Factoring the Cubic Expression

Oct 13, 2025 by sorusgentrification.com 111 views

Introduction

In mathematics, solving cubic equations is a fundamental problem that has been studied for centuries. Given a cubic function $f(x) = x^3 + 3x^2 + 4x - 12$ , we are tasked with rewriting the equation $f(x) = 0$ in a specific form and finding the constants $a$ and $b$ . This problem requires a deep understanding of algebraic manipulations and the properties of cubic equations.

The Cubic Function

The given cubic function is $f(x) = x^3 + 3x^2 + 4x - 12$ . To rewrite this equation in the desired form, we need to factorize the cubic expression. We can start by factoring out the greatest common factor (GCF) of the terms.

f(x) = x^3 + 3x^2 + 4x - 12

Factoring the Cubic Expression

We can factor out the GCF, which is $x - 3$ , from the first three terms.

f(x) = (x - 3)(x^2 + 6x + 4)

Now, we need to factorize the quadratic expression $x^2 + 6x + 4$ . We can use the quadratic formula to find the roots of this expression.

x^2 + 6x + 4 = (x + 3)^2 - 5

Substituting this expression back into the factored form of $f(x)$ , we get:

f(x) = (x - 3)((x + 3)^2 - 5)

Rewriting the Equation

Now, we can rewrite the equation $f(x) = 0$ in the desired form. We start by setting the factored expression equal to zero:

(x - 3)((x + 3)^2 - 5) = 0

This equation is true if either $(x - 3) = 0$ or $((x + 3)^2 - 5) = 0$ . We can solve the first equation to find the root $x = 3$ .

x - 3 = 0 \Rightarrow x = 3

For the second equation, we can expand the squared term and simplify:

(x + 3)^2 - 5 = 0

Expanding the squared term, we get:

x^2 + 6x + 9 - 5 = 0

Simplifying, we get:

x^2 + 6x + 4 = 0

This is a quadratic equation, and we can solve it using the quadratic formula:

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

In this case, $a = 1$ , $b = 6$ , and $c = 4$ . Substituting these values into the quadratic formula, we get:

x = \frac{-6 \pm \sqrt{6^2 - 4(1)(4)}}{2(1)}

Simplifying, we get:

x = \frac{-6 \pm \sqrt{36 - 16}}{2}

Simplifying further, we get:

x = \frac{-6 \pm \sqrt{20}}{2}

Simplifying again, we get:

x = \frac{-6 \pm 2\sqrt{5}}{2}

Simplifying once more, we get:

x = -3 \pm \sqrt{5}

Now, we can rewrite the equation $f(x) = 0$ in the desired form:

x = \sqrt{\left(\frac{a(b-x)}{(3+x)}\right)}

where $x \neq -3$ . To find the constants $a$ and $b$ , we can equate the two expressions:

\sqrt{\left(\frac{a(b-x)}{(3+x)}\right)} = -3 \pm \sqrt{5}

Squaring both sides, we get:

\frac{a(b-x)}{(3+x)} = (-3 \pm \sqrt{5})^2

Simplifying, we get:

\frac{a(b-x)}{(3+x)} = 9 \pm 6\sqrt{5} + 5

Simplifying further, we get:

\frac{a(b-x)}{(3+x)} = 14 \pm 6\sqrt{5}

Now, we can solve for $a$ and $b$ by equating the two expressions:

a(b-x) = (14 \pm 6\sqrt{5})(3+x)

Expanding the right-hand side, we get:

a(b-x) = 42 + 14x \pm 18\sqrt{5} + 6x\sqrt{5}

Simplifying, we get:

a(b-x) = (42 + 20x) \pm (18\sqrt{5} + 6x\sqrt{5})

