Factoring a Quadratic Polynomial, Part 2
In this section, we're going to look at the "general" quadratic equation. In our first discussion of this kind of polynomial, we looked only at the case where the first or "leading" coefficient was one. That's a special case that's relatively easy to handle. In this lesson, we'll look at the more general case we're the leading coefficient can be any number.
There are two basic approaches to this sort of problem. The first is similar to the method we used in earlier discussion: You look at all the factors of the first coefficient and all the factors of the last coefficient and try to find a pair that, when you multiply them together and then add the results, you get the middle term. There are often a lot of possibilities that you have to consider with that approach so it quickly becomes very messy. The approach we're going to discuss and suggest that you use, is much more straight forward and relies on the "factoring by grouping" method that we discussed in the last lesson. The procedure works like this:
Factor:  6x^{2} + 19x + 10 

6 · 10 = 60 

15 · 4 = 60 15 + 4 = 19 

6x^{2} + 15x + 4x + 10 

3x(x + 5) + 2(2x + 5) (3x + 2)(2x + 5) 
Example 1
Factor 2x^{2} + 21x + 10.
We need to start by multiplying the first and last numbers in the polynomial together. That means 2 · 10 = 20. Now we need to find two numbers that multiply to give 20 but add up to 21 (the coefficient of the x term). A little trial and error shows us that we're looking for 20 and 1 since 20 · 1 = 20 and 20 + 1 = 21.
The next step is to rewrite the middle term of the original polynomial, 21x, in terms of 20 and 1. That would mean
21x = 20x + 1x
Doing that in the original polynomial gives us
2x^{2} + 21x + 10 = 2x^{2} + 20x + 1x + 10
The expression on the right hand side is one we can factor by grouping to get our final answer.
2x^{2} + 20x + 1x + 10
2x(x + 10) + (x + 10)
(2x + 1)(x + 10)
Example 3
Factor 15x^{2}  26x + 8.
In this example, we start off with 15 · 8 = 120 which has a lot of factors. With a number as big as this, it may be helpful for you to make a complete list like we did in Example 2. If you do, be sure that you consider the negative possibilities as well as the positive ones, e.g. 60 · 2. If you make the list and sort thorugh it you'll find that 20 · 6 = 120 and 20 + 6 = 26 which is the coefficient of the x term in our example.
Splitting the 26x into 20x and 6x gives us
15x^{2}  20x  6x + 8
Which we can factor to
5x(3x  4)  2(3x  4)
(3x  4)(5x  2)
Example 2
Factor 4x^{2} + 17x  15.
This example follows the same pattern as the Example 1. First we need to multiply together the first and last coefficients: 4 · 15 = 60. Now we need two numbers that multiply to equal 60 but that add up to 17. It might be obvious to you that +20 and 3 do that. If it isn't then write out all the numbers that multiply out to 60 like we did in the "Factoring Trinomials 1" lesson and add them all up:
1 x 60  2 x 30  3 x 20 
4 x 15  5 x 12  6 x 10 
1 x 60  2 x 30  3 x 20 
4 x 15  5 x 12  6 x 10 
Now we'll find the sums of each pair.
1  60 = 59  2  30 = 28  3  20 = 17 
4  15 = 11  5  12 = 7  6  10 = 4 
1 + 60 = 59  2 + 30 = 28  3 + 20 = 17 
4 + 15 = 11  5 + 12 = 7  6 + 10 = 4 
Look over the bottom list and you'll see that 20 and 3 are the only pair that add up to 17 so those are the ones that we need to rewrite 17x in terms of:
17x = 20x  3x
Doing that in the original polynomial gives us something we can factor by grouping.
4x^{2} + 20x  3x  15
4x(x + 5)  3(x + 5)
(4x  3)(x + 5)
Example 4
Factor 2x^{2} + x + 28.
This one may look a little confusing because of the negative sign in front of the x^{2} term but it works the same as all of the others. First we look at 2 · 28 = 56 and then we notice that 8 · 7 = 56 and 8 + 7 = 1. (Remember that if there's no number written out in front of a variable then its coefficient is 1.)
Now we can rewrite x as 8x  7x to get
2x^{2} + 8x  7x + 28
2x(x  4)  7(x  4)
(x  4)(2x  7)
Video Lectures
Dynamic Practice  Factoring a Quadratic Polynomial
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