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Solution 2.1:3e

From Förberedande kurs i matematik 1

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m (Lösning 2.1:3e moved to Solution 2.1:3e: Robot: moved page)
Current revision (08:37, 23 September 2008) (edit) (undo)
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Both terms contain ''x'', which can therefore be taken out as a factor (as can 2),
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<!--center> [[Image:2_1_3e.gif]] </center-->
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Both terms contain <math>x</math>, which can therefore be taken out as a factor (as can 2).
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:<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x). </math>
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{{Displayed math||<math>18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.}</math>}}
The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule
The remaining second-degree factor <math> 9-x^2 </math> can then be factorized using the conjugate rule
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:<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x), </math>
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{{Displayed math||<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,</math>}}
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which can also be written as <math> -2x(x+3)(x-3).</math>
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which can also be written as <math>-2x(x+3)(x-3).</math>
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Current revision

Both terms contain x, which can therefore be taken out as a factor (as can 2),

18x−2x3=2x9−2xx2=2x(9−x2).

The remaining second-degree factor 9−x2 can then be factorized using the conjugate rule

2x(9−x2)=2x(32−x2)=2x(3+x)(3−x)

which can also be written as −2x(x+3)(x−3)

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