Problem 67 - Precalculus Student Solutions Manual 5th

Question

Solve each cubic equation using factoring and the quadratic formula. $$x^{3}-8=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 2 \), \( x = -1 + i\sqrt{3} \), and \( x = -1 - i\sqrt{3} \).

Step 1: Recognize the form of the equation

The given equation is a cubic equation in the form of \( x^3 - 8 = 0 \). This can be identified as a difference of cubes.

Step 2: Apply the difference of cubes formula

Recall that the difference of cubes formula is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). For the given equation, \( a = x \) and \( b = 2 \).

Step 3: Factor the cubic equation

Using the difference of cubes formula, factor the equation as follows: \[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0 \].

Step 4: Solve the linear factor

Set the linear factor equal to zero and solve for \( x \):\( x - 2 = 0 \implies x = 2 \).

Step 5: Solve the quadratic factor using the quadratic formula

To solve the quadratic factor \( x^2 + 2x + 4 = 0 \), use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = 4 \).

Step 6: Calculate the discriminant

First, calculate the discriminant:\[ b^2 - 4ac = 2^2 - 4(1)(4) = 4 - 16 = -12 \].

Step 7: Find the roots of the quadratic equation

Since the discriminant is negative, the roots are complex:\[ x = \frac{-2 \pm \sqrt{-12}}{2 \times 1} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3} \].

Step 8: Combine all solutions

The solutions to the cubic equation are \( x = 2 \), \( x = -1 + i\sqrt{3} \), and \( x = -1 - i\sqrt{3} \).

Key Concepts

Difference of Cubes

The difference of cubes is a way to express a cubic polynomial in a factorable form. It follows the formula: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]. This formula helps simplify solving cubic equations.

When given a problem like \( x^3 - 8 = 0 \), recognize that it can be written as \( (x)^3 - (2)^3 \). Here, \( a \) is \( x \) and \( b \) is \( 2 \). Applying the difference of cubes formula, you can factor the equation into products of polynomials.

• The first factor is \( (x - 2) \)
• The second factor combines terms: \( x^2 + 2x + 4 \)

This process is key as it breaks down a complex cubic equation into simpler parts you can solve step-by-step.

Factoring

Factoring involves breaking down an expression into simpler expressions, or factors, that when multiplied give back the original equation. In cubic equations, you can start by seeing patterns like the difference of cubes.

For \( x^3 - 8 = 0 \):

• Recognize it as a difference of cubes and write it as \( (x)^3 - (2)^3 \).
• Apply the formula to factor it: \( (x - 2)(x^2 + 2x + 4) = 0 \).

Next, solve each factor separately.

• The linear factor is easy: \( x - 2 = 0 \), so \( x = 2 \).
• The quadratic factor \( x^2 + 2x + 4 = 0 \) needs further steps, often involving the quadratic formula.

Factoring makes complex equations more manageable.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

For our problem, the quadratic factor is \( x^2 + 2x + 4 \). Here:

• \( a = 1 \)
• \( b = 2 \)
• \( c = 4 \)

Substituting these into the quadratic formula, we find: \[ x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} \].

The discriminant \( b^2 - 4ac \) is \( -12 \), which is negative. This means the solutions are complex. The quadratic formula helps us find these complex roots.

Complex Roots

Complex roots come into play when the discriminant (\( b^2 - 4ac \)) in the quadratic formula is negative. These roots have both real and imaginary parts.

For \( x^2 + 2x + 4 = 0 \), we calculated the discriminant to be \( -12 \). Using this in the quadratic formula: \[ x = \frac{-2 \pm \sqrt{-12}}{2} \]. The square root of a negative number includes the imaginary unit \( i \), where \( i = \sqrt{-1} \).

So, \( \sqrt{-12} = 2i\sqrt{3} \). This gives us:

• \( x = \frac{-2 + 2i\sqrt{3}}{2} = -1 + i\sqrt{3} \)
• \( x = \frac{-2 - 2i\sqrt{3}}{2} = -1 - i\sqrt{3} \)

Thus, the complex roots of the cubic equation \( x^3 - 8 = 0 \) are \( -1 + i\sqrt{3} \) and \( -1 - i\sqrt{3} \). Combining these with the real root \( x = 2 \), we have all solutions.