Back To CourseRemedial Algebra I
25 chapters | 248 lessons | 1 flashcard set
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Amy has a master's degree in secondary education and has taught math at a public charter high school.
What are cubic equations? Cubic equations are those equations whose highest degree is 3, meaning the highest power or exponent is 3. Why 3? Well, think of a cube. How do you find the volume of a cube? Because all the sides are the same length, you cube one of the sides, meaning you take the side to the third power.
So if we had a cube that measured 6 inches on every side, then our volume would be 6 cubed, or 6 to the third power (6^3). See that little 3? Focus on the little. When you think of cubic equations, remember this little 3 that is always there for finding the volume of a cube. An example of a cubic equation is the equation: x^3 + 8x^2 + 19x + 12 = 0. Do you see the little 3?
What you are about to learn in this video about solving these kinds of equations will help you as you keep growing in your math skills. You will come across cubic equations in your problems and when trying to solve real-world physics problems. Once you are done watching this video lesson, you will have a useful method of solving the cubic equations that you will come across in life.
The process that you are about to learn does require that you know how to perform synthetic division or long division on your polynomials. If you don't already know it, take some time right now to brush up on your polynomial division skills.
The method that I am showing you uses what is called the rational roots test, which tells you that possible solutions or roots to a polynomial can be found by dividing a factor of the constant term by a factor of the number associated with the first term. I know this might sound a little confusing right now. But let me show you how it works and I am sure it will make more sense.
We are going to solve the cubic equation: x^3 + 8x^2 + 19x + 12 = 0. We start by applying the rational roots test to get a list of our possible solutions or roots. We first locate our constant term, which is 12, and our first term, which is x^3.
What is the number that is associated with the x^3? Since we don't see a number, we know that it is 1. Now we are going to make a list of our possible factors for each of the terms.
Our factors of 12 are 1, 2, 3, 4, 6, and 12. We can't forget the negative versions of all these numbers as well. So, our complete list of factors of 12 is +/- 1, +/- 2, +/- 3, +/- 4, +/- 6, and +/- 12.
Next, we need to find the factors of 1, the number associated with our first term. What numbers divide evenly into 1? There's only one number, and that is 1. So, our factors of 1 are +/- 1.
Now, according to the rational roots test, our possible solutions can be found by dividing the factors of our constant term by the factors of the number associated with the first term. So, we are going to take each of our factors of 12 and divide it by the factors of 1.
If we have more than one factor from our first term, we will also divide our factors of our constant term by that number. For example, if the number associated with our first term is 2, then the possible factors of 2 are +/- 1 and +/- 2. We would then divide each of our factors of this constant term first by +/- 1 and then again by +/- 2.
We only have the +/- 1 to worry about, so our list of possible solutions is this. These numbers can be simplified since dividing by 1 gives us our numerator. So, our simplified list is the positive and negative versions of 1, 2, 3, 4, 6, and 12.
Now that we have our list, our job is to now start plugging each of these into our equation to see which one will give us 0 as our answer. Yes, this is trial and error. Remember that we need to try both the positive and the negative versions. We only need to keep going until we find one that works.
Let's try the number 1 and see what it gives us. Plugging this into our equation we get 1^3 + 8 * 1^2 + 19 * 1 + 12 = 40. Nope! That doesn't work. It doesn't equal 0. So, we need to try another number. Well, 40 is not particularly close to 0 and I am adding everything up. So that tells me I need a negative somewhere, so let's try the number -3 next.
Plugging in the -3 I get (-3)^3 + 8 * (-3)^2 + 19 * (-3) + 12 = -27 + 8 * 9 - 57 + 12 = -27 + 72 - 57 + 12 = 0. A-ha! That works! I get a 0 for my answer. So, that tells me that -3 is one of my solutions. Remember that cubic equations will always have 3 solutions, but they may not all be real solutions.
Now that we've found one solution, let's find the other two solutions. For this, we need to divide our cubic equation by our first solution. We do this by turning our solution into a factor by writing (x - k), where k is our solution. Since -3 is our answer, we will write (x + 3). We get the plus sign since we are subtracting a negative, which turns into a positive.
So, we need to divide x^3 + 8x^2 + 19x + 12 by (x + 3). For this, we are going to use either synthetic division or long division. Using synthetic division I get this.
Recalling that when we use synthetic division, we use the solution from the factor we are dividing by. We are dividing by x + 3, so our solution is -3. So, we write this number to the far left.
The final result from dividing by synthetic division is x^2 + 5x + 4. So, to find the other two solutions we solve this new quadratic. We can solve it using any method that we are comfortable using. We can use the quadratic formula or we can solve by factoring. Let's factor this to solve.
Factoring x^2 + 5x + 4 we get (x + 4)(x + 1). From these factors, I finish solving to get x = -4 and x = -1 for my other two solutions. I am now done. My complete answer is my three solutions of x = -3, x = -4, and x = -1.
What have we learned? We learned that there is a systematic way of solving cubic equations, equations whose highest degree is 3.
The rational roots test, which tells us that possible solutions or roots to a polynomial can be found by dividing a factor of the constant term by a factor of the number associated with the first term. We apply the rational roots test to our cubic equation to find possible answers. And then we plug in each possible solution into our cubic equation until we find one that gives us a zero for our answer.
Once we've found such an answer, we take this root and use synthetic division to divide our cubic equation by this root. The result should be a quadratic, which we can finish solving by using what we know about quadratic equations. We should end up with a total of three solutions.
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Back To CourseRemedial Algebra I
25 chapters | 248 lessons | 1 flashcard set