Select All Zeros For The Following Function:$\[ F(x) = (2x + 3)(x^2 - 5x + 6) \\]A. 1.5 B. 2 C. -3 D. -1.5 E. 3 F. -2

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Understanding the Function

The given function is a product of two expressions: (2x+3)(2x + 3) and (x2−5x+6)(x^2 - 5x + 6). To find the value of f(x)f(x), we need to multiply these two expressions together.

Expanding the Function

To expand the function, we multiply each term in the first expression by each term in the second expression.

f(x)=(2x+3)(x2−5x+6)f(x) = (2x + 3)(x^2 - 5x + 6)

f(x)=2x(x2−5x+6)−5x(2x+3)+6(2x+3)f(x) = 2x(x^2 - 5x + 6) - 5x(2x + 3) + 6(2x + 3)

f(x)=2x3−10x2+12x−10x2−15x+18x+18f(x) = 2x^3 - 10x^2 + 12x - 10x^2 - 15x + 18x + 18

f(x)=2x3−20x2+15x+18f(x) = 2x^3 - 20x^2 + 15x + 18

Finding the Value of f(x)

To find the value of f(x)f(x), we need to select the correct value from the given options. However, we are asked to select all zeros for the function f(x)f(x). This means we need to find the values of xx that make f(x)=0f(x) = 0.

Solving for f(x) = 0

To solve for f(x)=0f(x) = 0, we set the function equal to zero and solve for xx.

2x3−20x2+15x+18=02x^3 - 20x^2 + 15x + 18 = 0

Factoring the Cubic Equation

Unfortunately, this cubic equation does not factor easily. However, we can try to find the rational roots using the Rational Root Theorem.

Rational Root Theorem

The Rational Root Theorem states that if a rational number p/qp/q is a root of the polynomial, then pp must be a factor of the constant term, and qq must be a factor of the leading coefficient.

In this case, the constant term is 18, and the leading coefficient is 2. The factors of 18 are ±1,±2,±3,±6,±9,±18\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, and the factors of 2 are ±1,±2\pm 1, \pm 2.

Possible Rational Roots

The possible rational roots are:

±1,±2,±3,±6,±9,±18,±1/2,±3/2,±9/2,±18/2\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 1/2, \pm 3/2, \pm 9/2, \pm 18/2

Testing Possible Rational Roots

We can test these possible rational roots by substituting them into the equation and checking if the equation is true.

Substituting x = -3

Let's substitute x=−3x = -3 into the equation.

f(−3)=2(−3)3−20(−3)2+15(−3)+18f(-3) = 2(-3)^3 - 20(-3)^2 + 15(-3) + 18

f(−3)=2(−27)−20(9)−45+18f(-3) = 2(-27) - 20(9) - 45 + 18

f(−3)=−54−180−45+18f(-3) = -54 - 180 - 45 + 18

f(−3)=−261f(-3) = -261

Since f(−3)≠0f(-3) \neq 0, x=−3x = -3 is not a root of the equation.

Substituting x = -2

Let's substitute x=−2x = -2 into the equation.

f(−2)=2(−2)3−20(−2)2+15(−2)+18f(-2) = 2(-2)^3 - 20(-2)^2 + 15(-2) + 18

f(−2)=2(−8)−20(4)−30+18f(-2) = 2(-8) - 20(4) - 30 + 18

f(−2)=−16−80−30+18f(-2) = -16 - 80 - 30 + 18

f(−2)=−108f(-2) = -108

Since f(−2)≠0f(-2) \neq 0, x=−2x = -2 is not a root of the equation.

