Section: 刻度表的聚合系数乘数 | 6.8 将夸度表的聚合物乘数乘法

Section outline

The volume of a rectangular prism is $\begin{align*}10x^3 - 25x^2 - 15x\end{align*}$ . What are the lengths of the prism's sides?
::矩形棱晶体积为 10x3 - 25x2 - 15x。棱晶面的长度是多少?

Factoring Polynomials in Quadratic Form
::刻度表的聚合系数乘数

The last type of factorable polynomial are those that are in quadratic form . Quadratic form is when a polynomial looks like a trinomial or binomial and can be factored like a quadratic. One example is when a polynomial is in the form $\begin{align*}ax^4+bx^2+c\end{align*}$ . Another possibility is something similar to the difference of squares , $\begin{align*}a^4-b^4\end{align*}$ . This can be factored to $\begin{align*}(a^2-b^2)(a^2+b^2)\end{align*}$ or $\begin{align*}(a-b)(a+b)(a^2+b^2)\end{align*}$ . Always keep in mind that the greatest common factors should be factored out first.
::最后一个可考虑的多元性类型是四面形的。二次曲线形式是多面形看起来像三边形或二进制, 并且可以像四方形那样进行计算。一个例子是多面形为x4+bx2+c。另一种可能性与正方形( a4-b4) 的差别相似, 可以与(a2-b2)(a2+b2) 或(a-b)(a2+b)(a2+b2) 或(a-b)(a2) +b) 或(b)(a2+b) 。始终要记住, 最共同的因素应该首先被考虑。

1. Factor the polynomial: $\begin{align*}2x^4-x^2-15\end{align*}$
::1. 多数值系数: 2x4-x2-15

This particular polynomial is factorable. First, $\begin{align*}ac = -30\end{align*}$ . The factors of -30 that add up to -1 are -6 and 5. Expand the middle term and then use factoring by grouping .
::此特定的多数值是因数。首先, ac\\\\ 30。 - 30 的乘数加到 - 1, 是 - 6 和 5 。扩大中期, 然后通过分组使用乘数。

$\begin{aligned} 2 x^{4} - x^{2} - 15 \\ 2 x^{4} - 6 x^{2} + 5 x^{2} - 15 \\ 2 x^{2} (x^{2} - 3) + 5 (x^{2} - 3) \\ (x^{2} - 3) (2 x^{2} + 5) \end{aligned}$
$\begin{align*}& 2x^4-x^2-15\\ & 2x^4-6x^2+5x^2-15\\ & 2x^2(x^2-3)+5(x^2-3)\\ & (x^2-3)(2x^2+5)\end{align*}$
::2x4 - x2 - 152x4 - 6x4 - 6x2+5x2 - 152x2(x2 - 3)+5(x2 - 3)(x2 - 3)(x2 - 3)(2x2+5)

Both of the factors are not factorable, so we are done.
::这两个因素都是不可考虑的因素,因此我们这样做了。

2. Factor the polynomial: $\begin{align*}81x^4-16\end{align*}$
::2. 多数值系数: 81x4-16

Treat this polynomial equation like a difference of squares.
::将这个多面方程式作为方形的差数处理。

$\begin{aligned} 81 x^{4} - 16 \\ (9 x^{2} - 4) (9 x^{2} + 4) \end{aligned}$
$\begin{align*}& 81x^4-16\\ & (9x^2-4)(9x^2+4)\end{align*}$
::81x4-16(9x2-4)(9x2+4)

Now, we can factor $\begin{align*}9x^2-4\end{align*}$ using the difference of squares a second time.
::现在,我们可以第二次使用方形的差数来乘以9x2-4。

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

$\begin{align*}9x^2+4\end{align*}$ cannot be factored because it is a sum of squares. This will have imaginary solutions.
::9x2+4 无法计算, 因为它是方形之和。这将包含假想的解决方案。

Now, let's find all the real-number solutions of $\begin{align*}6x^5-51x^3-27x = 0\end{align*}$ .
::现在,让我们找到所有 6x5 -51x3 -27x=0的真实数字解决方案。

First, pull out the GCF among the three terms .
::首先,在三个条件中拿出绿色气候基金。

$\begin{aligned} 6 x^{5} - 51 x^{3} - 27 x & = 0 \\ 3 x (2 x^{4} - 17 x^{2} - 9) & = 0 \end{aligned}$
$\begin{align*}6x^5-51x^3-27x &= 0\\ 3x(2x^4-17x^2-9) &= 0\end{align*}$
::6x5 - 51x3 - 27x=03x(2x4 - 17x2 - 9)=0

