mathispower4u 提供一个例子,说明如何列出多面函数可能的理性根或零。

Example 2
::例2

Find the rational solutions to $\begin{aligned} f (x) = 6 x^{4} - 43 x^{3} + 66 x^{2} - 3 x - 10 \end{aligned}$ .
::查找 f( x) = 6x4 - 43x3+66x2 - 3x- 10 的合理解决方案。

Solution: Let's start our search with the positive integers and use the Factor Theorem.
::解答:让我们从正整数开始搜索, 并使用因子定理。

\begin{aligned} f (1) & = 6 (1)^{4} - 43 (1)^{3} + 66 (1)^{2} - 3 (1) - 10 = 16 not a zero \\ f (2) & = 6 (2)^{4} - 43 (2)^{3} + 66 (2)^{2} - 3 (2) - 10 = 0 a zero \\ f (5) & = 6 (5)^{4} - 43 (5)^{3} + 66 (5)^{2} - 3 (5) - 10 = 0 a zero \end{aligned}

::f(1)=6(1)4-4-4-43(1)3+66(1)2-3(1)-10=16无零f(2)=6(2)4-443(2)3+66(2)2-3(2)-10=0f(5)=6(5)4-44(5)3+66(5)2-3(5)-10=0a

Now we know that 2 and 5 are zeros. If we divide out the factors associated with these zeros, we can reduce this problem to factoring a quadratic expression.
::现在我们知道2和5是零。如果我们把与这些零相关的因素分开,我们就可以减少这个问题,将之作为四面形表达方式的因子。

\begin{aligned} \underline{2} | 6 - 43 66 - 3 - 10 \\ \underline{↓ 12 - 62 8 10} \\ 6 - 31 4 5 0 \\ \underline{5} | 6 - 31 4 5 \\ \underline{↓ 30 - 5 - 5} \\ 6 - 1 - 1 0 \end{aligned}

To find the last two zeros, we can continue to test the numbers above , or we can factor the remaining polynomial, $\begin{aligned} 6 x^{2} - x - 1. a c = - 6 \end{aligned}$ and the factors of -1 that add up to -6 are -3 and 2. Expand the $\begin{aligned} x - \end{aligned}$ term and .
::为了找到最后两个零,我们可以继续测试以上数字,或者我们可以将剩余的多数值、6x2-x-1-1.ac=6乘以系数,加上 -6的-1系数为 -3和2,扩大x-期和2。

\begin{aligned} \underset{⏟}{6 x^{2} - 3 x} + \underset{⏟}{2 x - 1} \\ 3 x (2 x - 1) + 1 (2 x - 1) \\ (2 x - 1) (3 x + 1) \end{aligned}

::6x2 - 3x%2x1 - 1x3x(2x-1)+1(2x-1)(2x-1)(3x+1)

Setting these two factors equal to zero, we have $\begin{aligned} x = \frac{1}{2} \end{aligned}$ and $\begin{aligned} - \frac{1}{3} \end{aligned}$ . Therefore, the solutions to this polynomial are 5, 2, $\begin{aligned} \frac{1}{2} \end{aligned}$ and $\begin{aligned} - \frac{1}{3} \end{aligned}$ .
::将这两个因素设定为零, 我们有 x=12 和 -13。因此, 这个多元的解决方案是 5, 2, 12 和 - 13 。

To check this work, multiply the factors together to see if the result is the original polynomial.
::要检查此工作, 请乘以各种因素, 来查看结果是否为原始的多元值。

\begin{aligned} \underset{⏟}{(2 x - 1) (3 x + 1)} \underset{⏟}{(x - 5) (x - 2)} \\ (6 x^{2} - x - 1) (x^{2} - 7 x + 10) \\ 6 x^{4} - 43 x^{3} + 66 x^{2} - 3 x - 10 \end{aligned}

2x-1)(3x+1) (x-5)(x-2) (6x2-x-1)(x2-7x+10) ×6x4-43x3+66x2-3x-10)

by Randy Anderson demonstrates what the Remainder and Factor Theorems are, and how they can be used to find the linear factorization of a polynomial.
::Randy Anderson展示了残骸和因子定理是什么, 以及如何利用它们来找到多面体的线性因子化。

The lists of possible rational zeros can be quite long. It would be helpful to have ways to limit the list. Luckily, there are two.
::可能的理性零点列表可能很长。找到限制列表的方法将会有所帮助。幸运的是,有两个。

