多边分会
Section outline
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Division of Polynomials
::多边分会A rational expression is formed by taking the quotient of two .
::理性表达方式由两个商数组成。Some examples of rational expressions are
::合理表达的一些例子
::2x2-14x2-3x+42x9x2+4x5x2+5x5x-12x32x+3Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator . In special cases we can simplify a rational expression by dividing the numerator by the denominator.
::和理性数字一样,顶部的表达式称为分子,底部的表达式称为分母。在特殊情况下,我们可以通过将分子除以分母来简化理性表达式。Divide a Polynomial by a Monomial
::将多面体除以单面体We’ll start by dividing a polynomial by a monomial . To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.
::我们首先将一个多名制用一个单名制来划分。 为此,我们将多名制的每个术语用一个单名制来划分。 当分子有多个术语时, 分数底部的单名制作为分子中所有术语的共同分母。Divide:
::除数 :a)
::a) 8x2-4x+162
::8x2-4x+162=8x22-4x2+162=4x2-2x+8b)
:b) 3x2+6x-1x
::3x3+6x-1x=3x3x3xx6x-1x3x6x1x3x2+6-1xc)
:c)-3x2-18x+69x
::- 3x2 - 18x+69x3x29x-18x9x+69x_x3 - 2+23xA common error is to cancel the denominator with just one term in the numerator.
::常见的错误是在分子中仅用一个词来取消分母。Consider the quotient .
::考虑商数 3x+44。Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.
::记住, 4 的分母对于分子中的这两个词都是共同的。 换句话说, 我们将分子中的这两个词除以数字 4 。The correct way to simplify is:
::简化的正确方式是:
::3x+44=3x4+44=3x4+1A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just . This is incorrect, because the entire numerator needs to be divided by 4.
::一个常见的错误是从分子和分母中划出数字4,只留下3x。这不正确,因为整个分子需要除以4。Dividing Polynomials by Monomials
::以荣誉形式分裂的多面体Divide .
::除以 5x3-10x2+x-25-5x2。
::5x3-10x2+x-25-5x2=5x3-5x2-10x2-5x2+x-5x2-2-25-5x2The negative sign in the denominator changes all the signs of the fractions:
::分母中的负符号会改变分数的所有符号 :
::-5x35x2+10x25x2_x5x2+255x2+255x2_x+2-15x+5x2Divide a Polynomial by a Binomial
::以二进制除以多式We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.
::我们用一种方法来划分多种族,这个方法与数字的长分法非常相似。 我们将举例来解释这个方法。Divide :
::除以 x2+4x+5x+3 :When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor .
::当我们执行分数时,分子中的表达方式被称为红利,分母中的表达方式称为分数。To start the division we rewrite the problem in the following form:
::为开始划分,我们以以下形式重写问题:
::x3+3x2+4x+5We start by dividing the first term in the dividend by the first term in the divisor: .
::我们首先将红利的第一个学期除以第一个学期,在divisor: x2x=x。We place the answer on the line above the term:
::我们把答案放在x术语上方的行上:
::x+3x2+4x+5xxxNext, we multiply the term in the answer by the divisor, , and place the result under the dividend, matching like terms . times is , so we put that under the divisor:
::接下来,在回答中将x值乘以 dvisor, x+3, 然后将结果置于红利之下, 匹配类似条件 。 x x x x 乘以 x2+3 乘以 x2+3x, 因此我们把它放在 divisor 下 :
::x3+3xx2+4x+5xxxx2+3xNow we subtract from . It is useful to change the signs of the terms of to and add like terms vertically :
::现在,我们从 x2+4x+5 中减去 x2+3x。 将 x2+3x 术语的符号改为 − x2 - 3x 并垂直添加类似术语 :
::x3+3xx2+4x+5xxxxx2 -3x_xNow, we bring down the 5, the next term in the dividend.
::现在,我们把红利的下一个任期,即5年降为5年。
::x+3x2+4x+5xxx -x2 -3x_ x+5And now we go through that procedure once more. First we divide the first term of by the first term of the divisor. divided by is 1, so we place this answer on the line above the constant term of the dividend:
::现在,我们再次经历这个程序。 首先,我们把第一个X+5任期除以第一个任期。 x除以x为一,所以我们把这个答案放在股息的固定期限之上:
::x3+3x2+4x+5x + 1 -x2 - 3x_ x5Multiply 1 by the divisor, , and write the answer below , matching like terms.
::乘以 divisor x+3, x+3, 并在 x+5 下写答案, 类似条件 。
::x3+3xx2+4x+5x + + 1 -x2 - 3x_ x+5 x+3Subtract from by changing the signs of to and adding like terms:
::将x+3的符号改为-x-3, 并添加类似条件, 从 x+5 中减去 x+3 :
::x+3x2+4x+5x + 1 -x2 - 3x_ x+5 - x3_ 2Since there are no more terms from the dividend to bring down, we are done. The quotient is and the remainder is 2.
::由于股息不再有任何条件可以降息,我们就完成了。 商数是 x+1, 其余为 2 。Remember that for a division with a remainder the answer is . So the answer to this division problem is .
