Solve The Polynomial Equation:$2x^3 + X^2 = 32x + 16$Select The Correct Answer From Each Drop-down Menu:$x = \square, \square, \text{ Or } \square$

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Introduction

Solving polynomial equations is a fundamental concept in algebra, and it requires a deep understanding of various mathematical techniques. In this article, we will focus on solving the polynomial equation $2x^3 + x^2 = 32x + 16$. This equation is a cubic equation, which means it has a degree of three. Solving cubic equations can be challenging, but with the right approach, we can find the solutions.

Understanding the Equation

The given equation is $2x^3 + x^2 = 32x + 16$. To solve this equation, we need to isolate the variable $x$. The first step is to move all the terms to one side of the equation, so we have:

$$2x^3 + x^2 - 32x - 16 = 0$$

This equation is a cubic equation, and it has three roots. We need to find these roots to solve the equation.

Factoring the Equation

One way to solve the equation is to factor it. Factoring a cubic equation can be challenging, but we can try to find a rational root using the Rational Root Theorem. The Rational Root Theorem states that if a rational number $p/q$ is a root of the polynomial, then $p$ must be a factor of the constant term, and $q$ must be a factor of the leading coefficient.

In this case, the constant term is $-16$, and the leading coefficient is $2$. The factors of $-16$ are $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$, and the factors of $2$ are $\pm 1, \pm 2$. Therefore, the possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$.

Using Synthetic Division

To find the rational roots, we can use synthetic division. Synthetic division is a method of dividing a polynomial by a linear factor. We can use synthetic division to divide the polynomial $2x^3 + x^2 - 32x - 16$ by each of the possible rational roots.

Let's start with the rational root $x = 2$. We can use synthetic division to divide the polynomial by $x - 2$:

  2  1  -32 -16
-2 | 2  4  16
  2  3  -16

The result of the synthetic division is $2x^2 + 3x - 16$. This means that $x - 2$ is a factor of the polynomial, and we can write the polynomial as:

$$2x^3 + x^2 - 32x - 16 = (x - 2)(2x^2 + 3x - 16)$$

Finding the Remaining Roots

Now that we have factored the polynomial, we can find the remaining roots. We can use the quadratic formula to find the roots of the quadratic equation $2x^2 + 3x - 16 = 0$.

The quadratic formula is:

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

In this case, $a = 2$, $b = 3$, and $c = -16$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-16)}}{2(2)}$$

Simplifying the expression, we get:

$$x = \frac{-3 \pm \sqrt{9 + 128}}{4}$$

$$x = \frac{-3 \pm \sqrt{137}}{4}$$

Therefore, the remaining roots are $x = \frac{-3 + \sqrt{137}}{4}$ and $x = \frac{-3 - \sqrt{137}}{4}$.

Conclusion

In this article, we solved the polynomial equation $2x^3 + x^2 = 32x + 16$. We factored the polynomial using synthetic division and found the rational root $x = 2$. We then used the quadratic formula to find the remaining roots. The solutions to the equation are $x = 2$, $x = \frac{-3 + \sqrt{137}}{4}$, and $x = \frac{-3 - \sqrt{137}}{4}$.

Final Answer

The final answer is:

• $x = \boxed{2}$
• $x = \boxed{\frac{-3 + \sqrt{137}}{4}}$
• $x = \boxed{\frac{-3 - \sqrt{137}}{4}}$
  

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Introduction

In our previous article, we solved the polynomial equation $2x^3 + x^2 = 32x + 16$. We factored the polynomial using synthetic division and found the rational root $x = 2$. We then used the quadratic formula to find the remaining roots. In this article, we will answer some common questions related to solving the polynomial equation.

Q: What is the degree of the polynomial equation $2x^3 + x^2 = 32x + 16$?

A: The degree of the polynomial equation $2x^3 + x^2 = 32x + 16$ is 3, which means it is a cubic equation.

Q: How do I factor a cubic polynomial?

A: Factoring a cubic polynomial can be challenging, but we can try to find a rational root using the Rational Root Theorem. We can also use synthetic division to divide the polynomial by a linear factor.

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, then $p$ must be a factor of the constant term, and $q$ must be a factor of the leading coefficient.

Q: How do I use synthetic division to divide a polynomial?

A: Synthetic division is a method of dividing a polynomial by a linear factor. We can use synthetic division to divide the polynomial by each of the possible rational roots.

Q: What is the quadratic formula?

A: The quadratic formula is a formula for finding the roots of a quadratic equation. It is given by:

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

Q: How do I find the remaining roots of a quadratic equation?

A: We can use the quadratic formula to find the roots of a quadratic equation. We can also use factoring or other methods to find the roots.

Q: What are the solutions to the polynomial equation $2x^3 + x^2 = 32x + 16$?

A: The solutions to the polynomial equation $2x^3 + x^2 = 32x + 16$ are $x = 2$, $x = \frac{-3 + \sqrt{137}}{4}$, and $x = \frac{-3 - \sqrt{137}}{4}$.

Q: How do I check my work when solving a polynomial equation?

A: We can check our work by plugging the solutions back into the original equation. If the solutions satisfy the equation, then we have found the correct solutions.

Q: What are some common mistakes to avoid when solving a polynomial equation?

A: Some common mistakes to avoid when solving a polynomial equation include:

• Not factoring the polynomial correctly
• Not using the correct method to find the roots
• Not checking the work
• Not considering all possible solutions

Conclusion

In this article, we answered some common questions related to solving the polynomial equation $2x^3 + x^2 = 32x + 16$. We covered topics such as factoring, synthetic division, the Rational Root Theorem, and the quadratic formula. We also discussed how to check our work and avoid common mistakes.

Final Answer

The final answer is:

• $x = \boxed{2}$
• $x = \boxed{\frac{-3 + \sqrt{137}}{4}}$
• $x = \boxed{\frac{-3 - \sqrt{137}}{4}}$