Descartes's Rule of Signs and the Bounds Theorem
::笛卡尔的标志规则和界界定理

Descartes's Rule of Sign
::笛卡尔的签署规则

The number of positive real zeros, n, is the number of sign changes of f(x), or an even number less, n-2, n-4, etc., until there are 1 or 0.
::正实际零数, n,是 f(x) 或更少的 n-2, n-4 等的符号更改数,直到有 1 或 0 。

The number of negative real zeros, n, is the number of sign changes of f(-x) or an even number less, n-2, n-4, etc., until there are 1 or 0.
::负实际零数, n,是 f( x) 或更少的 n-2, n-4 等的符号更改数,直到有 1 或 0 。

Bounds Theorem
::弹出定理

$\begin{aligned} c \end{aligned}$ is an upper bound for the real zeros if, when you divide by $\begin{aligned} x - c \end{aligned}$ , the coefficients of the quotient and the remainder all have the same sign.
::c 为实际零数的上限,如果除以 x-c,则商数系数和剩余系数都有相同的符号。

$\begin{aligned} c \end{aligned}$ is a lower bound for the real zeros if, when you divide by $\begin{aligned} x - c \end{aligned}$ , the coefficients of the quotient and the remainder have alternating signs.
::c 是指实际零数的下限,如果在除以 x-c时,商数的系数和其余的系数有交替的符号。

This video by CK-12 demonstrates how to use Descartes's Rule of Signs to determine the type and number of zeros of a polynomial function.
::CK-12的这段影片展示了如何使用笛卡尔的标志规则来确定多面函数零的种类和数量。

by George Engel demonstrates how to use the Bounds Theorem to determine the upper and lower bounds for polynomials.
::George Engel展示了如何使用Bounds定理来确定多面体的上下界。

Example 3
::例3

Find all the real solutions to $\begin{aligned} f (x) = x^{3} + 2 x^{2} - 10 x - 8 \end{aligned}$ .
::查找 f( x) =x3+2x2- 10x- 8的所有真实解决方案。

Solution: T he Rational Root Theorem helps us determine all possible rational roots.
::解答:理性根理论帮助我们确定所有可能的理性根。

\begin{aligned} \frac{factors of -8}{factors of 1} = \frac{\pm 8, \pm 4, \pm 2, \pm 1}{\pm 1} \end{aligned}

::-8因子 -8 8,8,4,2,1

If we want to limit the list, we can use Descartes's Rule of Signs and the Bounds Theorem. First, using Descartes's Rules of Signs, we have
::如果我们想要限制列表, 我们可以使用笛卡尔的标志规则和“ 界标定论 ” 。首先, 我们使用笛卡尔的标志规则,

\begin{aligned} f (x) & = x^{3} + 2 x^{2} - 10 x - 8 \\ f (- x) & = (- x)^{3} + 2 (- x)^{2} - 10 (- x) - 8 \\ = - x^{3} + 2 x^{2} + 10 x - 8 \end{aligned}

x)=x3+2x2-10x-8f(x)=(-x)3+2(-x)2-10(x)8=-x3+2x2+10x-8=-x3+2x2+10x-8)

There is one sign change in $\begin{aligned} f (x) \end{aligned}$ , so there is one positive real zero. There are two sign changes in $\begin{aligned} f (- x) \end{aligned}$ , so there are two or zero negative real zeros.
::f( x) 有1个符号变化, 所以有1个正正实际零。 f( x) 有2个符号变化, 所以有2个或0个负实际零。

Let's try to find the positive real zero first.
::让我们先找出正数真正的零。

\begin{aligned} f (1) = (1)^{3} + 2 (1)^{2} - 10 (1) - 8 = - 15 & f (2) = (2)^{3} + 2 (2)^{2} - 10 (2) - 8 = - 12 \\ \underline{1} | 1 2 - 10 - 8 & \underline{2} | 1 2 - 10 - 8 \\ \underline{↓ 1 3 - 7} & \underline{↓ 2 8 - 16} \\ 1 3 - 7 - 15 & 1 4 - 2 - 24 \\ f (4) = (4)^{3} + 2 (4)^{2} - 10 (4) - 8 = 32 \\ \underline{4} | 1 2 - 10 - 8 \\ \underline{↓ 6 32 88} \\ 1 8 22 80 \end{aligned}

::f(1)=(1)3+2(1)-2-10(1)-8=-15f(2)=(2)3+2(2)-2-2-2-10(2)-8=-121__1-2-10-81-2-10-8__1-2-10-8__1-8__1 3-7_ _2-2-8-16_1 3-7-151 4-2-2-24f(4)(4)3+2(4)(4)-44-2(4)-210(4)-8=324__1-2-10-8__6 32 88_1 8 80

