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Chapter 5: Problem 28

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ 2 x^{3}+x^{2}-7 x-6=0 $$

Short Answer

Expert verified
The real solutions are \(x = -1\), \(x = -\frac{3}{2}\), and \(x = 2\).

Step by step solution

01

Identify Possible Rational Zeros

The Rational Zero Theorem states that any rational solution, expressed as a fraction \( \frac{p}{q} \), is such that \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. For the equation \( 2x^3 + x^2 - 7x - 6 = 0 \), the constant term is \(-6\) and the leading coefficient is \(2\). The factors of \(-6\) are \(\{卤1, 卤2, 卤3, 卤6 \}\) and the factors of \(2\) are \(\{卤1, 卤2 \}\). Thus, the possible rational zeros are \(\{ 卤1, 卤2, 卤3, 卤6, 卤\frac{1}{2}, 卤\frac{3}{2} \}\).
02

Test Possible Zeros Using Synthetic Division

We test the possible rational zeros by substituting them into the polynomial using synthetic division and checking for a remainder of zero. We start with \( x = 1 \) and continue testing other possible values until a zero is found.
03

Use Synthetic Division with \( x = -1 \)

Using synthetic division to test \( x = -1 \), we organize the coefficients \( [2, 1, -7, -6] \) and proceed:1. Bring down the \( 2 \).2. Multiply \( 2 \) by \(-1\) and add to \( 1 \) to get \(-1\).3. Multiply \(-1\) by \(-1\) and add to \(-7\) to get \(-6\).4. Multiply \(-6\) by \(-1\) and add to \(-6\), resulting in \(0\).The process results in no remainder, indicating that \( x = -1 \) is a zero.
04

Form the Quotient Polynomial

The synthetic division with \( x = -1 \) reduces the polynomial to \( 2x^2 - x - 6 \). This is the quotient polynomial that we now work with to find additional zeros.
05

Factor the Quadratic Polynomial

Next, factor the quadratic \( 2x^2 - x - 6 \). We look for two numbers whose product is \( -12 \) (from \( 2 \times -6 \)) and sum to \(-1\). These numbers are \(-3\) and \(4\). Rewrite the polynomial as \( 2x^2 - 3x + 2x - 6 \). Factoring by grouping gives \((2x^2 - 3x) + (2x - 6)\) which factors to \((2x + 3)(x - 2)\).
06

Find the Remaining Real Zeros

Set each factor from Step 5 equal to zero. From \(2x + 3 = 0\), solving gives \(x = -\frac{3}{2}\), and from \(x - 2 = 0\), solving gives \(x = 2\). These are the remaining real zeros.

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Key Concepts

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

Synthetic Division
Synthetic Division is a handy, streamlined way of dividing a polynomial by a linear expression of the form \( x - c \). This method simplifies the process compared to traditional long division.
  • First, we write down the coefficients of the polynomial. For the polynomial \(2x^3 + x^2 - 7x - 6\), the coefficients are \([2, 1, -7, -6]\).
  • We then select a possible zero to test. For instance, if we choose \(x = -1\), we place it on the left.
  • Begin by bringing down the first coefficient unchanged. Multiply it by the selected value \(x = -1\) and add to the next coefficient. Repeat this multiplication and addition process across all coefficients.
This method reveals whether the number is a zero by checking if the final remainder is zero. It's much faster compared to long-hand division, particularly when dealing with complex polynomials.
Factors of Polynomial
When breaking down a polynomial, finding its factors is crucial, as these divisors tell us more about its roots or zeros.
  • In the given polynomial \(2x^3 + x^2 - 7x - 6\), after using synthetic division with \(x = -1\), we found a factor: \(x + 1\).
  • This process simplifies the polynomial to \(2x^2 - x - 6\), a quadratic that can be further factored.
  • For the quadratic, we look for two numbers that multiply to the product of the leading coefficient and the constant term, here \(2 \times -6 = -12\), while also adding to the middle term coefficient, \(-1\).
These factors are significant because they can be directly set to zero to find polynomial's zeros. Factorization makes solving equations faster and more systematic.
Real Zeros
Finding the real zeros of a polynomial involves determining the values of \(x\) where the polynomial evaluates to zero, effectively solving for the roots.
  • Begin by applying the Rational Zero Theorem that gives potential numbers to verify as zeros. If a number results in a remainder of zero upon using synthetic division, it's a confirmed zero.
  • For the quadratic \(2x^2 - x - 6\), setting each factor equal to zero gives real zeros. From \((2x + 3)(x - 2)\), we solve \(2x + 3 = 0\) to get \(x = -\frac{3}{2}\), and \(x - 2 = 0\), which gives \(x = 2\).
  • These steps identify all real solutions for the polynomial equation.
Understanding and finding real zeros is vital for graphing functions, optimizing problems, and solving real-world applications where these mathematical models apply.

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Most popular questions from this chapter

For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given. $$ f(x)=x^{3}+x-2, y=0,1,2 $$

For the following exercises, find the inverse of the function and graph both the function and its inverse. $$ f(x)=(x-4)^{2}, \quad x \geq 4 $$

For the following exercises, use the given information to find the unknown value. \(y\) varies directly as the cube root of \(x\). When \(x=125,\) then \(y=15\). Find \(y\) when \(x=1,000\).

For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies inversely as the square of \(x\) and when \(x=3, \quad y=2\).

For the following exercises, use a calculator to graph the equation implied by the given variation. \(y\) varies inversely with \(x\) and when \(x=6, y=2\).
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