# Thread: Quadratic Equation

1. ## Quadratic Equation

can someone help me on this?
solve the quadratic equations by factoring

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

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

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

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

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

3. 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:
$\displaystyle (x+2)^3 = x^3 + 3 \cdot 2x^2 + 3 \cdot 2^2x + 2^3 = x^3 + 6x^2 + 12x + 8$

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

$\displaystyle 6x^2 + 12 x = 0$

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

etc.

-Dan