Factor what is inside the parenthesis like a quadratic equation . $\begin{align*}ac = -18\end{align*}$ and the factors of -18 that add up to -17 are -18 and 1. Expand the middle term and then use factoring by grouping.
::括号内的系数是- 17。 ac\\\ 18 和 - 18 的系数是 - 17是 - 18 和 1. 扩大中期,然后通过分组使用乘数。

$\begin{aligned} 6 x^{5} - 51 x^{3} - 27 x & = 0 \\ 3 x (2 x^{4} - 17 x^{2} - 9) & = 0 \\ 3 x (2 x^{4} - 18 x^{2} + x^{2} - 9) & = 0 \\ 3 x [2 x^{2} (x^{2} - 9) + 1 (x^{2} - 9)] & = 0 \\ 3 x (x^{2} - 9) (2 x^{2} + 1) & = 0 \end{aligned}$
$\begin{align*}6x^5-51x^3-27x &= 0\\ 3x(2x^4-17x^2-9) &= 0\\ 3x(2x^4-18x^2+x^2-9) &= 0\\ 3x[2x^2(x^2-9)+1(x^2-9)] &= 0\\ 3x(x^2-9)(2x^2+1) &= 0\end{align*}$
::6x5-51-51x3-27x=03x(2x4-17x2-9)=03x(2x4-18x2+x2-9)=03x[2x2(x2-9)+1(x2-9)]=03x(x2-9)(2x2+1)=0

Factor $\begin{align*}x^2-9\end{align*}$ further and solve for $\begin{align*}x\end{align*}$ where possible. $\begin{align*}2x^2+1\end{align*}$ is not factorable.
::2x2+1 无法计算。

$\begin{aligned} 3 x (x^{2} - 9) (2 x^{2} + 1) & = 0 \\ 3 x (x - 3) (x + 3) (2 x^{2} + 1) & = 0 \\ x & = - 3, 0, 3 \end{aligned}$
$\begin{align*}3x(x^2-9)(2x^2+1) &= 0\\ 3x(x-3)(x+3)(2x^2+1) &= 0\\ x &= -3, 0, 3\end{align*}$
::3x(x2- 9) (2x2+1) =03x(x-3)(x+3)(2x2+1) =0x*3,0,3)

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the lengths of the prism's sides.
::早些时候,有人要求你找出棱镜两面的长度。

To find the lengths of the prism's sides, we need to factor $\begin{align*}10x^3 - 25x^2 - 15x\end{align*}$ .
::要找到棱镜侧的长度, 我们需要乘以 10x3 - 25x2 - 15x。

First, pull out the GCF among the three terms.
::首先,在三个条件中拿出绿色气候基金。

$\begin{aligned} 10 x^{3} - 25 x^{2} - 15 x \\ 5 x (2 x^{2} - 5 x - 3) \end{aligned}$
$\begin{align*}10x^3 - 25x^2 - 15x\\ 5x(2x^2 - 5x - 3)\end{align*}$
::10x3 - 25x2 - 15x5x(2x2 - 5x-3)

Factor what is inside the parenthesis like a quadratic equation. $\begin{align*}ac = -6\end{align*}$ and the factors of -6 that add up to -5 are -6 and 1.
::括号内的系数,如二次方程。 ac6, 加上 -5 的 -6 系数为 -6 和 1。

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

Therefore , the lengths of the rectangular prism's sides are $\begin{align*}5x\end{align*}$ , $\begin{align*}2x + 1\end{align*}$ , and $\begin{align*}x - 3\end{align*}$ .
::因此,矩形棱柱两侧的长度分别为5x、2x+1和x-3。

Example 2
::例2

Factor: $\begin{align*}3x^4+14x^2+8\end{align*}$ .
::系数:3x4+14x2+8。

$\begin{align*}ac = 24\end{align*}$ and the factors of 24 that add up to 14 are 12 and 2.
::ac=24,24系数加到14是12和2。

$\begin{aligned} 3 x^{4} + 14 x^{2} + 8 \\ 3 x^{4} + 12 x^{2} + 2 x^{2} + 8 \\ 3 x^{2} (x^{2} + 4) + 2 (x^{4} + 4) \\ (x^{2} + 4) (3 x^{2} + 2) \end{aligned}$
$\begin{align*}& 3x^4+14x^2+8\\ & 3x^4+12x^2+2x^2+8\\ & 3x^2(x^2+4)+2(x^4+4)\\ & (x^2+4)(3x^2+2)\end{align*}$
::3x4+14x2+83x4+12x2+2x2+832x2(x2+4)+2(x4+4(x2+4)(x2+4)(3x2+2)

