I have to figure out how x^3+216=0 is factored, and what potential solutions might be. I factored it like this: (x+3)(x+3)(x+24), resulting in the solutions being -3 (double route) and -24. But my book says these are not correct answers. Why not? Can you show me how to do it the right way? Thanks!
Use
A³+;B³=(A+B)(A²-AB+B²)
A³-;B³=(A-B)(A²+AB+B²)
and note that 216=6³
To solve the equation x^3 + 216 = 0, we need to first simplify the equation and then factor it in the correct way.
1. Start by subtracting 216 from both sides of the equation:
x^3 = -216
2. Now, we can rewrite -216 as (-6)^3 since (-6) cubed is equal to -216:
x^3 = (-6)^3
3. So, we have x^3 = (-6)^3, which means that x = -6 is one of the solutions.
4. Now, let's factor the left side of the equation using the polynomial factoring technique, rather than the incorrect factoring you used.
We'll use the "difference of cubes" formula, which states that a^3 - b^3 = (a - b)(a^2 + ab + b^2).
Applying this formula, we can factor x^3 + 216 as follows:
x^3 + 216 = (x + 6)(x^2 - 6x + 36)
5. By factoring, we have (x + 6)(x^2 - 6x + 36) = 0.
Now, we can solve for x using each factor:
a) Setting (x + 6) = 0, we find x = -6 as we derived earlier. So, -6 is indeed one of the solutions.
b) Now, let's solve for x using the second factor:
(x^2 - 6x + 36) = 0
This is a quadratic equation that can be factored or solved using the quadratic formula.
However, if we try to factor it further, we'll find that x^2 - 6x + 36 cannot be factored using real numbers. It doesn't have any real solutions.
Therefore, the correct solution to the equation x^3 + 216 = 0 is x = -6 (a single root of multiplicity 1) since the quadratic factor does not contribute any real solutions.
Hence, the book was not considering the incorrect factorization you provided but emphasized finding the correct factorization using the difference of cubes formula.