According to the Bounds Theorem, 4 is an upper bound for the zeros. Trying negative numbers, we have
::根据《Bounds Theorem》,4是零的上限。尝试负数,我们有

\begin{aligned} f (- 1) = (- 1)^{3} + 2 (- 1)^{2} - 10 (- 1) - 8 = 3 \\ f (- 2) = (- 2)^{3} + 2 (- 2)^{2} - 10 (- 2) - 8 = 12 \\ f (- 4) = (- 4)^{3} + 2 (- 4)^{2} - 10 (- 4) - 8 = 0 \end{aligned}

::f(-1)=(-1)=(-1)3+2(2)-1-2-10(-1)-8=3f(-2)=(-2)3+2(-2)-2-10(-2)-8=12f(4)=(4)3+2(-4)-2-10(4)-8=0

$\begin{aligned} x + 4 \end{aligned}$ is a factor of the polynomial. Let's divide it out.
::x+4 是多面性的一个因素。让我们将其分割出来。

\begin{aligned} \underline{- 4} | 1 2 - 10 - 8 \\ \underline{↓ - 4 8 8} \\ 1 - 2 - 2 0 \end{aligned}

At the end of the synthetic division , the remaining polynomial is $\begin{aligned} x^{2} - 2 x - 2, \end{aligned}$ which is not factorable. Therefore, to find the last two solutions, we use the Quadratic Formula .
::在合成分区的结尾处,剩下的多面体为x2-2x-2,这是不可考虑的。因此,为了找到最后两种解决办法,我们使用二次公式。

\begin{aligned} x & = \frac{2 \pm \sqrt{(- 2)^{2} - 4 (1) (- 2)}}{2 (1)} \\ = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2 \sqrt{3}}{2} = 1 \pm \sqrt{3} \approx 2.73, - 0.73 \end{aligned}

::x=2(-2-2)2-4(1)(-1-2)2(1)=24+82=2122=2232=132.73,-0.73

The roots, or zeros, of $\begin{aligned} f (x) = x^{4} + 6 x^{3} - 2 x^{2} - 48 x - 32 \end{aligned}$ are -4 (twice), 2.73, and -0.73.
::f(x) =x4+6x3-2x2-48x-32的根或零是-4(两次)、 2.73 和 - 0.73。

Example 4
::例4

A testing station collects data over a 10-month period about the number of mosquito eggs laid indoors. It was modeled by $\begin{aligned} y = 0.025 x (x^{3} - 42 x^{2} + 332 x + 840) \end{aligned}$ , where x is the month and y is the number of mosquito eggs. When does the model predict there will be no mosquito eggs? We cover how to find zeros of a polynomial function in this section.
::一个测试站在10个月期间收集了室内投放的蚊子数量的数据,其模型为 y= 0.025x(x3-42x2+332x+840) , x 是月份, y 是蚊子蛋的数量。模型预测何时不会有蚊子蛋 ? 我们如何在本节中找到一个多功能的零。

Solution: The possible rational solutions of the polynomial inside " data-term="Parentheses" role="term" tabindex="0"> parentheses are the factors of 840:
::解决方案:圆括号内多面括号中可能的合理解决方案是840因素:

\begin{aligned} \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7, \pm 8, \pm 10, \pm 12, \pm 14, \pm 15, \pm 20, \pm 21, \pm 24, \\ \pm 28, \pm 30, \pm 35, \pm 40, \pm 42, \pm 56, \pm 60, \pm 70, \pm 84, \\ \pm 105, \pm 120, \pm 140, \pm 168, \pm 210, \pm 280, \pm 420, \pm 840 \end{aligned}

Using the Factor Theorem, $\begin{aligned} (- 2)^{3} - 42 (- 2)^{2} + 332 (- 2) + 840 = - 8 - 168 - 664 + 840 = 0 \end{aligned}$ . -2 is a root.
::使用系数定理, (-2) 3-42(-2) 2+332(-2)+840=-8-168-664+840=0。 -2是根。

If we use synthetic division, we can reduce this polynomial to a quadratic.
::如果我们使用合成分裂, 我们可以把这个多面体减为四面形。

\begin{aligned} - 2 | 1 - 42 332 840 \\ \underline{- 2 88 - 840} \\ 1 - 44 420 0 \end{aligned}

