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

1. shubhamsrg

x^4 (x+2) + 4x^2(x+2) + 16(x+2)

2. klimenkov

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

3. klimenkov

Oh.. I forgot 32. $$S=(x^2+2 x+4)( x+2)(x^2-2 x+4)$$.

4. sirm3d

$x^5+2x^4+4x^3+8x^2+16x+32=\frac{x^6-32}{x-2}\\=\frac{(x^3-8)(x^3+8)}{x-2}=\frac{(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)}{x-2}\\=(x+2)(x^2+2x+4)(x^2-2x+4)$

5. sirm3d

oopd, that should be $\frac{x^6-64}{x-2}$

6. zugzwang

Nice one, @sirm3d, the answer (and process) I was hoping for :)