Substituting x = -1.5

Let's substitute x=−1.5x = -1.5 into the equation.

f(−1.5)=2(−1.5)3−20(−1.5)2+15(−1.5)+18f(-1.5) = 2(-1.5)^3 - 20(-1.5)^2 + 15(-1.5) + 18

f(−1.5)=2(−3.375)−20(2.25)−22.5+18f(-1.5) = 2(-3.375) - 20(2.25) - 22.5 + 18

f(−1.5)=−6.75−45−22.5+18f(-1.5) = -6.75 - 45 - 22.5 + 18

f(−1.5)=−56.25f(-1.5) = -56.25

Since f(−1.5)≠0f(-1.5) \neq 0, x=−1.5x = -1.5 is not a root of the equation.

Substituting x = 1.5

Let's substitute x=1.5x = 1.5 into the equation.

f(1.5)=2(1.5)3−20(1.5)2+15(1.5)+18f(1.5) = 2(1.5)^3 - 20(1.5)^2 + 15(1.5) + 18

f(1.5)=2(3.375)−20(2.25)+22.5+18f(1.5) = 2(3.375) - 20(2.25) + 22.5 + 18

f(1.5)=6.75−45+22.5+18f(1.5) = 6.75 - 45 + 22.5 + 18

f(1.5)=2.25f(1.5) = 2.25

Since f(1.5)≠0f(1.5) \neq 0, x=1.5x = 1.5 is not a root of the equation.

Substituting x = 2

Let's substitute x=2x = 2 into the equation.

f(2)=2(2)3−20(2)2+15(2)+18f(2) = 2(2)^3 - 20(2)^2 + 15(2) + 18

f(2)=2(8)−20(4)+30+18f(2) = 2(8) - 20(4) + 30 + 18

f(2)=16−80+30+18f(2) = 16 - 80 + 30 + 18

f(2)=−16f(2) = -16

Since f(2)≠0f(2) \neq 0, x=2x = 2 is not a root of the equation.

Substituting x = 3

Let's substitute x=3x = 3 into the equation.

f(3)=2(3)3−20(3)2+15(3)+18f(3) = 2(3)^3 - 20(3)^2 + 15(3) + 18

f(3)=2(27)−20(9)+45+18f(3) = 2(27) - 20(9) + 45 + 18

f(3)=54−180+45+18f(3) = 54 - 180 + 45 + 18

f(3)=−63f(3) = -63

Since f(3)≠0f(3) \neq 0, x=3x = 3 is not a root of the equation.

Conclusion

Unfortunately, we were unable to find any rational roots for the equation f(x)=2x3−20x2+15x+18=0f(x) = 2x^3 - 20x^2 + 15x + 18 = 0. This means that the equation does not have any rational solutions.

However, we can try to find the approximate solutions using numerical methods. One such method is the Newton-Raphson method.

Newton-Raphson Method

The Newton-Raphson method is an iterative method that uses the formula:

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

to find the roots of the equation.

Finding the Root

Let's use the Newton-Raphson method to find the root of the equation.

We start with an initial guess of x0=0x_0 = 0.

x1=x0−f(x0)f′(x0)x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

f(x0)=2(0)3−20(0)2+15(0)+18f(x_0) = 2(0)^3 - 20(0)^2 + 15(0) + 18

f(x0)=18f(x_0) = 18

f′(x0)=6x2−40x+15f'(x_0) = 6x^2 - 40x + 15

f′(x0)=6(0)2−40(0)+15f'(x_0) = 6(0)^2 - 40(0) + 15

f′(x0)=15f'(x_0) = 15

x1=x0−f(x0)f′(x0)x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

x1=0−1815x_1 = 0 - \frac{18}{15}

x1=−1.2x_1 = -1.2

We repeat the process with x1=−1.2x_1 = -1.2.

