Solving for x in the first factor

Oct 13, 2025 by leolavanderia.com.br 70 views

Part 1: Factoring and Zero Product Property

Introduction

Polynomial equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a polynomial equation by factoring and then using the zero product property. We will break down the process into two parts, with this part covering the factoring of the polynomial equation.

The Polynomial Equation

The given polynomial equation is:

${ 75x - 25 = 3x^3 - x^2 }$

Our goal is to rewrite this equation in factored form, which will allow us to use the zero product property to find the solutions.

Step 1: Rearrange the Equation

To begin, we need to rearrange the equation to have all the terms on one side. We can do this by subtracting $75x$ from both sides and adding $25$ to both sides:

${ 3x^3 - x^2 - 75x + 25 = 0 }$

Step 2: Factor Out Common Terms

Now, we need to factor out any common terms from the equation. In this case, we can factor out a $3$ from the first two terms and a $-25$ from the last two terms:

${ 3(x^3 - \frac{1}{3}x^2) - 25(x - 1) = 0 }$

Step 3: Factor the Quadratic Expression

Next, we need to factor the quadratic expression $x^3 - \frac{1}{3}x^2$ . We can do this by factoring out an $x^2$ term:

${ x^2(x - \frac{1}{3}) - 25(x - 1) = 0 }$

Step 4: Factor the Difference of Squares

Now, we need to factor the difference of squares $x^2 - 25$ . We can do this by factoring it as $(x + 5)(x - 5)$ :

${ (x + 5)(x - 5)(x - \frac{1}{3}) - 25(x - 1) = 0 }$

Step 5: Factor the Final Expression

Finally, we need to factor the final expression $(x - 5)(x - \frac{1}{3}) - 25(x - 1)$ . We can do this by factoring it as $(x - 5)(x - \frac{1}{3} - 25)$ :

${ (x - 5)(x - \frac{1}{3} - 25)(x - 1) = 0 }$

Conclusion

In this part, we have successfully factored the polynomial equation $75x - 25 = 3x^3 - x^2$ into the form $(x - 5)(x - \frac{1}{3} - 25)(x - 1) = 0$ . This will allow us to use the zero product property to find the solutions in the next part.

Part 2: Using the Zero Product Property

In the next part, we will use the zero product property to find the solutions of the equation $(x - 5)(x - \frac{1}{3} - 25)(x - 1) = 0$ . We will set each factor equal to zero and solve for $x$ .

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 this case, we have three factors:

$(x - 5) = 0$
$(x - \frac{1}{3} - 25) = 0$
$(x - 1) = 0$

We will set each of these factors equal to zero and solve for $x$ .

Solving for x

Solving for x in the first factor

We set the first factor equal to zero:

${ x - 5 = 0 }$

Solving for $x$ , we get:

${ x = 5 }$

Solving for x in the second factor

We set the second factor equal to zero:

${ x - \frac{1}{3} - 25 = 0 }$

Solving for $x$ , we get:

${ x = \frac{1}{3} + 25 }$

${ x = \frac{76}{3} }$

Solving for x in the third factor

We set the third factor equal to zero:

${ x - 1 = 0 }$

Solving for $x$ , we get:

${ x = 1 }$

Conclusion

Part 2: Using the Zero Product Property

Q&A: Solving Polynomial Equations

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.

Q: How do I apply the zero product property to solve a polynomial equation?

A: To apply the zero product property, you need to set each factor equal to zero and solve for x. This will give you the solutions to the polynomial equation.

Q: What are some common mistakes to avoid when using the zero product property?

A: Some common mistakes to avoid when using the zero product property include:

Not setting each factor equal to zero
Not solving for x in each factor
Not checking for extraneous solutions

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you need to plug each solution back into the original equation and check if it is true. If the solution does not satisfy the original equation, then it is an extraneous solution.

Q: What are some tips for factoring polynomial equations?

A: Some tips for factoring polynomial equations include:

Look for common factors
Look for differences of squares
Look for perfect squares
Use the distributive property to expand the equation

Q: How do I know if a polynomial equation can be factored?

A: A polynomial equation can be factored if it can be written as a product of two or more binomials. If the equation cannot be written as a product of binomials, then it cannot be factored.

Q: What are some common types of polynomial equations?

A: Some common types of polynomial equations include:

Linear equations
Quadratic equations
Cubic equations
Quartic equations

Q: How do I solve a polynomial equation with multiple solutions?

A: To solve a polynomial equation with multiple solutions, you need to set each factor equal to zero and solve for x. This will give you the solutions to the polynomial equation.

Conclusion

In this article, we have provided a step-by-step guide to solving polynomial equations using the zero product property. We have also answered some common questions about solving polynomial equations, including how to apply the zero product property, how to check for extraneous solutions, and how to factor polynomial equations.

Additional Resources

Practice Problems

Solve the polynomial equation $x^2 + 5x + 6 = 0$ using the zero product property.
Factor the polynomial equation $x^3 + 2x^2 - 7x - 12 = 0$ .
Solve the polynomial equation $x^4 - 4x^3 + 4x^2 - 4x + 1 = 0$ using the zero product property.

Answer Key

$x = -2$ or $x = -3$
$(x + 1)(x - 2)(x + 3) = 0$
$x = 1$