Algebra: Factoring Polynomials Part 2
Factoring Polynomials Part 2
This algebra video tutorial is the second part of two videos on factoring polynomials. In part 1, we explained the basic ideas of factorization, defined some terminology, and demonstrated the method for factoring monic, quadratic polynomials. In part 2, we demonstrate the method for factoring non-monic quadratic polynomials, give an example of how to apply our product formulas, and use the idea of partial factoring to demonstrate how to factor general polynomials where no obvious method exists.
For the method of factoring non-monic quadratics, we first look at the general product of two linear polynomials. The methology for factoring comes out of this through the process of reversing the direction of the product formula. To do this, we note that the linear coefficient is the sum of two terms whose product is equal to the product of the quadratic coefficient and the constant term.
The relationship that we noted above, suggests the methodolgy for factoring non-monic quadratic polynomials, which is comprised of six steps:
- Multiply the quadratic coefficient and the constant term to get the product abcd.
- Find the pairs of integers that whose product is abcd.
- If possible, find the integer pair that has its sum equal to the linear coeffient. If the pair does not exist, the polynomial is irreducible over the integers.
- If an integer pair exists, break up the linear term into a sum with each of the integers as a coefficient.
- Then factor the first two terms and the last two terms separately.
- The sum in each of the prior factors is the same and can be factored out via distributivity to finish the factorization.
As an example of non-monic quadratic factoring, consider the polynomial 6x² + 19x + 15. Note that we can not factor out any part of the quadratic coefficient 6 to reduce the polynomial. So, we multiply the quadratic coefficient 6 and the constant term 15 to get 90. Next, we look at the integer pairs that have the product 90. For this, we have
(1, 90), (2, 45), (3, 30), (5, 18), (6, 15), (9, 10)
and the negative of these
(-1, -90), (-2, -45), (-3, -30), (-5, -18), (-6, -15), (-9, -10)
These pairs have the sums 91, 47, 33, 23, 21, and 19 and the negative of these values. From these, we select the pair that has its sum equal to the linear coefficient 19. This is (9, 10). So, we split the linear term into 9 + 10 to get 6x² + 9x + 10x + 15. Then we factor the first two terms and the last two terms to get 3x(2x + 3) + 5(2x + 3). Finally, we factor out (2x + 3) to get (3x + 5)(2x + 3).
You might be wondering, "How do I know which term to write first when I break up the linear coefficient?" The anwer is that it does not matter. If we had written 6x² + 10x + 9x + 15 above instead, then we would factor the pairs of terms as 2x(3x + 5) + 3(3x + 5), which would then factor as (2x + 3)(3x + 5). This is the same thing that we had, except that the positions of the terms are swapped.
In addition to the methods that we demonstrated so far, we can use any of our product formulas to factor polynomials. For example, if we have the polynomial 9x² - 25, we could factor this by using the non-monic quadratic method that we just explained. However, it is much easier to notice that this is a difference of squares. So, it can be factored using the diffeerence of squares formula, where the terms that are squared are 3x and 5. Hence, we have (3x + 5)(3x - 5)
We have talked about some specific methods of factoring and the product formulas, but when these methods do not work, it may be difficult to see how to factor a polynomial. In these cases, we can try to factor a few terms at a time and see if a pattern emerges.
For example, if we have the polynomial 15x³ + 12x² + 10x + 8, we can try to factor the first two and last two terms, separately. This gives us 3x²(5x + 4) + 2(5x + 4). Now we can see that there is a common (5x + 4) factor. Taking this out, we have (3x² + 2)(5x + 4).
We could try to factor the quadratic term 3x² + 2 further, using the non-monic quadratic method. However, a quick check will tell you that this polynomial is irredicible.