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Chapter 0: Problem 42

Find all real zeros of the polynomial. $$ (x+1)^{2}-36 $$

Short Answer

Expert verified
The real zeros of the polynomial are \(x = -7\) and \(x = 5\).

Step by step solution

01

Rewrite the Polynomial

Rewrite the polynomial equation as a difference of squares: \((x+1)^2 - 36 = 0\) can be written as \((x+1+6)(x+1-6) = 0\), since \(36 = 6^2\).
02

Apply the Zero Product Property

The zero product property states that if \(a \times b = 0\), then either \(a = 0\), \(b = 0\), or both. So, we can say \(x+1+6=0\) or \(x+1-6=0\).
03

Solve the Equations

Solve the equations in step 2: for \(x+1+6=0\), we find \(x = -1-6 = -7\). For \(x+1-6=0\), we find \(x = -1+6 = 5\).

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Key Concepts

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

Difference of Squares
The difference of squares is a technique used to simplify certain polynomial expressions. It applies when you have an expression in the form of \(a^2 - b^2\). This can be factored into the product \((a - b)(a + b)\). This property is handy because it transforms a quadratic expression into a product of two binomials, which can then be solved more easily.

In the original exercise, we started with \((x+1)^2 - 36\). Here, \( (x+1)^2\) and \(36\) are both perfect squares.
  • \((x+1)^2\) is the square of \(x+1\),
  • \(36\) is the square of \(6\) (since \(6^2 = 36\)).
By recognizing this pattern, we can rewrite this equation as \((x+1+6)(x+1-6)\), demonstrating the use of the difference of squares to simplify and solve the equation.
Zero Product Property
The zero product property is a fundamental concept in algebra used to solve polynomial equations. This property states that if the product of two factors is zero, then at least one of the factors must be zero. This is expressed as:
  • If \(a \times b = 0\), then either \(a = 0\), \(b = 0\), or both.
This property is essential when dealing with factored forms of equations.

For the equation derived in our original problem, \((x+1+6)(x+1-6)=0\), you can apply the zero product property by saying:
  • \(x+1+6 = 0\) or
  • \(x+1-6=0\)
Solving each equation leads to possible solutions for \(x\). This method simplifies finding zeros because you deal with simpler linear equations rather than the original quadratic.
Real Zeros
Finding the real zeros of a polynomial is equivalent to finding the values of \(x\) for which the polynomial equals zero. Zeros are also known as roots or solutions of the polynomial equation. In our example, by using the difference of squares and zero product property, we reduced a quadratic equation to two simpler linear equations.

We solved:
  • \(x+1+6 = 0\), which becomes \(x = -7\)
  • \(x+1-6 = 0\), which becomes \(x = 5\)
These are the real zeros of the original polynomial \((x+1)^2 - 36\). Real zeros are the values where the graph of the polynomial crosses or touches the x-axis, representing intercepts in a Cartesian plane. Knowing real zeros is crucial for sketching accurate graphs and understanding the behavior of polynomial functions.

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Most popular questions from this chapter

evaluate the expression for the given value of x. $$ \left(x^{2 / 3}\right)^{3} \quad x=10 $$

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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: \(A=P(1+r)(\)

simplify each expression by factoring. $$ 2 x(x-1)^{5 / 2}-4(x-1)^{3 / 2} $$

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