Chapter 8: Problem 61
Find all of the rational zeros for each function. $$ g(x)=2 x^{3}-9 x^{2}+7 x+6 $$
Short Answer
Expert verified
The rational zeros of the polynomial are \( x = 2, x = 3, \) and \( x = -\frac{1}{2} \).
Step by step solution
01
Determine Possible Rational Zeros
To find the rational zeros of the polynomial \( g(x) = 2x^3 - 9x^2 + 7x + 6 \), we first use the Rational Root Theorem. According to this theorem, the possible rational zeros are the fractions \( \frac{p}{q} \) where \( p \) is a factor of the constant term (6) and \( q \) is a factor of the leading coefficient (2). Thus, we list the factors:\[ p = \pm 1, \pm 2, \pm 3, \pm 6 \]\[ q = \pm 1, \pm 2 \]And the possible rational zeros are: \( \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \).
02
Test Possible Zeros Using Synthetic Division
We will test the potential zeros using synthetic division. We start with \( x = 1 \):\[\begin{array}{r|rrrr}1 & 2 & -9 & 7 & 6 \ & & 2 & -7 & 0 \\hline & 2 & -7 & 0 & 6 \\end{array}\]Residual is 6, thus \( x = 1 \) is not a zero. We proceed with \( x = 2 \):\[\begin{array}{r|rrrr}2 & 2 & -9 & 7 & 6 \ & & 4 & -10 & -6 \\hline & 2 & -5 & -3 & 0 \\end{array}\]Residual is 0, thus \( x = 2 \) is a zero.
03
Use Result to Simplify Polynomial
Since \( x = 2 \) is a zero, we have found it is a factor, and the quotient is \( 2x^2 - 5x - 3 \). We continue solving \( 2x^2 - 5x - 3 = 0 \) for other zeros. This can be factored or solved using the quadratic formula.
04
Use Quadratic Formula
Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to \( 2x^2 - 5x - 3 \), we have \( a = 2, b = -5, c = -3 \). Calculating:\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{4} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} \]\[ x = \frac{5 \pm 7}{4} \]This gives two solutions: \( x = 3 \) and \( x = -\frac{1}{2} \).
05
Conclusion
Thus, the rational zeros of the polynomial \( g(x) = 2x^3 - 9x^2 + 7x + 6 \) are \( x = 2, x = 3, \) and \( x = -\frac{1}{2} \). We verified them by checking that each makes the original polynomial equal to zero.
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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 simplified method for dividing polynomials, particularly useful when you need to evaluate potential rational zeros, as seen in polynomial functions. It is a shortcut to the more cumbersome polynomial long division and is ideal for dividing a polynomial by a linear expression of the form \( x - c \). Here's how synthetic division streamlines the process:
To perform synthetic division, follow these steps:
To perform synthetic division, follow these steps:
- Write down the coefficients of the polynomial in order.
- Place the potential zero (from the list of rational zeros) to the left.
- Drop down the leading coefficient.
- Multiply the dropped number by the potential zero and add this to the next coefficient.
- Continue this multiply-and-add process for all coefficients.
- The final number in this operation is the remainder. If it is zero, the tested number is a root of the polynomial.
Quadratic Formula
The quadratic formula is an essential tool in algebra, used to find the zeros of quadratic equations, given in the form \( ax^2 + bx + c = 0 \). The solutions for \( x \) can be found using:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula works for all quadratic equations and can also prove useful when factoring seems challenging or impossible.
In the step-by-step solution, once we reduced the polynomial using synthetic division to the quadratic \( 2x^2 - 5x - 3 \), we applied the quadratic formula to find the remaining zeros. With \( a = 2 \), \( b = -5 \), and \( c = -3 \), substituting these into the formula confirmed the zeros as \( x = 3 \) and \( x = -\frac{1}{2} \).
The quadratic formula not only provides solutions directly but also underscores the discriminant \( b^2 - 4ac \), helping determine the nature of the roots (real or complex). In our problem, a positive discriminant indicated two real solutions.
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula works for all quadratic equations and can also prove useful when factoring seems challenging or impossible.
In the step-by-step solution, once we reduced the polynomial using synthetic division to the quadratic \( 2x^2 - 5x - 3 \), we applied the quadratic formula to find the remaining zeros. With \( a = 2 \), \( b = -5 \), and \( c = -3 \), substituting these into the formula confirmed the zeros as \( x = 3 \) and \( x = -\frac{1}{2} \).
The quadratic formula not only provides solutions directly but also underscores the discriminant \( b^2 - 4ac \), helping determine the nature of the roots (real or complex). In our problem, a positive discriminant indicated two real solutions.
Factoring Polynomials
Factoring polynomials involves rewriting them as a product of simpler polynomials. By finding factors, you can solve for zeros efficiently. When the polynomial \( 2x^2 - 5x - 3 \) is factored, it breaks down into simpler linear components, making it easier to find its zeros.
Here's a brief approach to factoring quadratics:
Here's a brief approach to factoring quadratics:
- Try to express the quadratic in the form \( ax^2 + bx + c \) as \( (px + q)(rx + s) \) such that \( p \cdot r = a \) and \( q \cdot s = c \).
- If possible, find two numbers that multiply to \( ac \) and add to \( b \).
- Use these numbers to split the middle term and factor by grouping.