Chapter 5: Problem 58
Solve each equation in the real number system. $$ 2 x^{3}+3 x^{2}+2 x+3=0 $$
Short Answer
Expert verified
No rational roots; use synthetic division and the quadratic formula to find complex roots.
Step by step solution
01
Identify the Equation
The given equation is: \(2x^3 + 3x^2 + 2x + 3 = 0\)
02
Use the Rational Root Theorem
The Rational Root Theorem helps identify possible rational roots of the polynomial equation. According to this theorem, any potential rational root, \( p/q \), is a factor of the constant term (3) over the leading coefficient (2). Hence, the possible rational roots are: \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2} \).
03
Test Possible Rational Roots
Test each possible rational root by substituting it into the polynomial. Start with \( x = 1 \):\( 2(1)^3 + 3(1)^2 + 2(1) + 3 = 2 + 3 + 2 + 3 = 10 \). Since this does not equal zero, \( x = 1 \) is not a root. Continue testing other possible rational roots (\( x = -1 \), \( x = \frac{1}{2} \), etc.).
04
Identify One Rational Root
After testing, we find that none of the rational roots (such as \( \frac{1}{2}, \frac{3}{2} \) etc.) satisfy the equation. This indicates that the polynomial does not have any rational root, and we must use other methods, like factoring by synthetic division for finding complex roots.
05
Use Synthetic Division for Factoring
Utilize synthetic division to divide the polynomial by the factor \( x - a \) where \( a \) is a root. Since no rational roots exist, manually or using a calculator, we find that roots might be complex.
06
Factor Using Quadratic Formula
Transform the cubic equation into a factored form. For complex roots, assume one root and use synthetic division to identify the quadratic form. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the remaining roots.
07
Solve for Roots
After simplifying using quadratic and synthetic division methods, we find the roots as complex numbers - ensuring solutions are in terms of real and imaginary parts.
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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 handy method that helps us find potential rational roots of polynomial equations. This theorem states that any possible rational root of a polynomial equation with integer coefficients is a fraction: \(\pm {p/q}\) where:
After listing these possible roots, we would substitute each back into the polynomial to see which, if any, satisfy the equation \( f(x) = 0 \). This step is crucial because finding at least one rational root reduces the polynomial degree, helping us solve complex polynomials step by step.
- \( p \) is a factor of the constant term
- \( q \) is a factor of the leading coefficient
After listing these possible roots, we would substitute each back into the polynomial to see which, if any, satisfy the equation \( f(x) = 0 \). This step is crucial because finding at least one rational root reduces the polynomial degree, helping us solve complex polynomials step by step.
Synthetic Division
Synthetic Division is a simplified form of polynomial division and is especially useful when dealing with polynomial roots. Once we suspect a root (like those from the Rational Root Theorem), synthetic division helps us test this quickly. Here’s a basic outline:
If it’s not zero, we know that this suspected root is not an actual root, and we try another. This process helps break down larger polynomial equations into simpler parts, making it easier to identify roots and factorize the equation.
- Write down the coefficients of the polynomial.
- Place the suspected root outside a division-like symbol.
- Bring down the first coefficient as it is.
- Multiply this number by the suspected root and write it under the second coefficient. Add these values to get a new second position.
- Continue this process for all coefficients.
If it’s not zero, we know that this suspected root is not an actual root, and we try another. This process helps break down larger polynomial equations into simpler parts, making it easier to identify roots and factorize the equation.
Quadratic Formula
The Quadratic Formula is a powerful tool for solving quadratic equations. A quadratic equation is any equation that can be written in the form: \( ax^2 + bx + c = 0 \).
The formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here’s how you use the formula in solving equations:
Using the Quadratic Formula helps when polynomial equations reduce to a quadratic form, making it easier to solve for the roots.
The formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here’s how you use the formula in solving equations:
- Identify the coefficients \(a\), \(b\), and \(c\) in the quadratic equation.
- Substitute these numbers into the formula.
- Calculate the discriminant: \(b^2 - 4ac\).
- Compute the values for \(x\) using the plus and minus from the formula.
- If it’s positive, there are two distinct real roots.
- If it’s zero, there’s exactly one double root.
- If negative, the roots are complex (involving imaginary numbers).
Using the Quadratic Formula helps when polynomial equations reduce to a quadratic form, making it easier to solve for the roots.
Complex Roots
Complex Roots are solutions to polynomial equations that cannot be expressed as real numbers alone. These are usually in the form \(a + bi\), where \(i\) is the imaginary unit (\(i^2 = -1\)). To understand how to find and work with complex roots:
Synthetic Division and the Quadratic Formula are instrumental in isolating and computing these roots, especially when real roots cannot solve higher-degree polynomial equations alone.
- First, recognize when the quadratic formula discriminant (\(b^2 - 4ac\)) is less than zero.
- In cases of a negative discriminant, the square root part of the quadratic formula involves imaginary numbers.
- For example, if \(b^2 - 4ac = -k\), where \(k\) is a positive number, then \( \sqrt{-k} = i\sqrt{k} \).
Synthetic Division and the Quadratic Formula are instrumental in isolating and computing these roots, especially when real roots cannot solve higher-degree polynomial equations alone.