The polynomial that remains is

\begin{aligned} x^{2} - 44 x + 420 \end{aligned}

.
::-21-42 332 840-2 88-840_1-44 420 0 残留的多面体为 x2-44x+420。

\begin{aligned} x^{2} - 44 x + 420 = 0 \\ (x - 14) (x - 30) = 0 \\ x - 14 = 0 x - 30 = 0 \\ x = 14 x = 30 \end{aligned}

::x2 - 44x+420=0(x- 14)(x- 30)=0 - 30(x- 30)=0 - 14=0 x- 30=0=0x=14x=30

The zeros of the polynomial functions are 0, -2, 14, and 30.
::多面函数的零为 0, - 2, 14, 30。

Summary
::摘要

To find the real zeros of a polynomial function, we can start with the Rational Root Theorem. The possible rational roots are the combinations of the factors of the constant term in the numerator and the factor of the leading coefficient in the denominator. We can reduce our search using the Bounds Theorem or Descartes's Rule of Signs.
::要找到多元函数的真正零, 我们可以从理性根理论开始。可能的理性根源是分子中的常数因数和分母中主要系数因数的组合。我们可以使用 Bounds Theorem 或 Descartes 的符号规则来减少搜索。

Review
::回顾

Find all the possible rational solutions for the polynomials below. Use the Rational Root Theorem.
::为下面的多边协议寻找所有可能的理性解决方案。请使用理性根理论。

1. $\begin{aligned} f (x) = x^{3} + 6 x^{2} - 18 x + 20 \end{aligned}$
::1. f(x)=x3+6x2-18x+20

2. $\begin{aligned} f (x) = 4 x^{4} + x^{2} - 15 \end{aligned}$
::2. f(x)=4x4+x2-15

3. $\begin{aligned} f (x) = - 2 x^{3} + 7 x^{2} - x + 8 \end{aligned}$
::3. f(x)%2x3+7x2 -x+8

4. $\begin{aligned} f (x) = x^{4} - 3 x^{3} - 4 x^{2} + 15 x + 9 \end{aligned}$
::4. f(x)=x4-3x3-4x2+15x+9

5. $\begin{aligned} f (x) = 8 x^{4} - 5 x^{3} + 16 x^{2} + 37 x - 24 \end{aligned}$
::5. f(x)=8x4-5x3+16x2+37x-24

Find all the real number zeros for each function below. Use any method you like.
::查找以下每个函数的所有实际数字零。使用任何您喜欢的方法。

6. $\begin{aligned} f (x) = 6 x^{3} - 17 x^{2} + 11 x - 2 \end{aligned}$
::6. f(x)=6x3-17x2+11x-2

7. $\begin{aligned} f (x) = x^{4} + 7 x^{3} + 6 x^{2} - 32 x - 32 \end{aligned}$
::7. f(x)=x4+7x3+6x2-32-32x-32

8. $\begin{aligned} f (x) = 16 x^{3} + 40 x^{2} - 25 x - 3 \end{aligned}$
::8. f(x)=16x3+40x2-25x-3

9. $\begin{aligned} f (x) = 2 x^{3} - 9 x^{2} + 21 x - 18 \end{aligned}$
::9. f(x) = 2x3- 9x2+21x-18

10. $\begin{aligned} f (x) = 4 x^{3} - 16 x^{2} + 39 x - 295 \end{aligned}$
::10. f(x)=4x3-16x2+39x-295

11. $\begin{aligned} f (x) = 18 x^{4} + 3 x^{3} - 17 x^{2} + 17 x - 55 \end{aligned}$
::11. f(x)=18x4+3x3-17x2+17x-55

12. $\begin{aligned} f (x) = x^{5} + 7 x^{4} - 3 x^{3} - 65 x^{2} - 8 x - 156 \end{aligned}$
::12. f(x)=x5+7x4-3x3-65x2-8x-156

13. $\begin{aligned} f (x) = 4 x^{4} + 20 x^{3} - 23 x^{2} - 120 x + 144 \end{aligned}$
::13. f(x) = 4x4+20x3 - 23x2 - 120x+144

14. $\begin{aligned} f (x) = 9 x^{4} - 226 x^{2} + 25 \end{aligned}$
::14. f(x)=9x4-226x2+25

Explore More
::探索更多

1. John is constructing the bottom portion of a jewelry box for his girlfriend. He begins with a sheet of metal that is 10 inches wide and 15 inches long. He cuts the same size square out of each of the four corners of the sheet of metal. Let $\begin{aligned} V (x) \end{aligned}$ represent the volume of the resulting (open-topped) box.
::1. John正在为其女友建造首饰盒底部,首先是10英寸宽和15英寸长的金属板,将金属板四角的每个角切割成相同大小的平方,让V(x)代表所产生的(打开的)盒子的体积。

a. Write $\begin{aligned} V (x) \end{aligned}$ as a product of linear factors.
::a. 编写V(x),作为线性因素的产物。

b. For which values of $\begin{aligned} x \end{aligned}$ is $\begin{aligned} V (x) = 0 \end{aligned}$ ?
::b. V(x)=0的x值是多少?

c. Which of your solutions from part b are physically possible?
::c. B部分的哪些解决办法在实际中是可行的?

