Course: 6.9 长期多民族分会

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长长的多配偶分会
The area of a rectangle is $\begin{align*}6x^3 - 12x^2 + 4x - 8\end{align*}$ . The width of the rectangle is $\begin{align*}2x - 4\end{align*}$ . What is the length?
::矩形区域为 6x3 - 12x2+4x- 8. 矩形宽度为 2x-4。长度是多少?

Long Division of Polynomials
::长长的多配偶分会

Even though it does not seem like it, factoring is a form of division . Each factor goes into the larger polynomial evenly, without a remainder.
::尽管似乎并不像,但保理是一种划分形式。每一个因素均匀地进入更大的多元性,没有剩余部分。

For example, take the polynomial $\begin{align*}2x^3-3x^2-8x+12\end{align*}$ . If we use factoring by grouping , we find that the factors are $\begin{align*}(2x - 3)(x - 2)(x + 2)\end{align*}$ . If we multiply these three factors together, we will get the original polynomial. So, if we divide by $\begin{align*}2x - 3\end{align*}$ , we should get $\begin{align*}x^2 - 4.\end{align*}$
::例如,采用多数值 2x3-3x2- 3x2-8x+12。如果我们按分组使用乘数,我们发现系数是 2x-3 (x-2)(x+2) 。如果我们将这三个系数乘在一起,我们就会得到原始的多数值。所以,如果我们除以 2x-3,我们就应该得到 x2- 4 。

$\begin{aligned} 2 x - 3 2 x^{3} - 3 x^{2} - 8 x + 12 \end{aligned}$
$\begin{align*}\require{enclose} 2x-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12} \end{align*}$
::2 - 32x3 - 3x2 - 8x+12

How many times does $\begin{align*}2x\end{align*}$ go into $\begin{align*}2x^3?\end{align*}$ Since $\begin{align*}2x(x^2)=2x^3,\end{align*}$ it goes in $\begin{align*}x^2\end{align*}$ times.
::自 2x( x2) = 2x3 以x2 乘以。

$\begin{aligned} x^{2} \\ 2 x - 3 2 x^{3} - 3 x^{2} - 8 x + 12 \\ \underline{2 x^{3} - 3 x^{2}} \\ 0 \end{aligned}$
$\begin{align*}\require{enclose} {x}^{2}\phantom{00000000000000}\\ 2x-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12}\\ \underline{2{x}^{3}-3{x}^{2}}\phantom{000000000}\\ 0\phantom{0000000000} \end{align*}$
::x2000000000002x - 32x3 - 3x2 - 8x+122x3 - 3x2 - 00000000000000000

Place $\begin{align*}x^2\end{align*}$ above the $\begin{align*}x^2\end{align*}$ term in the polynomial.
::将 x2 移动到多义中的 x2 术语之上。

Multiply $\begin{align*}x^2\end{align*}$ by both terms in the divisor ( $\begin{align*}2x\end{align*}$ and -3) and place them under their like terms . Subtract from the dividend $\begin{align*}(2x^3-3x^2-8x+12).\end{align*}$ Pull down the next two terms and repeat.
::乘以乘数2, 在 divisor ( 2x 和 3) 中用两个词来表示, 并将其置于相同的条件之下。从股息( 2x3-3x2 - 8x2 - 8x+12) 中扣除。

$\begin{aligned} x^{2} - 4 \\ 2 x - 3 2 x^{3} - 3 x^{2} - 8 x + 12 \\ \underline{2 x^{3} - 3 x^{2}} \\ - 8 x + 12 \\ \underline{- 8 x + 12} \end{aligned}$
$\begin{align*}\require{enclose} {x}^{2}\phantom{00000}-4\phantom{00000}\\ {\color{red}{2x}}-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12}\\ \underline{2{x}^{3}-3{x}^{2}\quad\quad\quad\phantom{00}}\\ {\color{red}-8x}+12\\ \underline{-8x+12}\end{align*}$
::x200000-4000002x-32-3x3-3x2-8x+122x3-3x200__BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_BAR_三三三三三三三三三三三x_三三三三三三三三x_三x_三x_三x_一__八_

Since $\begin{align*}2x(\text{-}4)=\text{-}8x,\end{align*}$ $\begin{align*}2x\end{align*}$ goes into $\begin{align*}-8x\end{align*}$ a total of -4 times.
::自2x( 4) = 8x 2x 进入- 8x 共 - 4 次。

