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Chapter 0: Problem 50

Find all real zeros of the polynomial. $$ x^{3}-216 $$

Short Answer

Expert verified
The real zeros of the polynomial are \(x = 6\).

Step by step solution

01

Identify coefficients and form the equation

To find the root(s) of the polynomial \(x^{3} - 216\), one must set the polynomial equal to zero and solve for \(x\). Thus, the task is to solve: \(x^{3} - 216 = 0.\)
02

Recognize the form and factorize the equation

The equation can be seen as a difference of two cubes - \(x^3 = 6^3\), where \(a = x\), and \(b = 6\). Using the formula for difference of cubes, it can be factored as: \((x-6)(x^2 + 6x + 36)\) . So the equation becomes: \((x-6)(x^2 + 6x + 36) = 0.\)
03

Solve for x

Once in factored form, set each factor equal to zero and solve for \(x\). Solving \((x-6) = 0\) gives \(x = 6\), and equation \(x^2 + 6x + 36 = 0\) cannot be factored further as it has no real roots. Therefore, the only real root of the equation is \(x = 6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Cubes
When encountering a polynomial expression such as \(x^3 - 216\), it's helpful to recognize patterns that make it easier to solve. This specific expression is an example of a "difference of cubes." The difference of cubes follows a specific formula that allows us to factor it into simpler components. The formula for factoring the difference of cubes is:
  • \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
In our expression, \(x^3 - 216\), we can think of \(a\) as \(x\) and \(b\) as \(6\) (since \(6^3 = 216\)).
This allows us to rewrite the expression using the formula as:
  • \((x - 6)(x^2 + 6x + 36)\)
This method transforms a cubic polynomial into a product of a linear and a quadratic polynomial.
Once in this form, it becomes simpler to find the roots.
Factoring Polynomials
Factoring is a crucial step in simplifying and solving polynomial equations. When a polynomial can be factored, it breaks down into two or more simpler expressions that can be set equal to zero to find the roots. For the polynomial \(x^3 - 216\), we use the difference of cubes formula to factor it into a linear and a quadratic expression: \((x - 6)(x^2 + 6x + 36)\).
  • The first factor \((x - 6)\) is straightforward and leads to a real root when \(x = 6\).
  • The second factor \((x^2 + 6x + 36)\) is a quadratic expression.
Upon attempting to factor the quadratic \(x^2 + 6x + 36\), we find it has no real roots, as its discriminant \(b^2 - 4ac\) is negative (\(6^2 - 4 \cdot 1 \cdot 36 = -108\)).
This means that the quadratic does not intersect the x-axis, confirming that it has no real solutions.
Finding Real Roots
The process of finding real roots involves setting the factored polynomial expressions to zero. The given polynomial \(x^3 - 216\) was factored as \((x - 6)(x^2 + 6x + 36)\).
Each factor can individually be set to zero to find potential solutions.
  • For the linear factor \((x - 6) = 0\): The solution is immediate: \(x = 6\).
  • For the quadratic factor \((x^2 + 6x + 36) = 0\): Solving reveals no real solutions due to the negative discriminant, indicating no real roots exist.
Thus, the only real root of the polynomial \(x^3 - 216\) is \(x = 6\).
This approach simplifies the original cubic polynomial problem and efficiently narrows down the real solutions.

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Most popular questions from this chapter

Use the Rational Zero Theorem as an aid in finding all real zeros of the polynomial. $$ 18 x^{3}-9 x^{2}-8 x+4 $$

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