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asambeis [7]
4 years ago
15

How can we factorise the expression: 2x^6+17x^3+8?​

Mathematics
1 answer:
LekaFEV [45]4 years ago
7 0

Answer:

(a + 2)( {a}^{2}  - 2a +  4 )(2 {x}^{ 3}  + 1)

Step-by-step explanation:

2x^6+17x^3+8 \\  = 2x^6+16x^3 +  {x}^{3} +8 \\  =  2{x}^{3} ( {x}^{3}  + 8) + 1( {x}^{3}  + 8) \\  = ( {x}^{3}  + 8)(2 {x}^{ 3}  + 1) \\  = ( {x}^{3}  +  {2}^{3} )(2 {x}^{ 3}  + 1)  \\  = (a + 2)( {a}^{2}  - 2a +  {2}^{2} )(2 {x}^{ 3}  + 1)   \\  = (a + 2)( {a}^{2}  - 2a +  4 )(2 {x}^{ 3}  + 1)  \\

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Which answer includes a function, its inverse, and a dashed line that verifies the relationship?
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the first one

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yellow line is an exponential graph

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Line through (0,−3) slanting down.

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3 years ago
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7 0
3 years ago
3^x= 3*2^x solve this equation​
kompoz [17]

In the equation

3^x = 3\cdot 2^x

divide both sides by 2^x to get

\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3

Take the base-3/2 logarithm of both sides:

\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}

Alternatively, you can divide both sides by 3^x:

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Then take the base-2/3 logarith of both sides to get

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(Both answers are equivalent)

8 0
3 years ago
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