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zugzwang Group Title

Let's have a little fun: Factor out x⁵ + 2x⁴ + 4x³ + 8x² + 16x + 32

  • one year ago
  • one year ago

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  1. shubhamsrg Group Title
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    x^4 (x+2) + 4x^2(x+2) + 16(x+2)

    • one year ago
  2. klimenkov Group Title
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    \(S=x^5+2x^4+4x^3+8x^2+16x+32=32\cdot\left(\frac{x^5}{32}+\frac{x^4}{16}+\frac{x^3}{8}+\frac{x^2}{4}+\frac{x}{2}+1\right)\). \(S\cdot\frac x 2=32\cdot\left(\frac{x^6}{64}+\frac{x^5}{32}+\frac{x^4}{16}+\frac{x^3}{8}+\frac{x^2}{4}+\frac{x}{2}\right)\). Subtract the first equation from the second: \(S\cdot(\frac x 2-1)=32\cdot\left(\frac{x^6}{64}-1\right)\). \(S=\frac{\left(\frac{x^3}{8}-1\right)\left(\frac{x^3}{8}+1\right)}{(\frac x 2-1)}=\frac1{(\frac x2-1)}\cdot(\frac x 2-1)(\frac{x^2}4+\frac x 2+1)(\frac x 2+1)(\frac{x^2}4-\frac x 2+1)=\) \(=(\frac{x^2}4+\frac x 2+1)(\frac x 2+1)(\frac{x^2}4-\frac x 2+1)\).

    • one year ago
  3. klimenkov Group Title
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    Oh.. I forgot 32. \(S=(x^2+2 x+4)( x+2)(x^2-2 x+4)\).

    • one year ago
  4. sirm3d Group Title
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    \[x^5+2x^4+4x^3+8x^2+16x+32=\frac{x^6-32}{x-2}\\=\frac{(x^3-8)(x^3+8)}{x-2}=\frac{(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)}{x-2}\\=(x+2)(x^2+2x+4)(x^2-2x+4)\]

    • one year ago
  5. sirm3d Group Title
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    oopd, that should be \[\frac{x^6-64}{x-2}\]

    • one year ago
  6. zugzwang Group Title
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    Nice one, @sirm3d, the answer (and process) I was hoping for :)

    • one year ago
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