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Magazine Chapter 4: Problem 15
Find all complex solutions of each equation. Do not use a calculator. $$3 x^{3}+2 x^{2}-3 x-2=0$$
Short Answer
Expert verified
The solutions are \(x = 1\), \(x = -\frac{2}{3}\), and \(x = -1\).
Step by step solution
01
Identify Possible Rational Roots
We start by applying the Rational Root Theorem, which suggests that any rational solution is a factor of the constant term divided by a factor of the leading coefficient. For the equation \(3x^3 + 2x^2 - 3x - 2 = 0\), the constant term is \(-2\) and the leading coefficient is \(3\). Therefore, the possible rational roots are \(\pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}\).
02
Test Possible Rational Roots
Substitute each possible rational root into the equation to see if it equals zero. Start with \(x = 1\):1. \(3(1)^3 + 2(1)^2 - 3(1) - 2 = 3 + 2 - 3 - 2 = 0\)2. Thus, \(x = 1\) is a root.Next, factor \(x - 1\) out from the cubic equation.
03
Factor the Cubic Polynomial
Using the fact that \(x = 1\) is a root, perform synthetic division or polynomial division on \(3x^3 + 2x^2 - 3x - 2\) by \(x - 1\). This reduces the polynomial:After the division, the quotient is \(3x^2 + 5x + 2\). So, \(3x^3 + 2x^2 - 3x - 2 = (x - 1)(3x^2 + 5x + 2)\).
04
Solve the Quadratic Equation
Solve the quadratic equation \(3x^2 + 5x + 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 5\), and \(c = 2\):1. Calculate the discriminant: \(b^2 - 4ac = 5^2 - 4 \times 3 \times 2 = 25 - 24 = 1\).2. Substitute into the quadratic formula: \[x = \frac{-5 \pm \sqrt{1}}{6} = \frac{-5 \pm 1}{6}.\]3. Thus, the solutions are \(x = -\frac{4}{6} = -\frac{2}{3}\) and \(x = -\frac{6}{6} = -1\).
05
Write All Solutions
Combine the solutions found from both steps. The solutions to the equation \(3x^3 + 2x^2 - 3x - 2 = 0\) are \(x = 1\), \(x = -\frac{2}{3}\), and \(x = -1\). None of these solutions are complex numbers, so there is no need to consider complex conjugates for this equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is a useful tool for finding potential rational solutions of polynomial equations. It tells us that if a polynomial has a rational root, that root must be a factor of the constant term divided by a factor of the leading term. In our polynomial, \(3x^3 + 2x^2 - 3x - 2\), the constant term is \(-2\) and the leading coefficient is \(3\).
Thus, the possible rational roots are:
Thus, the possible rational roots are:
- \(\pm 1\)
- \(\pm 2\)
- \(\pm \frac{1}{3}\)
- \(\pm \frac{2}{3}\)
Quadratic Equation
A quadratic equation is any polynomial equation of the form \(ax^2 + bx + c = 0\). The quadratic equation in our exercise was obtained after factoring the original cubic polynomial via division: \(3x^2 + 5x + 2 = 0\).
To solve this quadratic equation, we utilize the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, \(a = 3\), \(b = 5\), and \(c = 2\). Solving this provides us with two distinct solutions: \(x = -\frac{2}{3}\) and \(x = -1\). The use of the quadratic formula is essential in finding the roots of quadratics, especially when factoring is not straightforward.
To solve this quadratic equation, we utilize the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, \(a = 3\), \(b = 5\), and \(c = 2\). Solving this provides us with two distinct solutions: \(x = -\frac{2}{3}\) and \(x = -1\). The use of the quadratic formula is essential in finding the roots of quadratics, especially when factoring is not straightforward.
Synthetic Division
Synthetic division is a shorthand method of polynomial division, particularly useful when dividing by linear factors like \(x - 1\). Compared to long division, synthetic division streamlines the process, employing only the coefficients of the polynomial.
To perform synthetic division for our polynomial \(3x^3 + 2x^2 - 3x - 2\) by \(x - 1\):
To perform synthetic division for our polynomial \(3x^3 + 2x^2 - 3x - 2\) by \(x - 1\):
- Set up the coefficients: \([3, 2, -3, -2]\)
- Use the root \(x = 1\) as the divisor
- Bring down the first coefficient \(3\).
- Continue through the process, multiplying and adding down.
Polynomial Division
Polynomial division is a technique by which polynomials are divided, one term at a time, similar to long division with numbers. It's essential for simplifying polynomials and finding factors.
In this context, polynomial division helps us reduce our given polynomial \(3x^3 + 2x^2 - 3x - 2\) by \(x - 1\), a known root, to obtain \(3x^2 + 5x + 2\).
Here's a quick refresher on the steps involved in polynomial division:
In this context, polynomial division helps us reduce our given polynomial \(3x^3 + 2x^2 - 3x - 2\) by \(x - 1\), a known root, to obtain \(3x^2 + 5x + 2\).
Here's a quick refresher on the steps involved in polynomial division:
- Divide the first term of the dividend by the first term of the divisor.
- Multiply the entire divisor by this result.
- Subtract that from the original polynomial.
- Repeat with the new resulting polynomial.
