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Magazine Chapter 3: Problem 34
Find all rational zeros of the polynomial. $$P(x)=6 x^{3}+11 x^{2}-3 x-2$$
Short Answer
Expert verified
The only rational zero is \( x = 1 \).
Step by step solution
01
Analyze the Rational Root Theorem
According to the Rational Root Theorem, any rational solution, expressed as \( \frac{p}{q} \), is derived from \( p \), which divides the constant term \(-2\), and \( q \), which divides the leading coefficient \(6\). The possible rational roots are the combinations of these divisors.
02
List the Possible Rational Roots
The divisors of \(-2\) are \( \pm 1, \pm 2 \), and the divisors of \(6\) are \( \pm 1, \pm 2, \pm 3, \pm 6 \). Therefore, the possible rational roots are: \( \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6} \).
03
Test Each Possible Root Using Synthetic Division
To find which of these candidates is a root, test each value using synthetic division. Begin by substituting \( x = 1 \). The synthetic division gives a remainder of \(0\), indicating \( x = 1 \) is a root.
04
Factor the Polynomial
Since \( x=1 \) is a root, factor \( (x-1) \) out of \( P(x) \). Using the result from synthetic division, the polynomial is expressed as \( (x-1)(6x^2+17x+2) \).
05
Solve the Quadratic Equation
Now, solve the quadratic \(6x^2+17x+2=0\) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a=6 \), \( b=17 \), and \( c=2 \).
06
Calculate the Quadratic Roots
Calculate: \( x = \frac{-17 \pm \sqrt{17^2-4 \times 6 \times 2}}{2 \times 6} = \frac{-17 \pm \sqrt{233}}{12} \). Since \( \sqrt{233} \) is not a perfect square, these roots are irrational.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Roots
Polynomial roots are values of the variable that make the polynomial equal to zero. Finding these roots helps us understand where the graph of the polynomial intersects with the x-axis. For a cubic polynomial like \( P(x) = 6x^3 + 11x^2 - 3x - 2 \), the Rational Root Theorem is an excellent tool to determine the possible rational roots.
The theorem suggests that if there's a rational root in the form of \( \frac{p}{q} \), then \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient. This gives a list of possible rational numbers to be tested as roots.
Tracking these roots allows us to deconstruct the polynomial into simpler factors, aiding in further algebraic manipulation such as finding zeroes or simplifying expressions.
The theorem suggests that if there's a rational root in the form of \( \frac{p}{q} \), then \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient. This gives a list of possible rational numbers to be tested as roots.
Tracking these roots allows us to deconstruct the polynomial into simpler factors, aiding in further algebraic manipulation such as finding zeroes or simplifying expressions.
Synthetic Division
Synthetic division is a simplified method for dividing a polynomial by a linear factor of the form \( x - r \). It鈥檚 a quick way to test potential roots without performing long division.
To use synthetic division, write down the coefficients of the polynomial. Then, choose a potential root to test (from those given by the Rational Root Theorem). For our polynomial, testing \( x = 1 \) led to a remainder of zero, confirming \( x = 1 \) as an actual root.
This process not only verifies roots but also helps construct the quotient polynomial, in this case reducing \( P(x) \) to \( (x - 1)(6x^2 + 17x + 2) \). This reduction is crucial for simplifying and solving the polynomial further.
To use synthetic division, write down the coefficients of the polynomial. Then, choose a potential root to test (from those given by the Rational Root Theorem). For our polynomial, testing \( x = 1 \) led to a remainder of zero, confirming \( x = 1 \) as an actual root.
This process not only verifies roots but also helps construct the quotient polynomial, in this case reducing \( P(x) \) to \( (x - 1)(6x^2 + 17x + 2) \). This reduction is crucial for simplifying and solving the polynomial further.
Quadratic Formula
The quadratic formula is a handy tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides the roots of the equation using the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our exercise, after confirming that \( x = 1 \) is a root of the cubic polynomial, we factored it down to a quadratic: \( 6x^2 + 17x + 2 \). To find its roots, we applied the quadratic formula. Substituting \( a = 6 \), \( b = 17 \), and \( c = 2 \), we calculated the discriminant \( b^2 - 4ac \), noting its square root is not a perfect square, indicating irrational roots.
The quadratic formula not only finds roots but also shows the nature of the roots 鈥� whether they're real, rational, or irrational, guiding us in understanding the behavior of the polynomial's graph.
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our exercise, after confirming that \( x = 1 \) is a root of the cubic polynomial, we factored it down to a quadratic: \( 6x^2 + 17x + 2 \). To find its roots, we applied the quadratic formula. Substituting \( a = 6 \), \( b = 17 \), and \( c = 2 \), we calculated the discriminant \( b^2 - 4ac \), noting its square root is not a perfect square, indicating irrational roots.
The quadratic formula not only finds roots but also shows the nature of the roots 鈥� whether they're real, rational, or irrational, guiding us in understanding the behavior of the polynomial's graph.
