Finding Roots Of Cubic Function Given A Factor A Step By Step Guide

Jul 11, 2025 by Admin 68 views

Finding All Roots of the Cubic Function f(x) = 5x³ + 5x² - 170x + 280 Given a Factor

In this article, we will explore how to find all the roots of the cubic function f(x) = 5x³ + 5x² - 170x + 280, given that one of its factors is (x + 7). This problem combines polynomial factorization, synthetic division, and solving quadratic equations. By understanding these techniques, you can efficiently solve similar problems in algebra and calculus. Let's dive into the step-by-step solution to uncover the roots of this function.

Understanding the Problem

Before we delve into the solution, let’s clarify the core concepts. The roots of a function f(x) are the values of x for which f(x) = 0. In other words, they are the x-intercepts of the function's graph. Given that (x + 7) is a factor of the cubic function f(x) = 5x³ + 5x² - 170x + 280, this means that x = -7 is one of the roots. Our task is to find the remaining roots. To do this, we will use synthetic division to reduce the cubic function to a quadratic function, and then solve the quadratic equation to find the other two roots. Understanding the relationship between factors and roots is crucial in polynomial algebra. The factor theorem states that if (x - a) is a factor of a polynomial f(x), then f(a) = 0, meaning a is a root of the polynomial. Conversely, if a is a root of f(x), then (x - a) is a factor. This theorem is the foundation for our solution. We will also leverage the quadratic formula to solve the resulting quadratic equation, which is a fundamental tool for finding the roots of any quadratic function in the form ax² + bx + c = 0. By mastering these concepts, you will be well-equipped to tackle various polynomial problems.

Step 1: Using Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form (x - a). In our case, we know that (x + 7) is a factor, so a = -7. We will use synthetic division to divide f(x) = 5x³ + 5x² - 170x + 280 by (x + 7). The coefficients of the cubic function are 5, 5, -170, and 280. We set up the synthetic division as follows:

-7 | 5   5   -170   280
   |______________________

Now, we perform the synthetic division:

Bring down the first coefficient (5).

-7 | 5   5   -170   280
   |______________________
     5

Multiply the number we brought down (5) by -7 and write the result (-35) under the next coefficient (5).

-7 | 5   5   -170   280
   |     -35
   |______________________
     5

Add the numbers in the second column (5 + (-35) = -30).

-7 | 5   5   -170   280
   |     -35
   |______________________
     5  -30

Multiply the result (-30) by -7 and write it under the next coefficient (-170). (-30 * -7 = 210)

-7 | 5   5   -170   280
   |     -35   210
   |______________________
     5  -30

Add the numbers in the third column (-170 + 210 = 40).

-7 | 5   5   -170   280
   |     -35   210
   |______________________
     5  -30   40

Multiply the result (40) by -7 and write it under the last coefficient (280). (40 * -7 = -280)

-7 | 5   5   -170   280
   |     -35   210   -280
   |______________________
     5  -30   40

Add the numbers in the last column (280 + (-280) = 0).

-7 | 5   5   -170   280
   |     -35   210   -280
   |______________________
     5  -30   40    0

The result of the synthetic division is 5, -30, and 40, which corresponds to the quadratic equation 5x² - 30x + 40. The remainder is 0, which confirms that (x + 7) is indeed a factor of the original cubic function.

Step 2: Forming the Quadratic Equation

After performing synthetic division, we have reduced the cubic function to a quadratic function. The coefficients from the synthetic division result give us the quadratic equation: 5x² - 30x + 40 = 0. This equation represents the quotient when the original cubic function f(x) = 5x³ + 5x² - 170x + 280 is divided by (x + 7). Solving this quadratic equation will give us the remaining roots of the original cubic function. To simplify the equation, we can divide all terms by the common factor of 5:

5x² - 30x + 40 = 0

Divide by 5:

x² - 6x + 8 = 0

Now we have a simplified quadratic equation that is easier to solve. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is a straightforward approach, but we will also explore the quadratic formula for completeness. Factoring involves finding two numbers that multiply to the constant term (8) and add up to the coefficient of the linear term (-6). These numbers will allow us to rewrite the quadratic expression as a product of two binomials. The quadratic formula, on the other hand, is a general solution that can be applied to any quadratic equation, regardless of whether it can be easily factored. Understanding both methods provides a comprehensive toolkit for solving quadratic equations.

Step 3: Solving the Quadratic Equation

We have the quadratic equation x² - 6x + 8 = 0. Let's solve this equation using two methods: factoring and the quadratic formula.

Method 1: Factoring

We need to find two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4, since (-2) * (-4) = 8 and (-2) + (-4) = -6. Therefore, we can factor the quadratic equation as follows:

(x - 2)(x - 4) = 0

Now, set each factor equal to zero and solve for x:

x - 2 = 0 or x - 4 = 0

x = 2 or x = 4

Thus, the roots of the quadratic equation are x = 2 and x = 4.

Method 2: Quadratic Formula

The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation x² - 6x + 8 = 0, a = 1, b = -6, and c = 8. Plugging these values into the quadratic formula, we get:

x = (-(-6) ± √((-6)² - 4 * 1 * 8)) / (2 * 1)

x = (6 ± √(36 - 32)) / 2

x = (6 ± √4) / 2

x = (6 ± 2) / 2

This gives us two solutions:

x = (6 + 2) / 2 = 8 / 2 = 4

x = (6 - 2) / 2 = 4 / 2 = 2

Again, we find that the roots are x = 2 and x = 4. Both factoring and the quadratic formula confirm the same roots for the quadratic equation. Factoring is often quicker when the quadratic equation can be easily factored, while the quadratic formula is a reliable method for any quadratic equation, especially those that are difficult to factor. Understanding both methods allows you to choose the most efficient approach for each problem.

Step 4: Combining the Roots

We have found the roots of the quadratic equation resulting from the synthetic division. These roots, combined with the root we were initially given, will provide all the roots of the original cubic function f(x) = 5x³ + 5x² - 170x + 280. We know that (x + 7) is a factor, which gives us the root x = -7. The quadratic equation x² - 6x + 8 = 0 yielded the roots x = 2 and x = 4. Therefore, the complete set of roots for the cubic function is x = -7, x = 2, and x = 4. To verify this, we can substitute each root back into the original function and confirm that the result is zero. For example, substituting x = -7 into f(x) should give us zero, since (x + 7) is a factor. Similarly, substituting x = 2 and x = 4 should also result in f(x) = 0, confirming that they are indeed roots of the function. This step of combining the roots from the linear factor and the quadratic equation is crucial in solving higher-degree polynomial equations. By systematically breaking down the problem using synthetic division and factoring (or the quadratic formula), we can find all the solutions and gain a deeper understanding of the polynomial's behavior.

Final Answer

The roots of the function f(x) = 5x³ + 5x² - 170x + 280 are x = -7, x = 2, and x = 4. Thus, the correct answer is B. x=-7, x=2, or x=4.

By following the steps outlined in this article, you can confidently solve similar polynomial problems. Remember to utilize synthetic division to reduce the polynomial's degree, form the resulting quadratic equation, and solve it using either factoring or the quadratic formula. Combining these techniques will allow you to efficiently find all the roots of a polynomial function.