How do you divide $\frac{8 x^{6} - 32 x^{5} + 4 x^{4}}{x + 2}$ $( 8x^6-32x^5+4x^4 )/(x+2)$ ?

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How do you divide $\frac{8 x^{6} - 32 x^{5} + 4 x^{4}}{x + 2}$ $( 8x^6-32x^5+4x^4 )/(x+2)$ ?

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Answer 1 · 2017-12-30T22:46:23

$f (x) = \frac{8 x^{6} - 32 x^{5} + 4 x^{4}}{x + 2} = \frac{4 x^{4} (2 x^{2} - 8 x + 1)}{x + 2}$ $f(x)={8x^6-32x^5+4x^4}/{x+2}={4x^4(2x^2-8x+1)}/{x+2}$

$2 x - 12$ $2x-12$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 2 x^{2} - 8 x + 1 ∣ x + 2$ $bar {2x^2-8x+1}|x+2$
$-$ $-$
$2 x^{2} + 4 x$ $2x^2+4x$
$=$ $=$
$0 x^{2} - 12 x + 1$ $0x^2-12x+1$
$-$ $-$
$0 x^{2} - 12 x - 24$ $0x^2-12x-24$
$=$ $=$
$0 x^{2} + 0 x - 23$ $0x^2+0x-23$

$\Rightarrow$ $=>$

$f (x) = 4 x^{4} [(2 x - 12) - \frac{23}{x + 2}]$ $f(x)=4x^4[(2x-12)-23/{x+2}]$

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How do you divide $\frac{8 x^{6} - 32 x^{5} + 4 x^{4}}{x + 2}$ $( 8x^6-32x^5+4x^4 )/(x+2)$ ?

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How do you divide $\frac{8 x^{6} - 32 x^{5} + 4 x^{4}}{x + 2}$ $( 8x^6-32x^5+4x^4 )/(x+2)$ ?