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VLD [36.1K]
3 years ago
10

How do I factor the following expression?

Mathematics
2 answers:
Crazy boy [7]3 years ago
6 0

\bf x^6-9x^4-81x^2+729\implies \stackrel{\textit{let's do some grouping}}{(x^6-9x^4)-(81x^2-729)} \\\\\\ \stackrel{\textit{some common factoring}}{x^4\boxed{(x^2-9)}-81\boxed{(x^2-9)}}\implies \stackrel{\textit{common factoring}}{\boxed{(x^2-9)}(x^4-81)} \\\\\\ \stackrel{\textit{difference of squares}}{[x^2-3^2][(x^2)^2-(3^2)^2]}\implies (x-3)(x+3)\stackrel{\textit{difference of squares}}{(x^2-3^2)}(x^2+3^2) \\\\\\ (x-3)(x+3)~(x-3)(x+3)~(x^2+9)

amid [387]3 years ago
6 0

On a hunch, I decided to check whether 729 is a 6th power, and found that it is:

729^(1/6) = 3.

Next, I decided to divide x-3 into x^6 + 0x^5 - 9x^4 + 0x^3 - 81x^2 + 0x + 729, through synthetic division and using 3 as my divisor. This left no remainder. The coefficients of the quotient were as follows: 3 3 0 0 -81 -243, which represents:

3x^5 + 3x^4 + 0x^3 + 0x^2 -81x -243

Next, I applied "factoring by grouping:"

3x^4(x+1) -81(x+3) = (x+3)(3x^4 - 81).

Note that 3x^4 - 81 is the same as 3(x^4-27).

Thus, the original polynomial factors into (x-3)(3)(x^4 - 27).

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Leave answer in the simplest radical form.
slega [8]

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7 0
2 years ago
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morpeh [17]

Answer:

hello your question lacks the required options but i will provide a general answer to your question

answer : The mean value  and the standard deviation would allow us use a Z-score

Step-by-step explanation:

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sample size = 40

The parameters that would be applicable while use Z-score are : The given mean value and the standard deviation

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+Zscore means that the score  is above the average value

-Zscore means that the score is below the average value

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3 years ago
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MAXImum [283]

Answer:

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3 years ago
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Answer:

Step-by-step explanation:

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2 years ago
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algol [13]

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7 0
3 years ago
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