Select The Correct Answer From Each Drop-down Menu.Consider This Polynomial Equation: 6 ( X − 3 ) ( X 2 + 4 ) ( X + 1 ) = 0 6(x-3)\left(x^2+4\right)(x+1)=0 6 ( X − 3 ) ( X 2 + 4 ) ( X + 1 ) = 0 Use The Equation To Complete This Statement.The Equation Has $\square$ Solutions. Its Real Solutions Are

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Understanding Polynomial Equations

A polynomial equation is an expression consisting of variables and coefficients combined using only addition, subtraction, and multiplication. In this article, we will focus on solving polynomial equations, specifically the given equation: 6(x−3)(x2+4)(x+1)=06(x-3)\left(x^2+4\right)(x+1)=0. This equation is a product of three factors, and we will use it to determine the number of solutions and the real solutions.

The Zero Product Property

The zero product property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. In the given equation, we have the product of three factors: 6(x−3)6(x-3), (x2+4)\left(x^2+4\right), and (x+1)(x+1). Since the product is equal to zero, at least one of these factors must be equal to zero.

Solving for the First Factor

The first factor is 6(x−3)6(x-3). To solve for this factor, we set it equal to zero and solve for xx:

6(x−3)=06(x-3) = 0

x−3=0x-3 = 0

x=3x = 3

Solving for the Second Factor

The second factor is (x2+4)\left(x^2+4\right). To solve for this factor, we set it equal to zero and solve for xx:

x2+4=0x^2+4 = 0

x2=−4x^2 = -4

x=±−4x = \pm \sqrt{-4}

Since the square root of a negative number is not a real number, this factor has no real solutions.

Solving for the Third Factor

The third factor is (x+1)(x+1). To solve for this factor, we set it equal to zero and solve for xx:

(x+1)=0(x+1) = 0

x=−1x = -1

Determining the Number of Solutions

We have found three solutions: x=3x = 3, x=−1x = -1, and x=±−4x = \pm \sqrt{-4}. However, the solution x=±−4x = \pm \sqrt{-4} is not a real solution, so we are left with two real solutions: x=3x = 3 and x=−1x = -1.

Conclusion

In conclusion, the given polynomial equation 6(x−3)(x2+4)(x+1)=06(x-3)\left(x^2+4\right)(x+1)=0 has two real solutions: x=3x = 3 and x=−1x = -1. The equation has a total of three solutions, but only two of them are real.

Discussion

  • What is the zero product property, and how is it used to solve polynomial equations?
  • How do you solve for the factors of a polynomial equation?
  • What is the difference between real and complex solutions?

Practice Problems

  1. Solve the polynomial equation: 2(x−2)(x+3)(x−1)=02(x-2)(x+3)(x-1)=0
  2. Solve the polynomial equation: (x+2)(x−4)(x+1)=0(x+2)(x-4)(x+1)=0
  3. Solve the polynomial equation: 3(x−1)(x+2)(x−3)=03(x-1)(x+2)(x-3)=0

Answer Key

  1. The equation has three solutions: x=2x = 2, x=−3x = -3, and x=1x = 1.
  2. The equation has three solutions: x=−2x = -2, x=4x = 4, and x=−1x = -1.
  3. The equation has three solutions: x=1x = 1, x=−2x = -2, and x=3x = 3.

Additional Resources

  • Khan Academy: Solving Polynomial Equations
  • Mathway: Solving Polynomial Equations
  • Wolfram Alpha: Solving Polynomial Equations

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Q: What is a polynomial equation?

A: A polynomial equation is an expression consisting of variables and coefficients combined using only addition, subtraction, and multiplication. It is a mathematical statement that can be written in the form of axn+bxn−1+…+cx+d=0ax^n + bx^{n-1} + \ldots + cx + d = 0, where aa, bb, cc, and dd are constants, and xx is the variable.

Q: How do I solve a polynomial equation?

A: To solve a polynomial equation, you need to find the values of the variable that make the equation true. This can be done by factoring the equation, using the zero product property, and solving for the variable.

Q: What is the zero product property?

A: The zero product property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. This property is used to solve polynomial equations by setting each factor equal to zero and solving for the variable.

Q: How do I factor a polynomial equation?

A: Factoring a polynomial equation involves expressing it as a product of simpler expressions, called factors. This can be done by finding the greatest common factor (GCF) of the terms, or by using techniques such as grouping, synthetic division, or the rational root theorem.

Q: What is the difference between real and complex solutions?

A: Real solutions are values of the variable that make the equation true and are not imaginary. Complex solutions, on the other hand, are values of the variable that make the equation true and are imaginary.

Q: How do I determine the number of solutions of a polynomial equation?

A: To determine the number of solutions of a polynomial equation, you need to examine the degree of the equation. If the degree is even, the equation may have two or more real solutions. If the degree is odd, the equation may have one real solution and one or more complex solutions.

Q: What are some common types of polynomial equations?

A: Some common types of polynomial equations include:

  • Linear equations: Equations of the form ax+b=0ax + b = 0, where aa and bb are constants.
  • Quadratic equations: Equations of the form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants.
  • Cubic equations: Equations of the form ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0, where aa, bb, cc, and dd are constants.
  • Quartic equations: Equations of the form ax4+bx3+cx2+dx+e=0ax^4 + bx^3 + cx^2 + dx + e = 0, where aa, bb, cc, dd, and ee are constants.

Q: How do I use technology to solve polynomial equations?

A: There are several software programs and online tools that can be used to solve polynomial equations, including:

  • Graphing calculators: These devices can be used to graph the equation and find the x-intercepts.
  • Computer algebra systems (CAS): These software programs can be used to solve polynomial equations and find the roots.
  • Online calculators: These websites can be used to solve polynomial equations and find the roots.

Discussion

  • What are some common mistakes to avoid when solving polynomial equations?
  • How do you determine the degree of a polynomial equation?
  • What are some real-world applications of polynomial equations?

Practice Problems

  1. Solve the polynomial equation: 2(x−2)(x+3)(x−1)=02(x-2)(x+3)(x-1)=0
  2. Solve the polynomial equation: (x+2)(x−4)(x+1)=0(x+2)(x-4)(x+1)=0
  3. Solve the polynomial equation: 3(x−1)(x+2)(x−3)=03(x-1)(x+2)(x-3)=0

Answer Key

  1. The equation has three solutions: x=2x = 2, x=−3x = -3, and x=1x = 1.
  2. The equation has three solutions: x=−2x = -2, x=4x = 4, and x=−1x = -1.
  3. The equation has three solutions: x=1x = 1, x=−2x = -2, and x=3x = 3.

Additional Resources

  • Khan Academy: Solving Polynomial Equations
  • Mathway: Solving Polynomial Equations
  • Wolfram Alpha: Solving Polynomial Equations