Now, we can equate the coefficients of the two expressions:

a(b-x) = 42 + 20x

Simplifying, we get:

ab - ax = 42 + 20x

Simplifying further, we get:

ab - ax - 20x = 42

Simplifying once more, we get:

ab - (a + 20)x = 42

Now, we can equate the coefficients of the two expressions:

a + 20 = 0

Solving for $a$ , we get:

a = -20

Substituting this value of $a$ back into the equation, we get:

-20b - (a + 20)x = 42

Simplifying, we get:

-20b - (-20 + 20)x = 42

Simplifying further, we get:

-20b = 42

Solving for $b$ , we get:

b = -\frac{42}{20}

Simplifying, we get:

b = -\frac{21}{10}

Therefore, the constants $a$ and $b$ are $a = -20$ and $b = -\frac{21}{10}$ .

Conclusion

Q: What is the main goal of rewriting the cubic equation?

A: The main goal of rewriting the cubic equation is to express it in a specific form, which is $x = \sqrt{\left(\frac{a(b-x)}{(3+x)}\right)}$ where $x \neq -3$ . This form is useful for solving the equation and finding the constants $a$ and $b$ .

Q: How do you factorize the cubic expression?

A: To factorize the cubic expression, we can start by factoring out the greatest common factor (GCF) of the terms. In this case, we can factor out the GCF, which is $x - 3$ , from the first three terms.

Q: What is the quadratic formula, and how is it used to solve the quadratic equation?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:

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

In this case, we can use the quadratic formula to solve the quadratic equation $x^2 + 6x + 4 = 0$ .

Q: How do you find the constants $a$ and $b$ ?

A: To find the constants $a$ and $b$ , we can equate the two expressions:

\sqrt{\left(\frac{a(b-x)}{(3+x)}\right)} = -3 \pm \sqrt{5}

Squaring both sides, we get:

\frac{a(b-x)}{(3+x)} = (-3 \pm \sqrt{5})^2

Simplifying, we get:

\frac{a(b-x)}{(3+x)} = 14 \pm 6\sqrt{5}

Now, we can solve for $a$ and $b$ by equating the two expressions:

a(b-x) = (14 \pm 6\sqrt{5})(3+x)

Q: What are the values of $a$ and $b$ ?

A: The values of $a$ and $b$ are $a = -20$ and $b = -\frac{21}{10}$ .

Q: What is the significance of the condition $x \neq -3$ ?

A: The condition $x \neq -3$ is important because it ensures that the expression $\frac{a(b-x)}{(3+x)}$ is defined. If $x = -3$ , then the denominator $(3+x)$ would be equal to zero, which would make the expression undefined.

Q: Can you provide an example of how to use the rewritten equation to solve a problem?

A: Yes, here is an example of how to use the rewritten equation to solve a problem:

Suppose we want to find the value of $x$ that satisfies the equation $f(x) = 0$ , where $f(x) = x^3 + 3x^2 + 4x - 12$ . We can use the rewritten equation to solve for $x$ :

x = \sqrt{\left(\frac{a(b-x)}{(3+x)}\right)}

Substituting the values of $a$ and $b$ , we get:

x = \sqrt{\left(\frac{-20\left(-\frac{21}{10}-x\right)}{(3+x)}\right)}

Simplifying, we get:

x = \sqrt{\left(\frac{420 + 20x}{3 + x}\right)}

Now, we can solve for $x$ by equating the two expressions:

\sqrt{\left(\frac{420 + 20x}{3 + x}\right)} = 0

Squaring both sides, we get:

\frac{420 + 20x}{3 + x} = 0

Simplifying, we get:

420 + 20x = 0

Solving for $x$ , we get:

x = -\frac{420}{20}

Simplifying, we get:

x = -21

Therefore, the value of $x$ that satisfies the equation $f(x) = 0$ is $x = -21$ .

Conclusion

In this article, we have provided a Q&A section to help readers understand the concepts and techniques used to rewrite the cubic equation. We have also provided an example of how to use the rewritten equation to solve a problem. We hope that this article has been helpful in clarifying the concepts and techniques used in rewriting the cubic equation.