::记住, 对于含有剩余部分的分区, 答案是 商数+remaindriddivisor 。 因此, 此分区问题的答案是 x2+4x+5x+3=x+1+2x+3 。Check
::支票支票支票支票To check the answer to a long division problem we use the fact that
::为了检查一个长期分裂问题的答案 我们用这个事实
:divisorx方位) +remainder=divid +remainder=divid (divisorx方位) +remainder=divid (divisorx方位) +remainder=divid =divid (divisorx方位) +remainder=divid (division) +remainder=d =dividd (divisorx方位) +remainder=d (divisorx方位) +remainder=d
For the problem above, here’s how we apply that fact to check our solution:
::对于上述问题, 以下是我们如何运用这个事实来检查我们的解决办法:
:x+3 (x+1)+2=x2+4x+3+2=x2+4x+5)
The answer checks out.
::答案检查出来。Example
::示例示例示例示例Example 1
::例1Divide .
::除以 x2+8x+17x+4。When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor .
::当我们执行分数时,分子中的表达方式被称为红利,分母中的表达方式称为分数。To start the division we rewrite the problem in the following form:
::为开始划分,我们以以下形式重写问题:
::x+4)x2+8x+17We start by dividing the first term in the dividend by the first term in the divisor: .
::我们首先将红利的第一个学期除以第一个学期,在divisor: x2x=x。We place the answer on the line above the term:
::我们把答案放在x术语上方的行上:
::x+4)x2+8x17xxNext, we multiply the term in the answer by the divisor, , and place the result under the dividend, matching like terms. times is , so we put that under the divisor:
::接下来,在回答中将x值乘以 dvisor, x+4, 然后将结果置于红利之下, 匹配类似条件 。 x x x x x x+4 乘以 x2+4x, 因此我们把它放在 divisor 下 :
::x+4)x2+8x+17xxxx2+4xNow we subtract from . It is useful to change the signs of the terms of to and add like terms vertically:
::现在我们从 x2+8x+17 中减去 x2+4x。 将 x2+4x 条件的符号更改为 − x2 - 4x 并垂直添加类似术语 :
::x+4)x2+8x+17xxxxx2-4x_4x_4xNow, we bring down the 17, the next term in the dividend.
::现在,我们降低17, 下个任期的红利。
::x+4) x2+8x+17xxxxx2-4x_4x_4x+17And now we go through that procedure once more. First we divide the first term of by the first term of the divisor. divided by is 4, so we place this answer on the line above the constant term of the dividend:
::现在,我们再次通过这个程序。首先,我们把4x+17的第一个任期除以第一个任期。4x除以x是4,所以我们把这个答案放在红利的固定期限之上:
::x+4) x2+8x+17x + 4-x2 - 4x_ x+17Multiply 4 by the divisor, , and write the answer below , matching like terms.
::乘以 4, x+4, 并在 4x+16 下写答案, 类似条件 。
::x+4) x2+8x+17x + 4-x2-4x_4x_4x+17 4x+16Subtract from by changing the signs of to and adding like terms:
::将4x+16从4x+17减为4x+17,将4x+16的符号改为-4x+16,并添加类似术语:
::x+4) x2+8x+17x + 4 - x2 - 4x_ x+17 - 4x - 16_ 1Since there are no more terms from the dividend to bring down, we are done. The quotient is and the remainder is 1.
::股息不再有任何条件可以减少,我们就完成了。商数为x+4,其余为1。Remember that for a division with a remainder the answer is . So the answer to this division problem is .
::记住, 对于含有剩余部分的分区, 答案是 商数+remaindriddivisor 。 因此, 此分区问题的答案是 x2+8x+17x+4=x+4+1x+4 。Check
::支票支票支票支票To check the answer to a long division problem we use the fact that
::为了检查一个长期分裂问题的答案 我们用这个事实
:divisorx方位) +remainder=divid +remainder=divid (divisorx方位) +remainder=divid (divisorx方位) +remainder=divid =divid (divisorx方位) +remainder=divid (division) +remainder=d =dividd (divisorx方位) +remainder=d (divisorx方位) +remainder=d
For the problem above, here’s how we apply that fact to check our solution:
::对于上述问题, 以下是我们如何运用这个事实来检查我们的解决办法:
:x+4)(x+4)+1=x2+8x+16+1=x2+8x+17)
The answer checks out.
::答案检查出来。Review
::回顾Divide the following polynomials:
::将下列多数值除以:-
::2x+42 -
::x - 4x -
::5x-355x -
::x2+2x-5xx -
::4x2+12x-36-4x -
::2x2+10x+72x2 -
::x3 - x - 2x2 -
::5x4-93x -
::x3 - 12x2+3x- 412x2 -
::3-6x+x3-9x3 3-6x+x3-9x3 -
::x2+3x+6x+1 -
::x2- 9x+6x- 1 -
::x2+5x+4x+4 -
::x2 - 10x+25x-5 -
::x2-20x+12x-3 -
::3x2-xx+5x-2 -
::9x2+2x-8x+4 -
::3x2 - 43x+1 -
::5x2+2x-92x-1 -
::x2 - 6x - 125x4
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 。 -