Quadratic Equation

**danielwu** · September 6th 2006, 11:14 PM

can someone help me on this?

solve the quadratic equations by factoring

(x + 2)^3 = x^3 + 8

**TD!** · September 6th 2006, 11:47 PM

If you know 8 = 2³, you see a sum of squares.

$ x^3 + 2^3 = \left( {x + 2} \right)\left( {x^2 - 2x + 4} \right) $

Everything to the left, factor (x+2) and simplify:

$ \begin{array}{l} \left( {x + 2} \right)^3 - \left( {x + 2} \right)\left( {x^2 - 2x + 4} \right) = 0 \\ \\ \left( {x + 2} \right)\left( {\left( {x + 2} \right)^2 - \left( {x^2 - 2x + 4} \right)} \right) = 0 \\ \\ \left( {x + 2} \right)\left( {6x} \right) = 0 \Leftrightarrow x = - 2 \vee x = 0 \\ \end{array} $

**topsquark** · September 7th 2006, 05:07 AM

Originally Posted by danielwu

can someone help me on this?

solve the quadratic equations by factoring

(x + 2)^3 = x^3 + 8

Or, less elegantly:
$(x+2)^3 = x^3 + 3 \cdot 2x^2 + 3 \cdot 2^2x + 2^3 = x^3 + 6x^2 + 12x + 8$

So
$x^3 + 6x^2 + 12x + 8 = x^3 + 8$

$6x^2 + 12 x = 0$

$(6x)(x+2) = 0$

etc.

-Dan

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