Factoring Quadratics When a is not 1, Part 1
As you saw in the first section on factoring, we will learn several ways to factor. That’s just the beginning, the minimal understanding required to move forward. As you get further in math, more advanced factoring methods will be introduced. But for now, these basics are most important.
One of the trickiest things is to remember the methods learned first. They’re often forgotten. In this section we will be combining two methods at once, as you’ll often be doing when factoring. In this section we will factor quadratics expressions where the leading coefficient is NOT 1, but there is a common factor between all terms.
Read through the notes, taking notes yourself. Then, watch the video and try the practice problems. Give yourself time to think and learn, don’t rush. Education is an investment in yourself for an unknown, future pay-off!
In this section, part 3, we do not introduce a new method, but instead a combination of the previous two methods learned so far. Our first method of factoring involved factoring out a constant, or a monomial, from a larger polynomial. Let’s see an example.
4x2 – 100
Here, both terms have a factor of 4. If we factor that out, we have the following.
4x2 – 100 = 4(x2 – 25)
The second thing we learned was about factoring quadratic expressions (ax2 + bx + c), where a = 1 (the leading coefficient is one). Because of how the patterns in multiplying binomials works, in order to factor we find the two values that multiply to make c and also add to b. Let’s see an example.
x2 – 25
Here, a = 1, b = 0, and c = – 25. We need to find the values that multiply to negative twenty-five and also add to zero. These values are the last terms in our binomials whose product is x2 – 25. The values are of course 5 and – 5.
x2 – 25 = (x + 5)(x – 5)
Today we will be combining those two processes. There is not a single factoring method that always works, nor a combination. However, one should always try to factor out a monomial or constant term first as it will make the terms smaller value, which is easier to manage. That’s what we’re doing here really.
Let’s re-examine our first example, as it contained our second example, as you probably realized.
4x2 – 100 = 4(x2 – 25)
4(x2 – 25) = 4(x + 5)(x – 5)
Putting it all together, we get:
4x2 – 100 = 4(x + 5)(x – 5)
This is a good time to remind yourself what you’re really doing when factoring.
Factoring breaks a polynomial down into its multiplicative parts.
You’re done factoring when you cannot break the polynomial down further. After learning the first method of factoring you would not be able to further factor 4(x2 – 25), but now you can.
You’re factoring is correct when you can re-distribute the factors and arrive at the original polynomial.
Let’s see one more example.
5x2 – 10x + 5
Here, we have a constant factor of 5, five can be divided evenly from each of the terms.
5(x2 – 2x + 1)
Our second term is a trinomial, and it is quadratic. This quadratic has a leading coefficient of 1. So, if we can find a pair of numbers whose product is c, (positive 1), and whose sum is b, (negative 2), then this can be further factored.
5(x + 1)(x – 1)
These two procedures cannot always be combined. Not all quadratic expressions are factorable. For that matter, it is NOT always the case that you can factor out a constant term when a ≠ 1. Do not get fooled into thinking the pattern always exists just because it does in the problems you’ll be practicing. The problems are designed to work so you get a feel for how this pattern does work. But the patterns change.
Coming Soon
