Course: 2.5 单元长分区和合成分区

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聚合长长分部和合成分部

While you may be experienced in factoring, there will always be that do not readily factor using basic or advanced techniques. How can you identify the roots of these polynomials?

Rational Roots and Dividing Polynomials
::理性根和分裂的多面性

There are numerous theorems that point out relationships between polynomials and their factors. For example there is a theorem that a polynomial of degree $\begin{align*}n\end{align*}$ must have exactly $\begin{align*}n\end{align*}$ solutions/factors that may or may not be real numbers. The Rational Root Theorem and the Remainder Theorem are two theorems that are particularly useful starting places when manipulating polynomials.
::有无数的理论指出多民族及其因素之间的关系。例如,有这样一个理论,即一个多民族程度的n必须具有可能是或不是真实数字的精确n 溶液/因素。理性根理论和残余理论是两个理论,它们是在操纵多民族运动时特别有用的起始点。

The Rational Root Theorem
::理性根理定理

The Rational Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction $\begin{align*}\left(x=\frac{p}{q}\right)\end{align*}$ , where $\begin{align*}p\end{align*}$ is an integer factor of the constant term and $\begin{align*}q\end{align*}$ is an integer factor of the leading coefficient .
::理性根理论指出,在多数值中,每一种合理解决办法都可以写成一个减数分数(x=pq),p是常数的整数系数,q是主要系数的整数系数。

Let's identify all the possible rational solutions of the following polynomial using the Rational Root Theorem.
::让我们用理性根理理论找出以下多面体的所有可能合理解决方案

$\begin{align*}12x^{18}-91x^{17}+x^{16}+\cdots+2x^2-14x+5=0\end{align*}$
::12x18-91x17+x16}%2x2-14x+5=0

The integer factors of 5 are 1, 5. The integer factors of 12 are 1, 2, 3, 4, 6 and 12. Since pairs of factors could both be negative, remember to include $\begin{align*}\pm\end{align*}$ .
::5的整数系数为1、5. 12的整数系数为1、2、3、4、6和12。由于一对因素可能都是负数,请记住包括++。

$\begin{align*}\pm \frac{p}{q}=\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, \frac{5}{1}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{6}, \frac{5}{12}\end{align*}$
::pq=11,12,13,14,16,112,51,52,53,54,56,512

T he possible solutions can be found from these 24 possible rational answers. If this question required you to find a solution, then the Rational Root Theorem would give you a great starting place. Once you have one root, you can use either polynomial long division or synthetic to divide the factor out and to keep reducing the expression . We will use the Rational Root Theorem in Example 3.
::从这24个可能的理性答案中可以找到可能的解决方案。如果这个问题需要您找到解决方案, 那么理性根理论会给您一个伟大的起点。一旦您有一个根, 您可以使用多等长的分隔或合成来分割系数并不断缩小表达式。我们将使用例 3 中的理性根理论。

Polynomial Long Division and the Remainder Theorem
::多元长分解和残余定理

Polynomial long division is identical to regular long division except the dividend and divisor are both polynomials instead of numbers.
::多元长的分割与常规长的分割完全相同,但红利和断层是多元的而不是数字。

The Remainder Theorem states that the remainder of a polynomial $\begin{align*}f(x)\end{align*}$ divided by a linear divisor $\begin{align*}(x-a)\end{align*}$ is equal to $\begin{align*}f(a)\end{align*}$ . The Remainder Theorem is only useful after you have performed polynomial long division because you are usually never given the divisor and the remainder to start. The main purpose of the Remainder Theorem in this setting is a means of double checking your application of polynomial long division.
::保存者论者称, 由线性对角( x - a) 除以线性对角( x - a) 的多元 f (x) 的剩余部分等于 f(a) 。保存者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者者论者论者论者论者论者者者者者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者论者者论者论者论者论者论者者者者者者者者者者者者者者者者者者者者, 。。者者者者者者者者者者者者者者者者论者论者论者论者者者者者者者者者者者者者者论者者者者者者者者者者论者论者论者者论者者者者者者论者论者论者论者者论者论者者者者者者论者

Let's put this knowledge to use and use polynomial long division to divide:
::让我们用这些知识来使用并使用多面长的分界线来分割:

$\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}$
::x3+2x2 - 5x+7x- 3

First note that it is clear that 3 is not a root of the polynomial because of the Rational Root Theorem and so there will definitely be a remainder. Start a polynomial long division question by writing the problem like a long division problem with regular numbers:
::首先要注意的是,由于有理的根理理论,显然3不是多元论的根,因此肯定还有剩余部分。开始一个多元长的分裂问题, 将问题写成常数的长分裂问题 :

$\begin{align*}\overline{}{x-3 \overline { ) {x^3+2x^2-5x+7}}}\end{align*}$
::X - 3x3x3+2x2 - 5x+7

Just like with regular numbers ask yourself “how many times does $\begin{align*}x\end{align*}$ go into $\begin{align*}x^3\end{align*}$ ?” which in this case is $\begin{align*}x^2\end{align*}$ .
::就像普通数字一样,问自己“x有多少次进入 x3?” 在这种情况下,是 x2。

$\begin{align*}\overset{x^2}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\end{align*}$
::x3x3+2x2 - 5x+7x2

Now multiply the $\begin{align*}x^2\end{align*}$ by $\begin{align*}x-3\end{align*}$ and copy below. Remember to subtract the entire quantity.
::现在将 x2 乘以 x- 3 和下面的复制件。记住要减去全部数量。

\begin{aligned} \overset{x^{2}}{x - 3 \bar{) x^{3} + 2 x^{2} - 5 x + 7}} \\ \underline{- (x^{3} - 3 x^{2})} \end{aligned}

$\begin{align*}& \overset{x^2}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\\ & \quad \ \underline{-(x^3-3x^2)}\end{align*}$
::x3x3+2x2 -5x+7x2 -(x3-3x2)_

Combine the rows, bring down the next number and repeat.
::合并行, 下调下一个数字并重复。

\begin{aligned} \overset{x^{2} + 5 x + 10}{x - 3 \bar{) x^{3} + 2 x^{2} - 5 x + 7}} \\ \underline{- (x^{3} - 3 x^{2})} \\ 5 x^{2} - 5 x \\ \underline{- (5 x^{2} - 15 x)} \\ 10 x + 7 \\ \underline{- (10 x - 30)} \\ 37 \end{aligned}

$\begin{align*}& \overset{x^2+5x+10}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\\ & \quad \ \underline{-(x^3-3x^2)}\\ & \qquad \qquad \ \ 5x^2-5x\\ & \qquad \quad \ \underline{-(5x^2-15x)}\\ & \qquad \qquad \qquad \quad \ 10x+7\\ & \qquad \qquad \qquad \underline{-(10x-30)}\\ & \qquad \qquad \qquad \qquad \qquad 37\end{align*}$
::x3 - 3x3+2x2 - 5x+7x2+5x10 - (x3 - 3x2) _ 5x2 - 5x5x - (5x2 - 15x) _ 10x+7 - (10x - 30)_ 37

The number 37 is the remainder. There are two things to think about at this point. First, interpret in an equation :
::37号是剩下的部分。在这一点上,有两件事需要考虑。首先,用方程式解释:

$\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}=(x^2+5x+10)+\frac{37}{x-3}\end{align*}$
::x3+2x2-2-5x+7x-3=(x2+5x+10)+37x-3

Second, check your result with the Remainder Theorem which states that the original function evaluated at 3 must be 37. Notice the notation indicating to substitute 3 in for $\begin{align*}x\end{align*}$ .
::其次,请检查您与保留论者之间的结果,该理论指出,原3点评价的功能必须是37。注意标记表示 x 中替换 3 。

$\begin{align*}(x^3+2x^2-5x+7)|_{x=3}=3^3+2 \cdot 3^2-5 \cdot 3+7=27+18-15+7=37\end{align*}$
:x3+2x2-5x+7)x=3=33+2=23+2=32-5}3+7=27+18+15+7=37)

Synthetic Division
::合成司

Synthetic division is an abbreviated version of polynomial long division where only the coefficients are used. Synthetic division is mostly used when the leading coefficients of the numerator and denominator are equal to 1 and the divisor is a first degree binomial . Let's use synthetic division to divide the same expression that we divided above with polynomial long division:
::合成分裂是一个缩略版的多元长分割法, 仅使用系数。当分子和分母的主要系数等于1, 和分母是第一级二进制的时, 通常使用合成分裂法。让我们使用合成分裂法来将上面的表达法和上面的多元长分割法分开 :

$\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}$
::x3+2x2 - 5x+7x- 3

Instead of continually writing and rewriting the $\begin{align*}x\end{align*}$ symbols, synthetic division relies on an ordered spacing.
::合成部分不是不断写作和重写x符号,而是依靠一个定的间距。

$\begin{align*}\underline{+3 \smash{\big|}} \ \ 1 \ \ 2 \ \ \text{-}5 \ \ \ 7\end{align*}$

Notice how only the coefficients for the denominator are used and the divisor includes a positive three rather than a negative three. The first coefficient is brought down and then multiplied by the three to produce the value which goes beneath the 2.
::注意只使用分母的系数和分母包括正数三,而不是负数三。第一个系数被降低,然后乘以三以产生低于2的值。

\begin{aligned} \underline{+ 3 |} & 1 2 - 5 7 \\ \underline{↓ 3} \\ 1 \end{aligned}

$\begin{align*}\underline{+3 \smash{\big|}} \ \ & 1 \ \ 2 \ \ \text{-}5 \ \ \ 7 \\ & \underline{\downarrow \, 3 \ \quad \quad \ }\\ & 1\end{align*}$

Next the new column is added. $\begin{align*}2+3=5\end{align*}$ , which goes beneath the $\begin{align*}2^{nd}\end{align*}$ column. Now, multiply $\begin{align*}5 \cdot +3=15\end{align*}$ , which goes underneath the -5 in the $\begin{align*}3^{rd}\end{align*}$ column. And the process repeats…
::下一列加入新的列。 2+3=5, 位于第二列下方。现在, 乘以 5+3=15, 在第三列的 - 5 下方。程序重复...

\begin{aligned} \underline{+ 3 |} & 1 2 - 5 7 \\ \underline{↓ 3 15 30} \\ 1 5 10 37 \end{aligned}

$\begin{align*}\underline{+3 \smash{\big|}} \ \ & 1 \ \ 2 \ \ \text{-}5 \ \ \ 7 \\ & \underline{\downarrow \, 3 \ \ 15 \ \ 30}\\ & 1 \ \ 5 \ \ 10 \ \ 37\\\end{align*}$

The last number, 37, is the remainder. When writing out the resulting expression, you will put this remainder over the divisor. The three other numbers represent the quadratic that is identical to the solution to the result from dividing the expression using polynomial long division. Note that when writing out the expression, you decrease the exponent of the leading coefficient of the original by 1.
::最后一个数字是 37, 是剩余部分。写入结果表达式时, 您将把剩余部分放在 divisor 上方。其他三个数字代表与使用多面长分隔表达式结果的解决方案完全相同的二次方。请注意, 在写入表达式时, 您会将原始系数的指数降低 1 。

$\begin{align*}(1x^2+5x+10)+\frac{37}{x-3}\end{align*}$
:1x2+5x+10)+37x-3

Examples
::实例

Example 1
::例1

Earlier, you were asked about identifying roots of polynomials that do not readily factor using the techniques you have learned so far. Identifying roots of polynomials by hand can be tricky business. The best way to identify roots is to use the rational root theorem to quickly identify likely candidates for solutions and then use synthetic or polynomial long division to quickly and effectively test them to see if their remainders are truly zero.
::早些时候,有人问您如何找出无法使用你迄今所学到的技术随时考虑因素的多元动物的根。亲手确定多元动物的根可能是棘手的事情。找到根源的最佳方法是利用理性根理论快速确定可能的解决办法的候选方,然后使用合成或多民族长的分解快速有效地测试它们的剩余部分是否真的为零。

Example 2
::例2

Divide the following polynomials.
::将下列多面体除以。

$\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}\end{align*}$
::x3+2x2 - 4x+8x-2

Since the leading coefficients of the numerator and denominator are both 1 and the denominator is a binomial, synthetic division is a good method to use here.
::由于分子和分母的主要系数为1,而分母是一种二元分解,合成分解是这里使用的一种好方法。

$\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}=x^2+4x+4+\frac{16}{x-2}\end{align*}$
::x3+2x2-4x+8x-2=x2+4x+4+4+16x-2

Example 3
::例3

Completely factor the following polynomial.
::完全系数如下:

$\begin{align*}x^4+6x^3+3x^2-26x-24\end{align*}$
::x4+6x3+3x2-26x-24

Notice that possible roots are $\begin{align*}\pm 1, 2, 3, 4, 6, 8, 24\end{align*}$ . Of these 14 possibilities, four will yield a remainder of zero. When you find one, use long division or synthetic division to factor out the root that you found. Then find another zero and repeat the process.
::请注意, 可能的根值是 +1, 2, 34, 6, 6, 8, 24。在这14种可能性中, 4 个将产生零的剩余值。当您发现一个可能性时, 使用长的分割或合成分割来计算您找到的根值。然后再找到一个零并重复此进程。

\begin{aligned} x^{4} + 6 x^{3} + 3 x^{2} - 26 x - 24 \\ = (x + 1) (x^{3} + 5 x^{2} - 2 x - 4) \\ = (x + 1) (x - 2) (x^{2} + 7 x + 12) \\ = (x + 1) (x - 2) (x + 3) (x + 4) \end{aligned}

$\begin{align*}& x^4+6x^3+3x^2-26x-24\\ & =(x+1)(x^3+5x^2-2x-4)\\ & =(x+1)(x-2)(x^2+7x+12)\\ &= (x+1)(x-2)(x+3)(x+4)\end{align*}$
::x4+6x3+3x2-2-2-26x-24=(x+1)(x3+5x2-2x-4)=(x+1)(x-2)(x-2)(x2+7x+12)=(x+1)(x-2)(x+3)(x+4)

The first zero found was -1. It was divided out of the original expression to find the remaining non-factored portion of the expression. The second zero found was 2 from the remaining piece and was divided out. Once you get down to a quadratic expression, you can use the other factoring techniques you know to factor the rest of the expression.
::发现的第一个零是 - 1。它被从原来的表达式中分离出来, 以找到表达式的剩余非因子部分。发现第二个零是剩余部分中的2个, 并且被分割出来。一旦您进入二次表达式, 您可以使用您知道的其他计算技术来计算表达式的其余部分。

Example 4
::例4

Divide the following polynomials.
::将下列多面体除以。

$\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}\end{align*}$
::3x5-2x2+10x-5x-1

Since the first coefficient of the numerator is not 1, polynomial long division is a good method to use here.
::因为分子的第一个系数不是1, 多数值长的分隔法是这里使用的一个好方法。

$\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}=3x^4+3x^3+3x^2+x+11+\frac{6}{x-1}\end{align*}$
::3x5-2x2+10x-5x-1=3x4+3x3+3x2+3x2+11+6x-1

Summary

The Rational Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction $\begin{align*}\frac{p}{q},\end{align*}$ where $\begin{align*}p\end{align*}$ is an integer factor of the constant term and $\begin{align*}q\end{align*}$ is an integer factor of the leading coefficient.
::理性根理论指出,在多数值中,每一种合理解决办法都可以写成一个减分pq,p是常数的整数系数,q是主要系数的整数系数。

Polynomial long division is a method for dividing polynomials, similar to regular long division with numbers.
::多元长的分割法是多面分割法,类似于带有数字的常规长的分割法。

The Remainder Theorem states that the remainder of a polynomial $\begin{align*}f(x)\end{align*}$ divided by a linear divisor $\begin{align*}(x-a)\end{align*}$ is equal to $\begin{align*}f(a).\end{align*}$
::遗迹论者称,多边 f(x) 的其余部分除以线性对角(x-a) 等于 f(a)。

Synthetic division is an abbreviated version of polynomial long division.
::合成分裂是一种缩略的多面长分裂。

Review
::回顾

Identify all possible rational solutions of the following polynomials using the Rational Root Theorem.
::使用“有理根理论”找出下列多边协议的所有可能的合理解决方案。

1. $\begin{align*}15x^{14}-12x^{13}+x^{12}+ \cdots+2x^2-5x+5=0\end{align*}$
::1. 15x14 - 12x13+x12*2x2 - 5x+5=0

2. $\begin{align*}18x^{11}+42x^{10}+x^9+\cdots+x^2-3x+7=0\end{align*}$
::2. 18x11+42x10+x9x2-3x+7=0

3. $\begin{align*}12x^{16}+11x^{15}+3x^{14}+\cdots+6x^2-2x+11=0\end{align*}$
::3. 12x16+11x15+3x14=6x2-2x+11=0

4. $\begin{align*}14x^7-7x^6+x^5+\cdots+x^2+6x+3=0\end{align*}$
::4. 14x7-7x6+x5x2+6x3=0

5. $\begin{align*}9x^9-10x^8+3x^7+\cdots+4x^2-2x+2=0\end{align*}$
::5. 9x9-10x8+3x7=4x2-2x+2=0

Completely factor the following polynomials.
::完全乘以下列多面体。

6. $\begin{align*}2x^4-x^3-21x^2-26x-8\end{align*}$
::6. 2x4 - x3 - 21x2 - 26x-8

7. $\begin{align*}x^4+7x^3+5x^2-31x-30\end{align*}$
::7. x4+7x3+5x2-31x-30

8. $\begin{align*}x^4+3x^3-8x^2-12x+16\end{align*}$
::8. x4+3x3-8x2 - 12x+16

9. $\begin{align*}4x^4+19x^3-48x^2-117x-54\end{align*}$
::9. 4x4+19x3-48x2-117x-54

10. $\begin{align*}2x^4+17x^3-8x^2-173x+210\end{align*}$
::10. 2x4+17x3-8x2-173x2+210

Divide the following polynomials.
::将下列多面体除以。

11. $\begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+4}\end{align*}$
::11. x4+7x3+5x2-31x-30x+4

12. $\begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+2}\end{align*}$
::12. x4+7x3+5x2-31x-30x+2

13. $\begin{align*}\frac{x^4+3x^3-8x^2-12x+16}{x+3}\end{align*}$
::13. x4+3x3-8x2 - 12x+16x+3

14. $\begin{align*}\frac{2x^4-x^3-21x^2-26x-8}{x^3-x^2-10x-8}\end{align*}$
::14. 2x4 - x3 - 21x2 - 26x-8x3 - x2 - 10x-8

15. $\begin{align*}\frac{x^4+8x^3+3x^2-32x-28}{x^3+10x^2+23x+14}\end{align*}$
::15. x4+8x3+3x2-3-3x2-32x-28x3+10x2+23x+14

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

聚合长长分部和合成分部

Rational Roots and Dividing Polynomials ::理性根和分裂的多面性

The Rational Root Theorem ::理性根理定理

Polynomial Long Division and the Remainder Theorem ::多元长分解和残余定理

Synthetic Division ::合成司

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Rational Roots and Dividing Polynomials
::理性根和分裂的多面性

The Rational Root Theorem
::理性根理定理

Polynomial Long Division and the Remainder Theorem
::多元长分解和残余定理

Synthetic Division
::合成司

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)