After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtracting, notice that everything cancels out. Therefore $\begin{align*}x^2 - 4\end{align*}$ is indeed a factor.
::在将两个条件乘以 4 后, 将两个条件乘以 4 后, 将您下调的条件置于您下调的条件之下。减去时, 通知所有条件都取消。因此 x2 - 4 确实是一个因素。

When , not every divisor will go in evenly to the dividend. If there is a remainder, write it as a fraction over the divisor.
::当不是每个小提琴家都能平均进入股息时。如果还有剩余部分,请将其写成比小的分数在小提琴上方。

Divide using long division: $\begin{align*}(2x^3-6x^2+5x-20) \div (x^2-5)\end{align*}$
::使用长除法除法的除法 : (2x3- 6x2+5x- 20) (x2- 5)

Set up the problem using a long division bar.
::使用长分隔栏设置问题。

$\begin{aligned} x^{2} - 5 2 x^{3} - 6 x^{2} + 5 x - 20 \end{aligned}$
$\begin{align*}\require{enclose} {x}^{2}-5\enclose{longdiv}{2{x}^{3}-6{x}^{2}+5x-20}\end{align*}$
::x2 - 52x3 - 6x2+5x-20

How many times does $\begin{align*}x^2\end{align*}$ go into $\begin{align*}2x^3?\end{align*}$ Since $\begin{align*}x^2(2x)=2x^3,\end{align*}$ it goes in $\begin{align*}2x\end{align*}$ times.
::x2 进入 2x3 多少次? 从 x2( 2x) = 2x3 开始, 它以 2x 倍。

$\begin{aligned} 2 x \\ x^{2} - 5 2 x^{3} - 6 x^{2} + 5 x - 20 \\ \underline{2 x^{3} - 10 x^{2}} \\ 4 x^{2} + 5 x - 20 \end{aligned}$
$\begin{align*}\require{enclose} \color{red}2x\phantom{00000000000000}\\ {\color{red}{x}^{2}}-5\enclose{longdiv}{{\color{red}2{x}^{3}}-6{x}^{2}+5x-20}\\ \underline{2{x}^{3}-10{x}^{2}\quad\quad\quad\phantom{0}}\\ 4{x}^{2}+5x-20\\\end{align*}$
::2x000000000x2 - 52x3 - 6x2+5x - 202x3 - 10x20 - 4x2+5x-20

Multiply $\begin{align*}2x\end{align*}$ by the divisor. Subtract that from the dividend.
::乘以乘以二乘法减去股息

Repeat the previous steps. Now, how many times does $\begin{align*}x^2\end{align*}$ go into $\begin{align*}4x^2\end{align*}$ ? It goes in 4 times.
::重复前几步。现在, x2 进入 4x2 多少次? 它使用 4 次。

$\begin{aligned} 2 x + 4 \\ x^{2} - 5 2 x^{3} - 6 x^{2} + 5 x - 20 \\ \underline{2 x^{3} - 10 x^{2}} \\ 4 x^{2} + 5 x - 20 \\ \underline{4 x^{2} - 20} \\ 5 x \end{aligned}$
$\begin{align*}\require{enclose} 2x\phantom{00}\color{red}+4\phantom{000000000}\\ {\color{red}{x}^{2}}-5\enclose{longdiv}{2{x}^{3}-6{x}^{2}+5x-20}\\ \underline{2{x}^{3}-10{x}^{2}\quad\quad\quad\phantom{0}}\\ {\color{red}4{x}^{2}}+5x-20\\ \underline{4{x}^{2}\phantom{0000}-20}\\ 5x\quad\phantom{00}\end{align*}$
::2x00+400000000x2-52x3-6x2+5x-202+5x20x3-10x20_4x2+5x-204x20000-20_5x00

This is the limit of this process . $\begin{align*}x^2\end{align*}$ cannot go evenly into $\begin{align*}5x\end{align*}$ because it has a higher degree . Therefore , $\begin{align*}5x\end{align*}$ is a remainder. The complete answer would be $\begin{align*}2x+4+\frac{5x}{x^2-5}.\end{align*}$
::这是这个过程的极限。 x2 无法平均进入 5x, 因为它具有更高的学位。因此, 5x 是剩余部分。完整答案是 2x+4+5x2-5 。

