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3 years ago

Factor out completely: 4a^2c^2-(a^2-b^2 c^2)^2

Mathematics

1 answer:

Alex17521 [72]3 years ago

6 0

The expression 4a^2c^2 - (a^2-b^2+c^2)^2 has to be factored.

4a^2c^2 - (a^2 - b^2 + c^2)^2

=> (2ac)^2 - (a^2 - b^2 + c^2)^2

=> (2ac - a^2 + b^2 - c^2)(2ac + a^2 - b^2 + c^2)

=> (b^2 - (a^2 - 2ac + c^2))((a^2 + 2ac + c^2) - b^2)

=> (b^2 - (a - c)^2)((a + c)^2 - b^2)

=> (b - a + c)(b + a - c)(a + b + c)(a - b + c)
<span>
The factorized form of 4a^2c^2 - (a^2-b^2+c^2)^2 is (b - a + c)(b + a - c)(a + b + c)(a - b + c)</span>

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The following data points represent the number of animal crackers in each kid's lunch box.

OLga [1]

Answer:

IQR = 7

Step-by-step explanation:

To get the interquartile range of the set of data, the following steps must be followed:

Datas 4, 4, 6 , 7, 10 , 11 , 12, 14 ,15

Step 1: Rearrange the datas in ascending or descending order if not arranged.

4, 4, 6 7, 10, 11, 12, 14, 15

Step 2:

We will locate the (N+1/4)the cumulative value of the data where N is the total frequency of the data.

N = 4 + 4 + 6 + 7 + 10 + 11 + 12 + 14 + 15

N = 83

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= 84/4

= 21st cumulative value

On cumulating the values up to the 21st value or more:

4 + 4 + 6 + 7

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Since the 21st value falls on 7, then the interquartile range of the set of data is 7

3 0

4 years ago

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Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago

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Answer:

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Step-by-step explanation:

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Snowcat [4.5K]

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dedylja [7]

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