Problem 33 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$(x+2)^{3}=x^{3}+8$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = -2\).

Step 1: Recognize the Equation Form

The given equation \((x+2)^3 = x^3 + 8\) is in the form of a difference of cubes with \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\). Notice that the original equation simplifies to this identity because \((x+2)^3 = x^3 + 6x^2 + 12x + 8\).

Step 2: Rewrite the Equation

Convert the equation into a standard polynomial form by expanding the left side: \((x+2)^3 = (x+2)(x+2)(x+2) = x^3 + 6x^2 + 12x + 8\). Therefore, our equation is \(x^3 + 6x^2 + 12x + 8 = x^3 + 8\).

Step 3: Simplify the Equation

Bring all terms to one side to simplify: \(x^3 + 6x^2 + 12x + 8 - x^3 - 8 = 0\). This simplifies to \(6x^2 + 12x = 0\).

Step 4: Factor the Polynomial

Factor out the greatest common factor from each term: \(6x(x+2) = 0\). The two factors are \(6x = 0\) and \(x+2 = 0\).

Step 5: Solve Each Factor for Zero

Set each factor equal to zero: For \(6x = 0\), divide by 6 to get \(x = 0\). For \(x+2 = 0\), subtract 2 from both sides to get \(x = -2\).

Step 6: Verify the Solutions

Substitute \(x = 0\) and \(x = -2\) back into the original equation to verify. For \(x = 0\), the equation becomes \((2)^3 = 0^3 + 8\), which is true. For \(x = -2\), the equation becomes \(0 = 0+8\), which is also true.

Key Concepts

Difference of Cubes

The 'Difference of Cubes' is a formula used when you come across expressions that look like two cubes being subtracted. Recognizing this pattern simplifies complex equations. The general formula is:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
In our exercise, the equation \((x+2)^3 = x^3 + 8\) can be related to cube identities. Essentially, you recognize that the equation simplifies by expanding the cube on the left side:

• When expanded, \((x+2)^3 = x^3 + 6x^2 + 12x + 8\), which does not directly appear as a difference of cubes.

While the given exercise is predominantly about a cube transformation, using a proper expansion technique helps bridge our understanding. Recognizing patterns like these aids in quickly identifying and solving polynomial equations without completely expanding every term from the outset.

Polynomial Factoring

Factoring polynomials involves breaking down a polynomial into simpler components, typically expressed as products of polynomials. In the simplified equation \(6x^2 + 12x = 0\), factoring comes into play for simplification.
Here's how you can efficiently factor a polynomial:

• Identify the greatest common factor (GCF) of all the terms. In our case, the GCF is \(6x\).
• Factor out the GCF from each term to simplify the equation. So, \(6x^2 + 12x\) factors into \(6x(x+2) = 0\).

Factoring simplifies complicated conditions into more manageable equations, making it easy to evaluate the variable solutions. You can think of this step as "unpacking" the polynomial into its simplest parts. It clears the path to focused solutions by allowing you to separately analyze each factor.

Quadratic Solutions

The final step of our exercise is solving the quadratic equation after it has been factored. We look at each factor from the factored polynomial equation \(6x(x+2) = 0\), solving them individually:

• For \(6x = 0\): divide by 6 to obtain \(x = 0\).
• For \(x+2 = 0\): subtract 2 from both sides to solve for \(x = -2\).

Both solutions, \(x = 0\) and \(x = -2\), verify against the original equation. Verification ensures these values satisfy the equation conditions.
Solving quadratic equations through factoring provides a clear path from complex algebra to simple solutions. Once factors are found, each factor equates to zero, and from which straightforward algebraic solutions arise. This process transforms a seemingly daunting equation into a series of more accessible arithmetic steps.