Divide using long division: $\begin{align*}(3x^4+x^3-17x^2+19x-6) \div (x^2-2x+1)\end{align*}$
::使用长除法除法除法 : (3x4+x3- 17x2+19x- 6) (x2-2x+1)

Determine if $\begin{align*}x^2-2x+1\end{align*}$ goes evenly into $\begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}$ . If so, try to factor the divisor and quotient further.
::确定 x2 - 2x+1 是否均匀地切入 3x4+x3 - 17x2+19x-6。如果是的话, 请尝试进一步乘以差数和商数。

First, do the long division. If $\begin{align*}x^2-2x+1\end{align*}$ goes in evenly, then the remainder will be zero.
::首先, 执行长的分隔。如果 x2-2x+1 平均进入, 那么剩下的将是零。

$\begin{aligned} 3 x^{2} + 7 x - 6 \\ x^{2} - 2 x + 1 3 x^{4} + x^{3} - 17 x^{2} + 19 x - 6 \\ \underline{3 x^{4} - 6 x^{3} 3 x^{2}} \\ 7 x^{3} - 20 x^{2} + 19 x \\ \underline{7 x^{3} - 14 x^{2} + 7 x} \\ - 6 x^{2} + 12 x - 6 \\ \underline{- 6 x^{2} + 12 x - 6} \\ 0 \end{aligned}$
$\begin{align*}\require{enclose} 3{x}^{2}+7x\phantom{000}-6\phantom{000000000}\\ {x}^{2}-2x+1\enclose{longdiv}{3{x}^{4}+{x}^{3}-17{x}^{2}+19x-6}\\ \underline{3{x}^{4}-6{x}^{3}\phantom{00}3{x}^{2}\phantom{000000000}}\\ 7{x}^{3}-20{x}^{2}+19x\phantom{000}\\ \underline{7{x}^{3}-14{x}^{2}+\phantom{0}7x\phantom{000}}\\ -6{x}^{2}+12x-6\\ \underline{-6{x}^{2}+12x-6}\\ 0\end{align*}$
::3x2+7x000 - 6000000000000x2 - 60000000x2 - 60000000000x2 - 2x+13x4+x3-17x2+19x63x4 - 6x2 - 6x3003x2000000 - 7x3 - 20x3 - 7x3 - 20x2+19x0007x3 - 14x2+07x000*_6x2+12x-6-6x2+12x6_0

This means that $\begin{align*}x^2-2x+1\end{align*}$ and $\begin{align*}3x^2+7x-6\end{align*}$ both go evenly into $\begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}$ . Let’s see if we can factor either $\begin{align*}x^2-2x+1\end{align*}$ or $\begin{align*}3x^2+7x-6\end{align*}$ further.
::这意味着 x2 - 2x+1 和 3x2+7x-6 均匀地进入 3x4+x3 - 17x2+19x-6 。让我们看看能不能进一步乘以 x2 - 2x+1 或 3x2+7x-6 。

$\begin{align*}x^2-2x+1=(x-1)(x-1)\end{align*}$ and $\begin{align*}3x^2+7x-6=(3x-2)(x+3)\end{align*}$ .
::x2-2x+1=(x-1)(x-1)和3x2+7x-6=(3x-2)(x+3)。

Therefore, $\begin{align*}3x^4+x^3-17x^2+19x-6=(x-1)(x-1)(x+3)(3x-2).\end{align*}$ You can multiply these to check the work. A binomial with a degree of one is a factor of a larger polynomial $\begin{align*}f(x),\end{align*}$ if it goes evenly into it. In this problem , $\begin{align*}(x-1)(x-1)(x+3)\end{align*}$ and $\begin{align*}(3x-2)\end{align*}$ are all factors of $\begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}$ . This indicates that 1, 1, -3, and $\begin{align*}\frac{2}{3}\end{align*}$ are all solutions of $\begin{align*}3x^4+x^3-17x^2+19x-6.\end{align*}$
::因此, 3x4+x3- 17x2+19x-6=(x-1)(x-1)(x-1)(x- 1)(x+3)(3x-2) 。您可以乘以这些来检查工作。具有一度的二进制是较大多圆性f(x) 的一个系数, 如果它处于均匀状态的话。在此问题上, (x-1)(x-1)(x-1)(x+3)(x+3) 和(3x-2) 3x4+x3- 17x2+19x-6) 的所有系数都是 3x4+x3- 172+19x-6。这表明, 1, 1, 3, 3 +x3- 3- 17x2+19x-6 的所有解决方案都是 3x4+x3- 17x2+19x-6 。

