This article explains how to factor a third degree polynomial. We will explore how to factor with recollection and with the factors of the known term.
Steps
Part 1 of 2: Factoring by collection
Step 1. Group the polynomial into two parts:
this will allow us to address each part separately.
Suppose we are working with the polynomial x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)
Step 2. In each part, find the common factor
- In the case of (x3 + 3x2), x2 is the common factor.
- In the case of (- 6x - 18), -6 is the common factor.
Step 3. Collect the common parts outside the two terms
- By collecting x2 in the first section, we will get x2(x + 3).
- Collecting -6, we will have -6 (x + 3).
Step 4. If each of the two terms contains the same factor, you can combine the factors together
This will give (x + 3) (x2 - 6).
Step 5. Find the solution by considering the roots
If you have x in the roots2, remember that both negative and positive numbers satisfy that equation.
The solutions are 3 and √6
Part 2 of 2: Factoring using the known term
Step 1. Rewrite the expression so that it is in the form aX3+ bX2+ cX+ d.
Suppose we work with the equation: x3 - 4x2 - 7x + 10 = 0.
Step 2. Find all the factors of d
The constant d is that number which is not associated with any variable.
Factors are those numbers that when multiplied together give another number. In our case, the factors of 10, or d, are: 1, 2, 5, and 10
Step 3. Find a factor that makes the polynomial equal to zero
We want to establish what is the factor that, substituted for x in the equation, makes the polynomial equal to zero.
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Let's start with the factor 1. We substitute 1 in all x of the equation:
(1)3 - 4(1)2 - 7(1) + 10 = 0
- It follows that: 1 - 4 - 7 + 10 = 0.
- Since 0 = 0 is a true statement, then we know that x = 1 is the solution.
Step 4. Fix things up a bit
If x = 1, we can change the statement a little to make it seem a little different without changing its meaning.
x = 1 is the same as saying x - 1 = 0 or (x - 1). We simply subtracted 1 from both sides of the equation
Step 5. Factor the root of the rest of the equation
Our root is "(x - 1)". Let's see if it's possible to collect it outside of the rest of the equation. Let's consider one polynomial at a time.
- It is possible to collect (x - 1) from x3? No, it's not possible. We can, however, take -x2 from the second variable; now we can factor it into factors: x2(x - 1) = x3 - x2.
- Is it possible to collect (x - 1) from what remains of the second variable? No, it's not possible. We need to take something from the third variable again. We take 3x from -7x.
- This will give -3x (x - 1) = -3x2 + 3x.
- Since we took 3x from -7x, the third variable will now be -10x and the constant will be 10. Can we factor that out? Yes, it is possible! -10 (x - 1) = -10x + 10.
- What we did was rearrange the variables so that we could collect (x - 1) across the equation. Here is the modified equation: x3 - x2 - 3x2 + 3x - 10x + 10 = 0, but it is the same as x3 - 4x2 - 7x + 10 = 0.
Step 6. Continue to substitute for the known term factors
Consider the numbers we factored using (x - 1) in step 5:
- x2(x - 1) - 3x (x - 1) - 10 (x - 1) = 0. We can rewrite to make factoring easier: (x - 1) (x2 - 3x - 10) = 0.
- Here we are trying to factor (x2 - 3x - 10). The decomposition will be (x + 2) (x - 5).
Step 7. The solutions will be the factored roots
To check if the solutions are correct, you can enter them one at a time in the original equation.
- (x - 1) (x + 2) (x - 5) = 0 The solutions are 1, -2, and 5.
- Insert -2 into the equation: (-2)3 - 4(-2)2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.
- Put 5 in the equation: (5)3 - 4(5)2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.
Advice
- A cubic polynomial is the product of three first-degree polynomials or the product of one first-degree polynomial and another second-degree polynomial that cannot be factored. In the latter case, to find the second degree polynomial, we use a long division once we have found the first degree polynomial.
- There are no non-decomposable cubic polynomials between real numbers, since every cubic polynomial must have a real root. Cubic polynomials such as x ^ 3 + x + 1 that have an irrational real root cannot be factored into polynomials with integer or rational coefficients. Although it can be factored with the cubic formula, it is irreducible as an integer polynomial.
