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Chapter 0: Problem 49

Find all real zeros of the polynomial. $$ x^{3}+64 $$

Short Answer

Expert verified
The real zero of the polynomial \(x^{3}+64\) is \(x=-4\).

Step by step solution

01

Identify the equation

We start with the given polynomial equation \(x^{3}+64=0\). The goal is to find all real solutions for x.
02

Express the polynomial as a sum of cubes

Next, recognize that the given polynomial is a sum of cubes: \(x^{3}+4^{3}\). Following the formula for the sum of cubes which is \((a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})\), where a = x and b = 4, this allows us to simplify the polynomial into this usable form.
03

Use the formula to break down the polynomial

Applying the formula for the sum of cubes on given polynomial, the equation becomes: \((x+4)(x^{2}-4x+16)=0\).
04

Set each factor equal to zero and solve

To find the zeros of the polynomial, we set each factor to zero and solve for x. Solving the first equation \(x+4=0\), we get \(x=-4\) which is the real solution. The second equation \(x^{2}-4x+16=0\) gives complex solutions which do not need to be considered in this case as we only want real zeros.

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Key Concepts

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

Sum of Cubes
The concept of the sum of cubes is fundamental when working with cubic polynomials. Recognizing these patterns can simplify seemingly complex expressions. In mathematics, a sum of cubes refers to an expression of the form \(a^3 + b^3\). These expressions can be factored using the formula:
  • \((a^3 + b^3) = (a + b)(a^2 - ab + b^2)\)
In the case of the polynomial \(x^3 + 64\), we recognize it as a sum of cubes, where \(a = x\) and \(b = 4\) (since \(4^3 = 64\)). This identification allows us to apply the sum of cubes formula to factor the expression efficiently.
Factoring Polynomials
Factoring polynomials involves expressing a polynomial as the product of its factors. Sometimes, by recognizing patterns, we can break down complex polynomials into manageable parts. Using the sum of cubes formula, \(x^3 + 64\) can be rewritten as follows:
  • \((x + 4)(x^2 - 4x + 16)\)
Here, we have factored the original cubic polynomial into the product of a linear polynomial \(x + 4\) and a quadratic polynomial \(x^2 - 4x + 16\). Factoring is a critical step because it simplifies the process of finding the zeros of the polynomial.
Solving Polynomial Equations
Once a polynomial is factored, solving the equation becomes much easier. We start with the factored form of the polynomial: \((x + 4)(x^2 - 4x + 16) = 0\). By setting each factor equal to zero, we find the solutions to the polynomial equation:
  • For \(x + 4 = 0\), we find \(x = -4\) as a real zero.
  • The quadratic \(x^2 - 4x + 16 = 0\) does not yield real solutions.
This process highlights how factoring simplifies solving equations by reducing the task to solving smaller, more manageable equations. In this case, the real zero of the polynomial is \(x = -4\). By understanding these steps, you're able to tackle similar polynomial equations with confidence.

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Most popular questions from this chapter

Rationalize the numerator or denominator and simplify. $$ \frac{13}{6+\sqrt{10}} $$

simplify each expression by factoring. $$ 5 x^{3 / 2}-x^{-3 / 2} $$

Use the Rational Zero Theorem as an aid in finding all real zeros of the polynomial. $$ x^{3}+2 x^{2}-5 x-6 $$

Use synthetic division to complete the indicated factorization. $$ 2 x^{3}-x^{2}-2 x+1=(x+1)(\quad) $$

Annuity A balance \(A,\) after \(n\) annual payments of \(P\) dollars have been made into an annuity earning an annual percentage rate of \(r\) compounded annually, is given by $$A=P(1+r)+P(1+r)^{2}+\cdots+P(1+r)^{n}$$ Rewrite this formula by completing the following factorization: \(A=P(1+r)(\)
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