Factor Theorem : A polynomial, $\begin{align*}f(x)\end{align*}$ , has a factor, $\begin{align*}(x - k)\end{align*}$ , if and only if $\begin{align*}f(k) = 0\end{align*}$ .
::系数定理: 多式, f(x), 有系数, (x-k) , 如果且仅在 f(k) =0 的情况下。

In other words, if $\begin{align*}k\end{align*}$ is a solution or a zero , then the factor, $\begin{align*}(x - k)\end{align*}$ divides evenly into $\begin{align*}f(x)\end{align*}$ .
::换句话说,如果 k是溶液或零,则系数(x-k)平均分为f(x)。

Now, let's determine if 5 is a solution of $\begin{align*}x^3+6x^2-8x+15\end{align*}$ .
::现在,让我们来决定 5 是否是 x3+6x2 - 8x+15 的解决方案。

To see if 5 is a solution, we need to divide the factor into $\begin{align*}x^3+6x^2-8x+15\end{align*}$ . The factor that corresponds with 5 is $\begin{align*}(x - 5)\end{align*}$ .
::要想确定 5 是否是一个解决方案,我们需要将系数除以 x3+6x2-8x+15。对应于 5 的系数是 (x-5) 。

$\begin{aligned} x^{2} + 11 x + 5 \\ x - 5 x^{3} + 6 x^{2} - 50 x + 15 \\ \underline{x^{3} - 5 x^{2}} \\ 11 x^{2} - 50 x \\ \underline{11 x^{2} - 55 x} \\ 5 x + 15 \\ \underline{5 x - 25} \\ 40 \end{aligned}$
$\begin{align*}\require{enclose} {x}^{2}+11x\phantom{0}+5\phantom{00000}\\ x-5\enclose{longdiv}{{x}^{3}+6{x}^{2}-50x+15}\\ \underline{{x}^{3}-\phantom{0}5{x}^{2}\phantom{000000000}}\\ 11{x}^{2}-\phantom{0}50x\phantom{0000}\\ \underline{11{x}^{2}-\phantom{0}55x\phantom{0000}}\\ 5x+15\\ \underline{5x-25}\\ 40\end{align*}$
::x2+11x0+500000x-5x3+6x2-50x+15x3-05x2000000_11x2-050x000011x2-055x0000_5x+155x-25_40

Since there is a remainder, 5 is not a solution.
::由于还有剩余部分,5项不是解决办法。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the length of the rectangle.
::早些时候,有人要求你找到矩形的长度。

First, do the long division.
::首先,做长期的划分。

$\begin{aligned} 3 x^{2} + 2 \\ 2 x - 4 6 x^{3} - 12 x^{2} + 4 x - 8 \\ \underline{6 x^{3} - 12 x^{2}} \\ 4 x - 8 \\ \underline{4 x - 8} \\ 0 \end{aligned}$
$\begin{align*}\require{enclose} 3{x}^{2}\phantom{0000000}+2\phantom{0000}\\ 2x-4\enclose{longdiv}{6{x}^{3}-12{x}^{2}+4x-8}\\ \underline{6{x}^{3}-12{x}^{2}\quad\quad\quad\phantom{0}}\\ 4x-8\\ \underline{4x-8}\\ 0\end{align*}$
::3x2000000+200002x-46x3-12x2+4x-86x3-12x20_4x-84x-8_0

This means that $\begin{align*}2x - 4\end{align*}$ and $\begin{align*}3x^2+2\end{align*}$ both go evenly into $\begin{align*}6x^3-12x^2+4x-8\end{align*}$ .
::这意味着 2x-4 和 3x2+2 均匀地进入 6x3 - 12x2+4x-8。

$\begin{align*}3x^2+2\end{align*}$ can't be factored further, so it is the rectangle's length.
::3x2+2 无法进一步计算, 所以它是矩形的长度。

Example 2
::例2

Divide: $\begin{align*}(5x^4+6x^3-12x^2-3) \div (x^2+3)\end{align*}$ .
::除法: (5x4+6x3- 12x2-3) (x2+3)。

Make sure to put a placeholder in for the $\begin{align*}x-\end{align*}$ term.
::确保 x- term 设置一个占位符。