2. A car manufacturer has designed a solar-powered car. There will be several costs associated with manufacturing the cars.
::2. 汽车制造商设计了一辆太阳能汽车,与汽车制造有关的费用很多。

Monthly cost of leasing the building	$72,000
Monthly depreciation of the equipment	$130,000
Monthly cost of paying the new employees	$480,000
C ost of other miscellaneous expenses	$47,000

a. If the cost of the materials for each car is $4,500, write a model for the cost, $\begin{aligned} C (x), \end{aligned}$ where x is the number of cars produced.
::a. 如果每辆汽车的材料费用为4 500美元,请填写成本模型C(x),其中x为所生产的汽车数量。

b. The monthly revenue from selling x cars is $\begin{aligned} R (x) = 27, 500 x \end{aligned}$ . How many cars must the company make each month in order to earn a profit?
::b. 出售x汽车的月收入为R(x)=27 500x.公司为赚取利润每月必须制造多少辆汽车?

c. If the manager decides to increase the number of employees and create two shifts resulting in an additional $300,000 in monthly employee salaries, with all other expenses remaining constant, how many cars would the company have to make in order to earn a profit?
::c. 如果经理决定增加雇员人数并实行两班制,导致每月雇员工资增加300 000美元,所有其他费用保持不变,公司为赚取利润必须制造多少辆汽车?

d. Suppose the factory at full capacity (1st and 2nd shift) can make a maximum of 75 cars per month. What is the minimum price the company can sell each car for to make a profit?
:d) 假设工厂满负荷(第1和第2轮轮班)每月最多可达75辆汽车,公司出售每辆汽车以盈利的最低价格是多少?

3. The population of mosquitoes in Cleveland, Mississippi, is modeled by the function $\begin{aligned} M (w) = 200 w^{2} - 0.01 w^{4} + 1, 200 \end{aligned}$ . $\begin{aligned} w \end{aligned}$ represents the the number of weeks since the city began spraying a chemical that kills the mosquitoes, and $\begin{aligned} M (w) \end{aligned}$ represents the number of living mosquitoes. How many weeks will it take for all the mosquitoes to perish?
::3. 密西西比州克利夫兰的蚊子人口以M(w)=200w2-0.01w4+1200的功能为模型,代表该市开始喷洒杀死蚊虫的化学品以来的周数,而M(w)则代表活蚊子的数量,所有蚊子需要多少周才能消亡?

4. The population of bears in a large forest is represented by the function $\begin{aligned} B (m) = 110 m^{2} - 0.35 m^{4} + 750 \end{aligned}$ , where $\begin{aligned} m \end{aligned}$ is the number of months the bears have been observed in the forest. Given this information, how many more months will there be bears in the forest?
::4. 大型森林中的熊人口以函数B(m)=110m2-0.35m4+750为代表,其中米为森林中观察到的熊月数。

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

Section outline

Polynomials ::多元数

Polynomials and Important Terminology ::聚合和重要术语

Zeros of Polynomial Functions ::多元函数零

Zeros of Polynomial Functions ::多元函数零

Rational Root Theorem ::有理根定理

Rational Root Theorem ::有理根定理

Example 1 ::例1

Example 2 ::例2

Descartes's Rule of Signs and the Bounds Theorem ::笛卡尔的标志规则 和界界定理

Example 3 ::例3

Example 4 ::例4

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

Polynomials
::多元数

Polynomials and Important Terminology
::聚合和重要术语

Zeros of Polynomial Functions
::多元函数零

Zeros of Polynomial Functions
::多元函数零

Rational Root Theorem
::有理根定理

Rational Root Theorem
::有理根定理

Example 1
::例1

Example 2
::例2

Descartes's Rule of Signs and the Bounds Theorem
::笛卡尔的标志规则和界界定理

Example 3
::例3

Example 4
::例4

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复