Solution 2.1:3e
From Förberedande kurs i matematik 1
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- | + | Both terms contain ''x'', which can therefore be taken out as a factor (as can 2), | |
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- | Both terms contain | + | |
- | + | {{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 | ||
- | + | {{Displayed math||<math> 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,,</math>}} | |
- | which can also be written as <math> -2x(x+3)(x-3).</math> | + | 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),
\displaystyle 18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x^2)\,\textrm{.} |
The remaining second-degree factor \displaystyle 9-x^2 can then be factorized using the conjugate rule
\displaystyle 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x)\,, |
which can also be written as \displaystyle -2x(x+3)(x-3).