$\begin{aligned} 5 x^{2} + 6 x - 27 \\ x^{2} + 3 5 x^{4} + 6 x^{3} - 12 x^{2} + 0 x - 3 \\ \underline{5 x^{4} + 15 x^{2}} \\ 6 x^{3} - 27 x^{2} + 0 x \\ \underline{6 x^{3} + 18 x} \\ - 27 x^{2} - 18 x - 3 \\ \underline{- 27 x^{2} - 81} \\ - 18 x + 78 \end{aligned}$
$\begin{align*}\require{enclose} 5{x}^{2}+6x\phantom{00}-27\phantom{000000000}\\ {x}^{2}+3\enclose{longdiv}{5{x}^{4}+6{x}^{3}-12{x}^{2}+{\color{red}0x}-3}\\ \underline{5{x}^{4}\phantom{00000}+15{x}^{2}\phantom{00000000}}\\ 6{x}^{3}-27{x}^{2}+{\color{red}0x}\quad\phantom{0}\\ \underline{6{x}^{3}\quad\phantom{000}+18x\phantom{000}}\\ -27{x}^{2}-18x-3\\ \underline{-27{x}^{2}\quad\phantom{00}-81}\\ -18x+78\end{align*}$
::5x2+6x00- 27000000000x2+354x4+6x3- 12x2+0x3-6x3- 12x2+0x3- 4x6x4- 40000+15x2000000_ 6x3- 15x200000_ 6x3- 27x2+0x2+0x0x3000+18x000_ 272_ 18x_ 3- 27x_ 18x_ 81_18x+78

The final answer is $\begin{align*}5x^2+6x-27-\frac{18x-78}{x^2+3}\end{align*}$ .
::最后一个答案是5x2+6x-27-18x-78x2+3。

Example 3
::例3

Is $\begin{align*}(x+4)\end{align*}$ a factor of $\begin{align*}x^3-2x^2-51x-108\end{align*}$ ? If so, find any other factors.
:x+4) 是 x3-2x2-51x-108的因数吗?如果是的话,请找到任何其他因数。

Divide $\begin{align*}(x + 4)\end{align*}$ into $\begin{align*}x^3-2x^2-51x-108\end{align*}$ and if the remainder is zero, it is a factor.
::将(x+4)除以x3-2x2-51x-108,如果其余为零,则是一个系数。

$\begin{aligned} x^{2} - 6 x - 27 \\ x + 4 x^{3} - 2 x^{2} - 51 x - 108 \\ \underline{x^{3} + 4 x^{2}} \\ - 6 x^{2} - 51 x \\ \underline{- 6 x^{2} - 24 x} \\ - 27 x - 108 \\ \underline{- 27 x - 108} \\ 0 \end{aligned}$
$\begin{align*}\require{enclose} {x}^{2}-6x-27\phantom{0000000}\\ x+4\enclose{longdiv}{{x}^{3}-2{x}^{2}-51x-108}\\ \underline{{x}^{3}+4{x}^{2}\phantom{00000000000}}\\ -6{x}^{2}-51x\phantom{000000}\\ \underline{-6{x}^{2}-24x\phantom{000000}}\\ -27x-108\\ \underline{-27x-108}\\ 0\end{align*}$
::x2 - 6x - 270000000x+4x3 - 2x2 - 51x - 108x3+4x2000000000 *6x2 - 51x000000 - 6x2 - 24x000 *27x - 108 - 27x - 108_ 0

$\begin{align*}x + 4\end{align*}$ is a factor. Let’s see if $\begin{align*}x^2 - 6x -27\end{align*}$ factors further. Yes, the factors of -27 that add up to -6 are -9 and 3. Therefore, the factors of $\begin{align*}x^3-2x^2-51x-108\end{align*}$ are $\begin{align*}(x + 4), (x - 9)\end{align*}$ , and $\begin{align*}(x + 3)\end{align*}$ .
::x+4 是一个因素。让我们看看 x2 - 6x - 27 因素是否进一步。是的, 27 因素加到 - 6 是 - 9 和 3 。因此, x3 - 2x2 - 51x- 108 因素是 (x+4) 、 (x- 9) 和 (x+3) 。

Example 4
::例4

What are the real-number solutions to Example 3?
::例3的实际数字解决办法是什么?