Example 3
::例3

Factor: $\begin{align*}36x^4-25\end{align*}$ .
::系数:36x4-25。

Factor this polynomial like a difference of squares.
::将这个多面形像平方形的差数乘以。

$\begin{aligned} 36 x^{4} - 25 \\ (6 x^{2} - 5) (6 x^{2} + 5) \end{aligned}$
$\begin{align*}& 36x^4-25\\ & (6x^2-5)(6x^2+5)\end{align*}$
::36x4-25(6x2-5)(6x2+5)

6 and 5 are not square numbers, so this cannot be factored further.
::6和5不是平方数字,因此无法进一步计算。

Example 4
::例4

Find all the real-number solutions of $\begin{align*}8x^5+26x^3-24x = 0\end{align*}$ .
::查找 8x5+26x3- 24x=0 的所有真实数字解决方案。

Pull out a $\begin{align*}2x\end{align*}$ from each term.
::从每个学期抽出2x

$\begin{aligned} 8 x^{5} + 26 x^{3} - 24 x & = 0 \\ 2 x (4 x^{4} + 13 x - 12) & = 0 \\ 2 x (4 x^{4} + 16 x^{2} - 3 x^{2} - 12) & = 0 \\ 2 x [4 x^{2} (x^{2} + 4) - 3 (x^{2} + 4)] & = 0 \\ 2 x (x^{2} + 4) (4 x^{2} - 3) & = 0 \end{aligned}$
$\begin{align*}8x^5+26x^3-24x &= 0\\ 2x(4x^4+13x-12) &= 0\\ 2x(4x^4+16x^2-3x^2-12) &= 0\\ 2x[4x^2(x^2+4)-3(x^2+4)] &= 0\\ 2x(x^2+4)(4x^2-3) &= 0\end{align*}$
::8x5+26x3-24x3-24x=02x(4x4+13x-12)=02x(4x4+16x2-3x2-12)=02x[4x2(x2+4)-3(x2+4)]=02x(x2+4)(4x2-3)=0

Set each factor equal to zero.
::设定每个系数等于零。

$\begin{aligned} 4 x^{2} - 3 = 0 \\ 2 x = 0 x^{2} + 4 = 0 \\ a n d x^{2} = \frac{3}{4} \\ x = 0 x^{2} = - 4 \\ x = \pm \frac{\sqrt{3}}{2} \end{aligned}$
$\begin{align*}& \qquad \qquad \qquad \qquad \qquad \qquad \ \ 4x^2-3 = 0\\ & 2x = 0 \quad x^2+4 = 0\\ & \qquad \quad \ \ \ \ \quad \qquad \qquad \ \ and \qquad \quad x^2 = \frac{3}{4}\\ & \ x = 0 \qquad \quad x^2 = -4\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \pm \frac{\sqrt{3}}{2}\end{align*}$
::4x2 - 3=02x=0x2+4=0和x2=0和x2=34x=0x2=4x4x4x_*4x_***32

Notice the second factor will give imaginary solutions.
::注意第二个因素将给出假想的解决方案。

Review
::回顾

Factor the following quadratics completely.
::系数如下二次方位完全。

$\begin{align*}x^4-6x^2+8\end{align*}$
::x4 - 6x2+8

$\begin{align*}x^4-4x^2-45\end{align*}$
::x4 - 4x2 - 45

$\begin{align*}x^4-18x^2+45\end{align*}$
::x4 - 18x2+45

$\begin{align*}4x^4-11x^2-3\end{align*}$
::4x4-11x2-3

$\begin{align*}6x^4+19x^2+8\end{align*}$
::6x4+19x2+8

$\begin{align*}x^4-81\end{align*}$
::x4 - 81

$\begin{align*}16x^4-1\end{align*}$
::16x4-1

$\begin{align*}6x^5+26x^3-20x\end{align*}$
::6x5+26x3-20x

$\begin{align*}4x^6-36x^2\end{align*}$
::4x6-36x2

$\begin{align*}625-81x^4\end{align*}$
::625-814x4

Find all the real-number solutions to the polynomials below.
::找到所有真实数字的解决方案来解决下面的多元分子问题。

$\begin{align*}2x^4-5x^2-12 = 0\end{align*}$
::2x4 - 5x2 - 12=0

$\begin{align*}x^4-16=0\end{align*}$
::x4 - 16=0

$\begin{align*}16x^4-49 = 0\end{align*}$
::16x4-49=0

$\begin{align*}12x^6+69x^4+45x^2 = 0\end{align*}$
::12x6+69x4+45x2=0

$\begin{align*}3x^4+17x^2 -6=0\end{align*}$
::3x4+17x2-6=0

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

Factoring Polynomials in Quadratic Form ::刻度表的聚合系数乘数

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)

Factoring Polynomials in Quadratic Form
::刻度表的聚合系数乘数

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)