# Solving a Quadratic Equation by Factoring Question

• May 17th 2010, 05:39 PM
dwatkins741
Solving a Quadratic Equation by Factoring Question
I have the problem (x+2)^3 = x^3+8
I use the cube formula (x+y)^(3) = x^(3) + 3x^(2)y + 3xy^(2) + y^(3) and get 6x^(2) + 12x = 0 and solve for x and get -2.

But the book says 0 is also an answer. Can someone please tell me why, and why sometimes 0 is not an answer?

Thanks.
• May 17th 2010, 05:46 PM
skeeter
Quote:

Originally Posted by dwatkins741
I have the problem (x+2)^3 = x^3+8
I use the cube formula (x+y)^(3) = x^(3) + 3x^(2)y + 3xy^(2) + y^(3) and get 6x^(2) + 12x = 0 and solve for x and get -2.

But the book says 0 is also an answer. Can someone please tell me why, and why sometimes 0 is not an answer?

Thanks.

$6x^2 + 12x = 0
$

factor ...

$6x(x + 2) = 0$

set each factor equal to 0 ...

$6x = 0$

$x+2 = 0$
• May 17th 2010, 06:52 PM
dwatkins741
Thanks
I forgot to factor before setting it equal to zero. Btw, how do you make your math text like that?