How to Factor Third Degree Polynomials
Third degree polynomials, also known as cubic polynomials, are a fundamental concept in algebra. These polynomials have the general form of ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and a is not equal to zero. Factoring third degree polynomials can be a challenging task, but with the right techniques and understanding, it can be achieved. In this article, we will explore various methods to factor third degree polynomials, including the Rational Root Theorem, synthetic division, and the cubic formula.
Using the Rational Root Theorem
The Rational Root Theorem is a valuable tool for finding possible rational roots of a cubic polynomial. According to the theorem, any rational root of the form p/q, where p is a factor of the constant term (d) and q is a factor of the leading coefficient (a), must be a root of the polynomial. By substituting these possible roots into the polynomial, we can determine if any of them are actual roots. Once we find a root, we can use synthetic division to factor out the corresponding linear factor.
Example: Factor the polynomial x^3 – 6x^2 + 11x – 6
To factor the polynomial x^3 – 6x^2 + 11x – 6, we first apply the Rational Root Theorem. The constant term is -6, and its factors are ±1, ±2, ±3, and ±6. The leading coefficient is 1, and its only factor is 1. Therefore, the possible rational roots are ±1, ±2, ±3, and ±6.
By substituting these values into the polynomial, we find that x = 1 is a root. Now, we can use synthetic division to factor out (x – 1):
1 | 1 -6 11 -6
| 1 -5 6
—————-
| 1 -5 6 0
The result shows that (x – 1) is a factor of the polynomial. Now we have a quadratic polynomial, x^2 – 5x + 6, which can be factored further.
Using Synthetic Division
Synthetic division is a simplified method for dividing a polynomial by a linear factor. By using synthetic division, we can easily factor out a linear factor from a cubic polynomial. To perform synthetic division, we list the coefficients of the polynomial in descending order, write the root as a divisor, and perform the division.
Example: Factor the polynomial x^3 – 6x^2 + 11x – 6 using synthetic division
Continuing with the previous example, we have found that x = 1 is a root of the polynomial x^3 – 6x^2 + 11x – 6. To factor out (x – 1), we perform synthetic division:
1 | 1 -6 11 -6
| 1 -5 6
—————-
| 1 -5 6 0
The result shows that (x – 1) is a factor of the polynomial. Now we have a quadratic polynomial, x^2 – 5x + 6, which can be factored further.
Using the Cubic Formula
The cubic formula is a general method for finding the roots of a cubic polynomial. It is a complex formula, but it can be used to find all three roots of a cubic polynomial, including irrational and complex roots. The cubic formula is as follows:
x = (-b ± √(b^2 – 4ac)) / (2a)
where a, b, and c are the coefficients of the cubic polynomial ax^3 + bx^2 + cx + d.
Example: Factor the polynomial x^3 – 6x^2 + 11x – 6 using the cubic formula
To factor the polynomial x^3 – 6x^2 + 11x – 6 using the cubic formula, we first identify the coefficients: a = 1, b = -6, and c = 11. Substituting these values into the formula, we get:
x = (-(-6) ± √((-6)^2 – 4(1)(11))) / (2(1))
x = (6 ± √(36 – 44)) / 2
x = (6 ± √(-8)) / 2
x = (6 ± 2√2i) / 2
x = 3 ± √2i
The cubic formula gives us three roots: x = 3 + √2i, x = 3 – √2i, and x = 1. By factoring out these roots, we can express the cubic polynomial as a product of linear factors:
x^3 – 6x^2 + 11x – 6 = (x – 1)(x – (3 + √2i))(x – (3 – √2i))
In conclusion, factoring third degree polynomials can be achieved through various methods, including the Rational Root Theorem, synthetic division, and the cubic formula. By understanding these techniques and applying them appropriately, we can successfully factor cubic polynomials and gain a deeper understanding of algebraic concepts.