x2=x1−f(x1)f′(x1)x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}

f(x1)=2(−1.2)3−20(−1.2)2+15(−1.2)+18f(x_1) = 2(-1.2)^3 - 20(-1.2)^2 + 15(-1.2) + 18

f(x1)=2(−1.728)−20(1.44)−18+18f(x_1) = 2(-1.728) - 20(1.44) - 18 + 18

f(x1)=−3.456−28.8f(x_1) = -3.456 - 28.8

f(x1)=−32.256f(x_1) = -32.256

f′(x1)=6(−1.2)2−40(−1.2)+15f'(x_1) = 6(-1.2)^2 - 40(-1.2) + 15

f′(x1)=6(1.44)+48+15f'(x_1) = 6(1.44) + 48 + 15

f′(x1)=8.64+48+15f'(x_1) = 8.64 + 48 + 15

f′(x1)=71.64f'(x_1) = 71.64

x2=x1−f(x1)f′(x1)x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}

x2=−1.2−−32.25671.64x_2 = -1.2 - \frac{-32.256}{71.64}

Q: What is the function f(x) and how is it defined?

A: The function f(x) is defined as the product of two expressions: (2x + 3) and (x^2 - 5x + 6). To find the value of f(x), we need to multiply these two expressions together.

Q: How do I expand the function f(x)?

A: To expand the function f(x), we multiply each term in the first expression by each term in the second expression.

Q: What are the possible rational roots of the equation f(x) = 0?

A: The possible rational roots are: ±1, ±2, ±3, ±6, ±9, ±18, ±1/2, ±3/2, ±9/2, ±18/2.

Q: How do I use the Newton-Raphson method to find the root of the equation f(x) = 0?

A: The Newton-Raphson method is an iterative method that uses the formula: x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} to find the roots of the equation.

Q: What is the initial guess for the Newton-Raphson method?

A: The initial guess for the Newton-Raphson method is x_0 = 0.

Q: How do I find the root of the equation f(x) = 0 using the Newton-Raphson method?

A: To find the root of the equation f(x) = 0 using the Newton-Raphson method, we repeat the process of finding x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} until we converge to a root.

Q: What is the approximate value of the root of the equation f(x) = 0?

A: The approximate value of the root of the equation f(x) = 0 is x ≈ -1.2.

Q: How do I determine if a value is a root of the equation f(x) = 0?

A: To determine if a value is a root of the equation f(x) = 0, we substitute the value into the equation and check if the result is equal to zero.

Q: What are some common mistakes to avoid when solving the equation f(x) = 0?

A: Some common mistakes to avoid when solving the equation f(x) = 0 include:

  • Not checking if the value is a root of the equation
  • Not using the correct method to find the root (e.g. Newton-Raphson method)
  • Not repeating the process of finding the root until convergence
  • Not checking if the approximate value is accurate

Q: How do I verify the accuracy of the approximate value of the root?

A: To verify the accuracy of the approximate value of the root, we can use numerical methods such as the Newton-Raphson method or other root-finding algorithms.

Q: What are some real-world applications of solving the equation f(x) = 0?

A: Some real-world applications of solving the equation f(x) = 0 include:

  • Finding the maximum or minimum of a function
  • Determining the stability of a system
  • Modeling population growth or decline
  • Solving optimization problems

Q: How do I choose the correct method to solve the equation f(x) = 0?

A: To choose the correct method to solve the equation f(x) = 0, we need to consider the type of equation, the number of roots, and the desired level of accuracy. Some common methods include:

  • Factoring
  • Quadratic formula
  • Newton-Raphson method
  • Other root-finding algorithms

Q: What are some common challenges when solving the equation f(x) = 0?

A: Some common challenges when solving the equation f(x) = 0 include:

  • Finding the correct method to solve the equation
  • Determining the number of roots
  • Ensuring the accuracy of the approximate value
  • Dealing with complex or irrational roots

Q: How do I overcome these challenges?

A: To overcome these challenges, we need to:

  • Choose the correct method to solve the equation
  • Use numerical methods or other root-finding algorithms
  • Verify the accuracy of the approximate value
  • Consider the type of equation and the desired level of accuracy

Q: What are some additional resources for learning more about solving the equation f(x) = 0?

A: Some additional resources for learning more about solving the equation f(x) = 0 include:

  • Online tutorials and videos
  • Textbooks and reference books
  • Online forums and communities
  • Professional development courses and workshops