Find All Complex Solutions (real And Non-real) Of The Equation $x^4-4x^3+6x^2-4x-15=0$. Provide Exact Answers.Real: $x=$ \$\square$[/tex\], $\square$ Non-real: $x=$ \$\square$

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Introduction

In this article, we will explore the process of finding complex solutions, both real and non-real, of a given quartic equation. The equation in question is $x4-4x3+6x^2-4x-15=0$. We will use various mathematical techniques to factorize the equation and find its roots.

Factorizing the Quartic Equation

To begin, we need to factorize the quartic equation. This can be done by using various methods such as grouping, synthetic division, or using the rational root theorem. In this case, we will use the rational root theorem to find the possible rational roots of the equation.

Rational Root Theorem

The rational root theorem states that if a rational number $p/q$ is a root of the polynomial $a_nxn+a_{n-1}x{n-1}+\cdots+a_1x+a_0$, then $p$ must be a factor of the constant term $a_0$, and $q$ must be a factor of the leading coefficient $a_n$.

Possible Rational Roots

Using the rational root theorem, we can find the possible rational roots of the equation. The factors of the constant term $-15$ are $\pm 1, \pm 3, \pm 5, \pm 15$, and the factors of the leading coefficient $1$ are $\pm 1$. Therefore, the possible rational roots of the equation are $\pm 1, \pm 3, \pm 5, \pm 15$.

Synthetic Division

We can use synthetic division to test these possible rational roots and find the actual roots of the equation. Synthetic division is a method of dividing a polynomial by a linear factor of the form $(x - r)$, where $r$ is a root of the polynomial.

Finding the Roots

Using synthetic division, we find that the equation can be factorized as follows:

x4−4x3+6x2−4x−15=(x−5)(x3+x2−1)x^4-4x^3+6x^2-4x-15=(x-5)(x^3+x^2-1)

We can further factorize the cubic equation $x3+x2-1$ as follows:

x3+x2−1=(x+1)(x2−1)x^3+x^2-1=(x+1)(x^2-1)

x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1)

Therefore, the factorized form of the quartic equation is:

(x−5)(x+1)(x−1)(x+1)=0(x-5)(x+1)(x-1)(x+1)=0

Finding the Complex Solutions

Now that we have the factorized form of the quartic equation, we can find the complex solutions of the equation. The complex solutions are the values of $x$ that make the equation equal to zero.

Real Solutions

The real solutions of the equation are the values of $x$ that make the equation equal to zero. In this case, the real solutions are $x=5$ and $x=1$.

Non-Real Solutions

The non-real solutions of the equation are the values of $x$ that make the equation equal to zero. In this case, the non-real solutions are $x=-1$ and $x=-1$.

Conclusion

In this article, we have found the complex solutions of a given quartic equation. The equation in question is $x4-4x3+6x^2-4x-15=0$. We have used various mathematical techniques to factorize the equation and find its roots. The real solutions of the equation are $x=5$ and $x=1$, and the non-real solutions are $x=-1$ and $x=-1$.

Final Answer

The final answer is:

  • Real solutions: $x=5$ and $x=1$
  • Non-real solutions: $x=-1$ and $x=-1$

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Introduction

In our previous article, we explored the process of finding complex solutions, both real and non-real, of a given quartic equation. The equation in question was $x4-4x3+6x^2-4x-15=0$. We used various mathematical techniques to factorize the equation and find its roots. In this article, we will answer some of the most frequently asked questions related to finding complex solutions of a quartic equation.

Q&A

Q: What is a quartic equation?

A: A quartic equation is a polynomial equation of degree four, which means that the highest power of the variable (in this case, $x$) is four.

Q: How do I factorize a quartic equation?

A: There are several methods to factorize a quartic equation, including grouping, synthetic division, and using the rational root theorem. In our previous article, we used the rational root theorem to find the possible rational roots of the equation.

Q: What is the rational root theorem?

A: The rational root theorem states that if a rational number $p/q$ is a root of the polynomial $a_nxn+a_{n-1}x{n-1}+\cdots+a_1x+a_0$, then $p$ must be a factor of the constant term $a_0$, and $q$ must be a factor of the leading coefficient $a_n$.

Q: How do I use synthetic division to find the roots of a polynomial?

A: Synthetic division is a method of dividing a polynomial by a linear factor of the form $(x - r)$, where $r$ is a root of the polynomial. To use synthetic division, you need to follow these steps:

  1. Write down the coefficients of the polynomial in a row.
  2. Write down the root of the polynomial in a column.
  3. Multiply the root by the first coefficient and write the result below the row.
  4. Add the result to the second coefficient and write the result below the row.
  5. Repeat steps 3 and 4 until you reach the last coefficient.
  6. The result is the quotient of the division.

Q: What are the real and non-real solutions of a quartic equation?

A: The real solutions of a quartic equation are the values of $x$ that make the equation equal to zero. The non-real solutions are the values of $x$ that make the equation equal to zero, but are not real numbers.

Q: How do I find the complex solutions of a quartic equation?

A: To find the complex solutions of a quartic equation, you need to factorize the equation and find its roots. You can use various mathematical techniques, including grouping, synthetic division, and using the rational root theorem.

Conclusion

In this article, we have answered some of the most frequently asked questions related to finding complex solutions of a quartic equation. We have discussed the rational root theorem, synthetic division, and the process of finding the real and non-real solutions of a quartic equation. We hope that this article has been helpful in understanding the process of finding complex solutions of a quartic equation.

Final Answer

The final answer is:

  • A quartic equation is a polynomial equation of degree four.
  • The rational root theorem states that if a rational number $p/q$ is a root of the polynomial $a_nxn+a_{n-1}x{n-1}+\cdots+a_1x+a_0$, then $p$ must be a factor of the constant term $a_0$, and $q$ must be a factor of the leading coefficient $a_n$.
  • Synthetic division is a method of dividing a polynomial by a linear factor of the form $(x - r)$, where $r$ is a root of the polynomial.
  • The real solutions of a quartic equation are the values of $x$ that make the equation equal to zero.
  • The non-real solutions of a quartic equation are the values of $x$ that make the equation equal to zero, but are not real numbers.