The solutions would be -4, 9, and 3; the opposite sign of each factor.
::解决办法是4、9和3;每个因素的相反标志。

Example 5
::例5

Determine if 6 is a solution to $\begin{align*}2x^3-9x^2-12x-24\end{align*}$ .
::确定 6 是 2x3- 9x2- 12x- 24 的解决方案。

To see if 6 is a solution, we need to divide $\begin{align*}(x - 6)\end{align*}$ into $\begin{align*}2x^3-9x^2-12x-24\end{align*}$ .
::看看6是否是一种解决办法,我们需要将(x-6)分为2x3-9x2-12x-24。

$\begin{aligned} 2 x^{2} + 3 x + 6 \\ x - 6 2 x^{3} - 9 x^{2} - 12 x - 24 \\ \underline{2 x^{3} - 12 x^{2}} \\ 3 x^{2} - 12 x \\ \underline{3 x^{2} - 18 x} \\ 6 x - 24 \\ \underline{6 x - 36} \\ 12 \end{aligned}$
$\begin{align*}\require{enclose} 2{x}^{2}+3x\phantom{0}+6\phantom{0}\phantom{00000}\\ x-6\enclose{longdiv}{2{x}^{3}-9{x}^{2}-12x-24}\\ \underline{2{x}^{3}-12{x}^{2}\phantom{000000000}}\\ 3{x}^{2}-12x\phantom{0000}\\ \underline{3{x}^{2}-18x\phantom{0000}}\\ 6x-24\\ \underline{6x-36}\\ 12\end{align*}$
::2x2+3x0+600000x-62x3-9x2-12x-242x3-12x12x2000000_3x2-12x2-12x0003x2-12x0003x2-18x0003x2-18x00_6x-246x-36_12

Because the remainder is not zero, 6 is not a solution.
::因为其余部分不是零,6项不是解决办法。

Review
::回顾

Divide the following polynomials using long division.
::使用长除法除以下列多数值。

$\begin{align*}(2x^3+5x^2-7x-6) \div (x+1)\end{align*}$
:2x3+5x2-7x-6)(x+1)

$\begin{align*}(x^4-10x^3+15x-30) \div (x-5)\end{align*}$
:x4-10x3+15x-30) (x-5)

$\begin{align*}(2x^4-8x^3+4x^2-11x-1) \div (x^2-1)\end{align*}$
:2x4-8x3+4x2-11x-1)(x2-1)

$\begin{align*}(3x^3-4x^2+5x-2) \div (3x+2)\end{align*}$
:3x3-4x2+5x-2)(3x+2)

$\begin{align*}(3x^4-5x^3-21x^2-30x+8) \div (x-4)\end{align*}$
:3x4-5x3-21x2-30x+8)(x-4)

$\begin{align*}(2x^5-5x^3+6x^2-15x+20) \div (2x^2+3)\end{align*}$
:2x5-5x3+6x2-15x+20)(2x2+3)

Determine all the real-number solutions to the following polynomials, given one factor.
::确定以下多边数的所有真实数字解决方案, 给一个系数。

$\begin{align*}x^3-9x^2+27x-15; (x+5)\end{align*}$
::x3- 9x2+27x-15; (x+5)

$\begin{align*}x^3+4x^2-9x-36; (x+4)\end{align*}$
::x3+4x2-9x-36; (x+4)

$\begin{align*}2x^3+7x^2-7x-30; (x-2)\end{align*}$
::2x3+7x2-7x2-7x-30; (x-2)

Determine all the real number solutions to the following polynomials, given one zero.
::确定以下多边数的所有真实数字解决方案, 给以零。

$\begin{align*}6x^3-37x^2+5x+6; 6\end{align*}$
::6x3-3-37x2+5x6;6

$\begin{align*}6x^3-41x^2+58x-15; 5\end{align*}$
::6x3-41x2+58x-15;5

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

Find the equation of a polynomial with the given zeros.
::查找给定零的多元分子方程式的方程式。

4, -2, and $\begin{align*}\frac{3}{2}\end{align*}$
::4,2-2和32

1, 0, and 3
::1, 0, 和 3

-5, -1, and $\begin{align*} \frac{3}{4}\end{align*}$
::5、5-1和34

Challenge Find two polynomials with the zeros 8, 5, 1, and -1.
::挑战发现两个多面体零点8,5,1和1

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

长长的多配偶分会

Long Division of Polynomials ::长长的多配偶分会

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Long Division of Polynomials
::长长的多配偶分会

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)