How Many Solutions Over The Complex Number System Does This Polynomial Have? 2 X 4 − 3 X 3 − 24 X 2 + 13 X + 12 = 0 2x^4 - 3x^3 - 24x^2 + 13x + 12 = 0 2 X 4 − 3 X 3 − 24 X 2 + 13 X + 12 = 0 Enter Your Answer As An Integer.

Mar 10, 2025 by ADMIN 223 views

**Solving the Polynomial Equation over the Complex Number System**

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Introduction

In this article, we will explore the solutions of a given polynomial equation over the complex number system. The polynomial equation is $2x^4 - 3x^3 - 24x^2 + 13x + 12 = 0$ . We will use various mathematical techniques to find the number of solutions of this equation over the complex number system.

Understanding the Complex Number System

The complex number system is an extension of the real number system, which includes all the real numbers and also all the imaginary numbers. An imaginary number is a number that, when squared, gives a negative result. The complex number system is denoted by the letter $C$ and is defined as the set of all numbers of the form $a + bi$ , where $a$ and $b$ are real numbers and $i$ is the imaginary unit, which satisfies the equation $i^2 = -1$ .

The Rational Root Theorem

The rational root theorem is a fundamental theorem in algebra that states that if a rational number $p/q$ is a root of a polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0$ , where $p$ and $q$ are integers and $q$ is non-zero, then $p$ must be a factor of the constant term $a_0$ and $q$ must be a factor of the leading coefficient $a_n$ .

Applying the Rational Root Theorem to the Given Polynomial

To find the possible rational roots of the given polynomial equation $2x^4 - 3x^3 - 24x^2 + 13x + 12 = 0$ , we need to find the factors of the constant term $12$ and the leading coefficient $2$ . The factors of $12$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ , and the factors of $2$ are $\pm 1, \pm 2$ .

Finding the Possible Rational Roots

Using the rational root theorem, we can find the possible rational roots of the given polynomial equation by dividing the factors of the constant term $12$ by the factors of the leading coefficient $2$ . This gives us the following possible rational roots:

$\pm 1$
$\pm 2$
$\pm 3$
$\pm 4$
$\pm 6$
$\pm 12$
$\pm 1/2$
$\pm 3/2$
$\pm 6/2$
$\pm 12/2$

Testing the Possible Rational Roots

To find the actual rational roots of the given polynomial equation, we need to test each of the possible rational roots by substituting them into the polynomial equation and checking if the equation is satisfied. We can use synthetic division or long division to divide the polynomial by each of the possible rational roots.

Finding the Actual Rational Roots

After testing each of the possible rational roots, we find that the following are the actual rational roots of the given polynomial equation:

$x = 2$
$x = -3$

Finding the Complex Roots

Since the given polynomial equation has degree $4$ , it must have four roots in total. We have already found two of the roots, which are $x = 2$ and $x = -3$ . To find the remaining two roots, we can use the quadratic formula to solve the quadratic equation obtained by dividing the polynomial by the linear factors corresponding to the two rational roots.

Using the Quadratic Formula

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

Finding the Complex Roots

Using the quadratic formula, we can find the complex roots of the given polynomial equation. After simplifying the expression, we get:

$x = \frac{3 \pm \sqrt{-63}}{2}$

Simplifying the Complex Roots

We can simplify the complex roots by expressing the square root of $-63$ in terms of complex numbers. We get:

$x = \frac{3 \pm i\sqrt{63}}{2}$

Conclusion

In this article, we have found the solutions of the given polynomial equation over the complex number system. We have used various mathematical techniques, including the rational root theorem, synthetic division, and the quadratic formula, to find the number of solutions of the equation. We have found that the polynomial equation has four roots in total, two of which are rational and two of which are complex.

Final Answer

The final answer is: $\boxed{4}$

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Q: What is the degree of the polynomial equation?

A: The degree of the polynomial equation is 4, which means it is a quartic equation.

Q: What is the leading coefficient of the polynomial equation?

A: The leading coefficient of the polynomial equation is 2.

Q: What is the constant term of the polynomial equation?

A: The constant term of the polynomial equation is 12.

Q: How many rational roots does the polynomial equation have?

A: The polynomial equation has 2 rational roots, which are x = 2 and x = -3.

Q: How many complex roots does the polynomial equation have?

A: The polynomial equation has 2 complex roots, which are x = (3 + i√63)/2 and x = (3 - i√63)/2.

Q: Can you explain the rational root theorem?

A: The rational root theorem states that if a rational number p/q is a root of a polynomial equation anxn + an-1xn-1 + ... + a1x + a0 = 0, where p and q are integers and q is non-zero, then p must be a factor of the constant term a0 and q must be a factor of the leading coefficient an.

Q: How do you find the possible rational roots of a polynomial equation?

A: To find the possible rational roots of a polynomial equation, you need to find the factors of the constant term and the leading coefficient, and then divide the factors of the constant term by the factors of the leading coefficient.

Q: What is the quadratic formula?

A: The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation.

Q: How do you use the quadratic formula to find the complex roots of a polynomial equation?

A: To use the quadratic formula to find the complex roots of a polynomial equation, you need to substitute the coefficients of the quadratic equation into the formula and simplify the expression.

Q: Can you explain the concept of complex numbers?

A: Complex numbers are numbers that have both a real part and an imaginary part. They are denoted by the letter C and are defined as the set of all numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit, which satisfies the equation i^2 = -1.

Q: How do you simplify complex numbers?

A: To simplify complex numbers, you need to express the square root of a negative number in terms of complex numbers. This can be done by using the formula √(-a) = i√a.

Q: Can you provide an example of a complex number?

A: An example of a complex number is 3 + 4i, where 3 is the real part and 4i is the imaginary part.

Q: How do you add and subtract complex numbers?

A: To add and subtract complex numbers, you need to add or subtract the real parts and the imaginary parts separately. For example, (3 + 4i) + (2 + 5i) = (3 + 2) + (4i + 5i) = 5 + 9i.

Q: Can you provide an example of a complex number in polar form?

A: An example of a complex number in polar form is 3 + 4i = √(3^2 + 4^2)e(iθ), where θ is the angle between the positive x-axis and the line connecting the origin to the point (3, 4) in the complex plane.

Q: How do you convert a complex number from rectangular form to polar form?

A: To convert a complex number from rectangular form to polar form, you need to find the magnitude (or modulus) and the angle (or argument) of the complex number. The magnitude is given by √(a^2 + b^2), where a and b are the real and imaginary parts of the complex number, respectively. The angle is given by θ = arctan(b/a).

Q: Can you provide an example of a complex number in exponential form?

A: An example of a complex number in exponential form is 3 + 4i = re^(iθ), where r is the magnitude and θ is the angle of the complex number.

Q: How do you convert a complex number from exponential form to rectangular form?

A: To convert a complex number from exponential form to rectangular form, you need to use Euler's formula, which states that e^(iθ) = cos(θ) + i sin(θ).