Factoring quadratics with one variable
Special pattern: difference of two perfect squares
Special pattern: perfect square trinomials
Factoring - when the leading coefficient is 1.
When the leading coefficient is not 1: the most general case, .
When I try to factor a trinomial like , I ask myself, "What is a pair of numbers that multiply to give me 5, and add to give me 3?" There is no such pair of real numbers, so I know that this trinomial can't be factored. We'll learn to use the quadratic formula to find its roots. In fact, we'll learn that they are a pair of complex conjugates - more on that later.
But now it's time to look at more general quadratics, what do we do when the leading coefficient isn't 1?
First check for common factors. looks hard, but if we notice that 3 is a factor of every term, we can turn this into a simpler problem by factoring it out: . Since 5•2 =10 and 5+2=7, I have as my factored form.
If the leading coefficient is negative, I always factor out a -1 first.
Let's look at and example that doesn't have a common factor: . If this can be factored into two binomials, we know that the first term of the trinomial comes from multiplying the first terms of the binomials. The factors could be or or . The 10 is going to come from multiplying the second terms of the two binomials, so the ?'s could be 10 and 1, 1 and10, 2 and 5, or 5 and 2. That's a lot of possibilities we need to try to find the one combination that gives us 19x for the middle term.
Fortunately, there is a simple trick we can use so that we don't have to guess and check. We start by multiplying the first and last coefficients. In our case, 6 • 10 = 60. Now we look for a pair of numbers that multiply to give us this number and add to give us 19.
60 |
60, 1 61
30,2 32
20,3 23
15,4 19
12,5 17
10,6 16
15 and 4 are the factors we need.
Now we write our trinomial as a quadranomial by splitting the middle term into these two numbers:
To finish, we are going to "factor by grouping." We'll look at the first two terms and the last two terms and factor out any common factors from both:
In this method, we will always have the same binomial. We now factor out that common binomial, and we have factored our trinomial completely:
We can quickly check our work by multiplying the two binomials to see that we get our original trinomial.
Here are some examples for you to try: