Course: 2.1 保理审查 | The Way To Learn

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保保权审查

To factor means to write an expression as a product instead of a sum. Factoring is particularly useful when solving equations set equal to zero because then logically at least one factor must be equal to zero. In PreCalculus, you should be able to factor even when there is no obvious greatest common factor or the difference is not between two perfect squares.
::系数是指将表达式写成一个产品而不是一个总和。当解答设定为等于零的方程式时,保理作用特别大,因为从逻辑上讲,至少一个因素必须等于零。在预考中,即使没有明显的最大共同因素,或者差异不是在两个完美的方形之间,你也应该能够计理。

How do you use the difference of perfect squares factoring technique on that don’t contain perfect squares and why would this be useful?
::你如何使用完美的平方计数技术的差别,

Factoring Functions
::保分函数

A polynomial is a sum of a finite number of terms . Each term consists of a constant that multiplies a variable . The variable may only be raised to a non-negative exponent . The letters $\begin{aligned} a, b, c \dots \end{aligned}$ in the following general polynomial expression stand for regular numbers like $\begin{aligned} 0, 5, - \frac{1}{4}, \sqrt{2} \end{aligned}$ and the $\begin{aligned} x \end{aligned}$ represents the variable.
::多式数是一个限定的术语总和。每个术语包含一个常数,可乘以变量。该变量只可升至非负式引言。以下通用多式表达式中的字母a、b、c...代表普通数字,如 0、5、-14、2, x 代表变量。

$\begin{aligned} a x^{n} + b x^{n - 1} + \dots + f x^{2} + g x + h \end{aligned}$
::axn+bxn- 1+...+fx2+gx+h

You have already learned many properties of polynomials. For example, you know the commutative property which states that terms of a polynomial can be rearranged to create an equivalent polynomial. When two polynomials are added, subtracted or multiplied the result is always a polynomial. This means polynomials are closed under addition , and is one of the properties that makes the factoring of polynomials possible. Polynomials are not closed under division because dividing two polynomials could result in a variable in the denominator, which is a rational expression (not a polynomial).
::您已经学会了多面性的许多特性。例如, 您知道有哪一种通量属性, 即多面性能可以重新排列, 以创建一个等效的多面性能。当添加两个多面性能时, 减去或乘以结果总是一个多面性能。这意味着多面性能在添加中是封闭的, 并且是使多面性能成为可能的参数的属性之一。多面性性能不是在分割下关闭的, 因为区分两个多面性能可能会在分母中产生变量, 这是一种理性表达( 不是多面性) 。

There are three methods for factoring that are essential to master.
::有三种保理方法对于掌握技术至关重要。

Greatest Common Factor Method
::最大共同系数法

The first method you should always try is to factor out the greatest common factor (GCF) of the expression.
::您应该尝试的第一种方法是将表达式的最大共同系数(GCF)计算出来。

To factor the following expression, first apply the GCF method:
::为了考虑到以下表述,首先采用全球合作框架方法:

$\begin{aligned} - \frac{1}{2} x^{4} + \frac{7}{2} x^{2} - 6 \end{aligned}$
::- 12x4+72x2-6

To find the GCF, it is common to try to factor out the $\begin{aligned} a \end{aligned}$ value. In this case, try factoring out $\begin{aligned} - \frac{1}{2} \end{aligned}$ .
::为了找到全球合作框架,通常会试图将价值考虑在内,在这种情况下,尝试将12作为因素来考虑。

$\begin{aligned} - \frac{1}{2} x^{4} + \frac{7}{2} x^{2} - 6 = - \frac{1}{2} (x^{4} - 7 x^{2} + 12) \end{aligned}$
::- 12x4+72x2-612(x4-7x2+12)

In order to check to see that this is an equivalent expression, you need to distribute the $\begin{aligned} - \frac{1}{2} \end{aligned}$ . When you distribute, the first coefficient matches because it just gets multiplied by 1, the second term becomes $\begin{aligned} \frac{7}{2} \end{aligned}$ and the third term becomes -6. Note that this expression is not completely factored yet but it is simplified as much as it can be with just the GCF method.
::为了检查是否这是一个相当的表达式, 您需要分配 - 12。当分配时, 第一个系数匹配, 因为它只是乘以1, 第二个系数匹配 72, 第三个系数匹配 - 6 。请注意, 这个表达式尚未完全计算出来, 但只要使用绿色气候基金的方法, 它就尽可能简化。

Factoring Into Binomials Method
::将系数乘以二元方法

The second method you should try is to see if you can factor the expression into the product of two binomials.
::您应该尝试的第二种方法是,看看您能否将表达式乘以两个二进制的产物。

To continue factoring the expression from the Greatest Common Factor section, factor the following expression into the product of two binomials and a constant:
::为了继续将 " 最大共同系数 " 部分的表述考虑在内,在两个二元制和一个常数的产物中考虑到以下表述:

$\begin{aligned} - \frac{1}{2} (x^{4} - 7 x^{2} + 12) \end{aligned}$
::- 12(x4-7x2+12)

Many students familiar with basic factoring may be initially stuck on a problem like this. However, you should recognize that beneath the $\begin{aligned} 4^{t h} \end{aligned}$ degree and the $\begin{aligned} - \frac{1}{2} \end{aligned}$ the problem boils down to being able to factor $\begin{aligned} u^{2} - 7 u + 12 \end{aligned}$ which is just $\begin{aligned} (u - 3) (u - 4) \end{aligned}$ .
::许多熟悉基本保理因素的学生最初可能陷入这样的问题。但是,你应该认识到,在第四学位和十二年级以下,问题归根结底就是能够将u2-7u+12作为(u-3)(u-4)因素。

Start by rewriting the problem: $\begin{aligned} - \frac{1}{2} (x^{4} - 7 x^{2} + 12) \end{aligned}$

::开始重写问题 : - 12( x4- 7x2+12)

Then choose a temporary substitution: Let $\begin{aligned} u = x^{2} \end{aligned}$ .
::然后选择一个临时替代:让 u=x2 。

Then substitute and factor away. Remember to substitute back at the end.
::然后换掉因子记得在最后换回来

$\begin{aligned} - \frac{1}{2} (u^{2} - 7 u + 12) & = - \frac{1}{2} (u - 3) (u - 4) \\ = - \frac{1}{2} (x^{2} - 3) (x^{2} - 4) \end{aligned}$

::-12(u2-7u+12) 12(u-3)(u-4) 12(x2-3)(x2-4)

This type of temporary substitution that enables you to see the underlying structure of an expression is very common in calculus. The expression is still not completely factored and since there are no more trinomials, you must apply the last method.
::这种类型的临时替代可以使您看到表达式的基本结构在微积分中非常常见。表达式仍然没有完全计算在内, 而且由于不再有三角关系, 您必须使用最后一种方法。

Difference of Squares Method
::平方法差异

The third method of basic factoring is the difference of squares . It is recognizable as one square monomial being subtracted from another square monomial.
::第三种基本保理法是平方之差,从另一平面单一保理法中减去一个平方单一保理法。

To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant:
::要完成乘数乘以分义部分所产生的表达式的乘数,请将表达式乘以四个线性系数和一个常数:

$\begin{aligned} - \frac{1}{2} (x^{2} - 3) (x^{2} - 4) \end{aligned}$
::- 12(x2-3)(x2-4)

Many students may recognize that $\begin{aligned} x^{2} - 4 \end{aligned}$ immediately factors by the difference of squares method to be $\begin{aligned} (x - 2) (x + 2) \end{aligned}$ . This problem asks for more because sometimes the difference of squares method can be applied to expressions like $\begin{aligned} x^{2} - 3 \end{aligned}$ where each term is not a perfect square . The number 3 actually is a square.
::许多学生可能认识到,由于平方法的差异, x2-4 立即成为(x-2)(x+2),这个问题要求更多,因为有时,平方法的差异可以适用于 x2-3 等表达式,其中每个术语不是完美的正方形,而数字3实际上是正方形。

$\begin{aligned} 3 = {(\sqrt{3})}^{2} \end{aligned}$

So the fully factored expression would be:
::因此,充分考虑的表述是:

$\begin{aligned} - \frac{1}{2} (x - \sqrt{3}) (x + \sqrt{3}) (x - 2) (x + 2) \end{aligned}$
::-12(xx-3)(x+3)(x-2)(x+2)(x+2)

Examples
::实例

Example 1
::例1

Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don'at contain perfect squares and why it would be useful. One reason why it might be useful to completely factor an expression like $\begin{aligned} - \frac{1}{2} (x^{4} - 7 x^{2} + 12) \end{aligned}$ into linear factors is if you wanted to find the roots of the function $\begin{aligned} f (x) = - \frac{1}{2} (x^{4} - 7 x^{2} + 12) \end{aligned}$ . The roots are $\begin{aligned} x = \pm \sqrt{3}, \pm 2 \end{aligned}$ .
::早些时候, 有人问您如何在包含完全正方形的多面体中使用完美的正方乘数技术的差别, 以及为什么它有用。将- 12( x4- 7x2+12) 等表达式完全纳入线性系数可能是有用的, 原因之一是如果您想要找到函数 f( x)\\\\ 12( x4- 7x2+12) 的根部, 根部是 x\3, 2 。原因之一。

You should recognize that $\begin{aligned} x^{2} - 3 \end{aligned}$ can still be thought of as the difference of perfect squares because the number 3 can be expressed as $\begin{aligned} {(\sqrt{3})}^{2} \end{aligned}$ . Rewriting the number 3 to fit a factoring pattern that you already know is an example of using the basic factoring techniques at a PreCalculus level.
::您应该认识到, x2-3 仍然可以被视为完美方形的区别, 因为3 可以用 (3)2 表示。重写数字 3 以适合您已经知道的乘数模式, 就是在预考水平上使用基本计数技术的例子。

Example 2
::例2

Factor the following expression into strictly linear factors if possible. If not possible, explain why.
::如果可能的话,将后面的表达方式乘以严格的线性因素。如果不可能,请解释原因。

\begin{aligned} \frac{x^{5}}{3} - \frac{11 x^{3}}{3} + 6 x = & \frac{1}{3} x (x^{4} - 11 x^{2} + 18) \\ = & \frac{1}{3} x (x^{2} - 2) (x^{2} - 9) \\ = & \frac{1}{3} x (x + \sqrt{2}) (x - \sqrt{2}) (x + 3) (x - 3) \end{aligned}

Example 3
::x53- 11xx33+6x=13x(x4- 11x2+18)=13x(x2-2)(x2- 9)=13x(x+2)(x-2)(x-2)(x-2)(x- 2)(x+3)(x-3)(x-3-3)例3

$\begin{aligned} - \frac{2}{7} x^{4} + \frac{74}{63} x^{2} - \frac{8}{63} \end{aligned}$
::- 27x4+7463x2-863

For $\begin{aligned} - \frac{2}{7} x^{4} + \frac{74}{63} x^{2} - \frac{8}{63} \end{aligned}$ , let $\begin{aligned} u = x^{2} \end{aligned}$ .
::对于- 27x4+7463x2- 863, let u=x2

\begin{aligned} = - \frac{2}{7} u^{2} + \frac{74}{63} u - \frac{8}{63} \\ = - \frac{2}{7} (u^{2} - \frac{37}{9} u + \frac{4}{9}) \end{aligned}

::* 27u2+7463u-86327(u2-379u+49)

Factoring through fractions like this can be extremely tricky. You must recognize that $\begin{aligned} - \frac{1}{9} \end{aligned}$ and -4 sum to $\begin{aligned} - \frac{37}{9} \end{aligned}$ and multiply to $\begin{aligned} \frac{4}{9} \end{aligned}$ .
::将这种分数乘以分数可能会非常困难。您必须认识到 - 19 和 - 4 和 - 379 和乘以 49 。

\begin{aligned} = - \frac{2}{7} (u - \frac{1}{9}) (u - 4) \\ = - \frac{2}{7} (x^{2} - \frac{1}{9}) (x^{2} - 4) \\ = - \frac{2}{7} (x - \frac{1}{3}) (x + \frac{1}{3}) (x - 2) (x + 2) \end{aligned}

::27(u-19)(u-4) 27(x2-19)(x2-4) 27(x-13)(x+13)(x-2)(x+2)

Example 4
::例4

$\begin{aligned} x^{4} + x^{2} - 72 \end{aligned}$
::x4+x2 - 72

$\begin{aligned} x^{4} + x^{2} - 72 \end{aligned}$ $\begin{aligned} = (x^{2} - 8) (x^{2} + 9) \end{aligned}$
::x4+x2-72=(x2-8)(x2+9)

Notice that $\begin{aligned} (x^{2} - 8) \end{aligned}$ can be written as the difference of perfect squares because $\begin{aligned} 8 = {(\sqrt{8})}^{2} = {(2 \sqrt{2})}^{2} \end{aligned}$ . On the other hand, $\begin{aligned} x^{2} + 9 \end{aligned}$ cannot be written as the difference between squares because the $\begin{aligned} x^{2} \end{aligned}$ and the 9 are being added not subtracted. This polynomial cannot be factored into strictly linear factors.
::注意 (x2- 8) 可以写成正方形的差, 因为 8=( 8) 2=( 222) 。另一方面, x2+9 不能写成正方形的差, 因为 x2 和 9 被添加时没有减去。此多数值不能作为纯线性因数来计算。

$\begin{aligned} x^{4} + x^{2} - 72 \end{aligned}$ $\begin{aligned} = (x - 2 \sqrt{2}) (x + 2 \sqrt{2}) (x^{2} + 9) \end{aligned}$
::x4+x2-72=(x-22)(x+22)(x2+9)

Summary

Factoring means to write an expression as a product instead of a sum, which is useful when solving equations set equal to zero.
::乘法是指将表达式写成产品而不是总和,这对于解决设定为零的方程式很有用。

A Polynomial is a sum of a finite number of terms.
::一夫多妻制是有限任期的总和。

Three methods for factoring are the Greatest Common Factor (GCF) method, Factoring Into Binomials method, and Difference of Squares method.
::保理的三种方法有:最大共同因数法、二进制保理法和广场差异法。

Review
::回顾

Factor each polynomial into strictly linear factors if possible. If not possible, explain why not.
::如果可能,将每个多数值乘以严格的线性系数。如果不可能,请解释为什么不能。

$\begin{aligned} x^{2} + 5 x + 6 \end{aligned}$
::x2+5x+6

$\begin{aligned} x^{4} + 5 x^{2} + 6 \end{aligned}$
::x4+5x2+6

$\begin{aligned} x^{4} - 16 \end{aligned}$
::x4 - 16

$\begin{aligned} 2 x^{2} - 20 \end{aligned}$
::2x2-20

$\begin{aligned} 3 x^{2} + 9 x + 6 \end{aligned}$
::3x2+9x+6

$\begin{aligned} \frac{x^{4}}{2} - 5 x^{2} + \frac{9}{2} \end{aligned}$
::x42-55x2+92

$\begin{aligned} \frac{2 x^{4}}{3} - \frac{34 x^{2}}{3} + \frac{32}{3} \end{aligned}$
::2x43-34x23+323

$\begin{aligned} x^{2} - \frac{1}{4} \end{aligned}$
::x2-14

$\begin{aligned} x^{4} - \frac{37 x^{2}}{4} + \frac{9}{4} \end{aligned}$
::x4-37x24+94

$\begin{aligned} \frac{3}{4} x^{4} - \frac{87}{4} x^{2} + 75 \end{aligned}$
::34x4-874x2+75

$\begin{aligned} \frac{1}{2} x^{4} - \frac{29}{2} x^{2} + 50 \end{aligned}$
::12x4-292x2+50

$\begin{aligned} \frac{x^{4}}{2} - \frac{5 x^{2}}{9} + \frac{1}{18} \end{aligned}$
::x42-55x29+118

$\begin{aligned} x^{4} - \frac{13}{36} x^{2} + \frac{1}{36} \end{aligned}$
::x4 - 1336x2+136

How does the degree of a polynomial relate to the number of linear factors?
::多元程度与线性系数的数目有何关系?

If a polynomial does not have strictly linear factors, what does this imply about the type of roots that the polynomial has?
::如果一个多民族体没有严格的线性因素,那么这对多民族体的根种意味着什么?

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

保保权审查

Factoring Functions ::保分函数

Greatest Common Factor Method ::最大共同系数法

Factoring Into Binomials Method ::将系数乘以二元方法

Difference of Squares Method ::平方法差异

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)