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3 years ago

How do you factor: x^3 - 8 (x cubed minus 8)

Mathematics

2 answers:

Firlakuza [10]3 years ago

6 0

$x^3 - 8 =x^3-2^3 = (x-2) (x^2 + 2x + 2^2)=(x-2) (x^2 + 2x + 4)\\ \\ \\a^3 - b^3 = (a - b) (a^2 + ab + b^2)$

gtnhenbr [62]3 years ago

3 0

Answer: (x - 2)(x² + 2x + 4)

Step-by-step explanation: In this problem, we're asked to factor x³ - 8.

Notice that x³ is a perfect cube and 8 is a perfect cube because 8 is 2 × 2 × 2 or 2³. So we have the difference of two cubes.

To factor the difference of two cubes, we use the following formula.

a³ - b³ can be factored as (a - b)(a² + ab + b²) and in this problem, since a³ is represented by x³, the value of a is x and since b³ is represented by 8, the value of b is 2.

So substituting x and 2 into the formula for <em>a</em> and <em>b</em>, we have

[(x) - (2)][(x)² + (x)(2) + (2)²] and notice that we changed the parentheses in the formula to brackets so that we're not dealing with too many sets of parentheses.

Next, simplifying inside the second set of brackets and changing the brackets back to parentheses, we have (x - 2)(x² + 2x + 4) which is our final answer.

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Select the correct answer.

mariarad [96]

The answer to this Question is A

What is Absolute Value Function or so called Modulus Function or Mod function?

It is a function that takes all real values as input and returns same values but +ve in nature eg if you give input as 2, output will be 2 and if you give input as -2, you will get output 2 again

Solution:

We have expression Ix-5I + 2

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1 year ago

Combine terms: 12a + 26b -4b – 16a.

Kryger [21]

Answer:

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Step-by-step explanation:

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2 years ago

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In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.

Irina-Kira [14]

Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

$c^{2} = a^{2} + b^{2} - 2abcosC$

Rearranging this equation, we can find the value ∠C as shown below.

$\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$
$C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})$

We can apply the same reasoning for finding the value of ∠B as shown.

$B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})$

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

$C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})$
$C \approx 37.19$

$B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})$
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Answer: ∠C = 37.19° and ∠B = 53.13°

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3 years ago

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Jason gets very nervous giving speeches in front of big crowds how can he combat his nervousness

gregori [183]

Answer: Jason can imagine everyone naked while on the stage, it will make him smile, and feel free.

Step-by-step explanation: I'ts pretty easy actually.

3 0

2 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

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2 years ago