Is there more than one way to factor this?
explain how to factor this trinomial: x2+bx+c and ax2+bx+c
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Are you asking if there is more than one way to factor trinomials or if there is more than one way to find a solution (value for x)?
If you are looking to find x, there are two basic routes that are commonly used to solve for x:
1. Factoring (not every equation is easy to factor)
- Finding and pulling out a GCF (greatest common factor)
- Splitting the trinomial up into two binomials that multiply together to get your original trinomial (commonly looks like: (x+5)(x-2) or even x(x+2))
2. Using the Quadratic Formula (will ALWAYS get you an answer for x, if there is a real number answer for x that can be found)
Basics of Factoring
Factoring is often used when attempting to solve for x in a trinomial that is equal to 0. The idea behind factoring is that you want to find two numbers that multiply together to equal c and whose sum/difference is equal to b. This gets more complicated if there is a number other than 1 in front of the x2 (if a≠1).
Situation 1: a=0
Well, you're left with something that looks like 0*x2+bx+c = 0 which can be rewritten as bx+c = 0. This is a simple binomial, and you can use basic algebra steps to solve for x.
Situation 2: a=1
Now you have x2+bx+c = 0. List the integer factors of c (Don't forget about negative numbers!). Which pair of factors can be added/subtracted to get b (If you can't find a pair of factors that add/subtract to b, then I would suggest using the quadratic formula.)? Place those factors in the correct spots in the following factoring form: (x + FACTOR 1)(x + FACTOR 2) = 0. Set each parenthesis set equal to 0, and use basic algebra steps to solve for x.
Situation 3: a is a positive or negative number other than 1 or 0.
This is very similar to Situation 2. However, you want to start out a little differently.
Start by looking for a GCF. You may be able to factor a out of each term, and then you're just left with situation 2. If you still have an a in front of x2 after factoring out the GCF (if there is one), it is probably easier to use the quadratic formula (see below). If you still want to continue with factoring, list some factors of c/a (if there are any). You are looking for a pair of factors whose sum/difference is b/a. Once you find a pair of factors that satisfy both conditions, place those factors in the correct spots in the following factoring form: (x + FACTOR 1)(x + FACTOR 2). From here, set each parenthesis set equal to 0 and use basic algebra steps to solve for x.
This is a quick way to find a solution for x if you are having difficulty with factoring or if you are working with a trinomial that is not easily factored. Plug the coefficients of your trinomial (a, b, and c) into the following formula:
x = (-b ± √(b2-4ac)) / 2a
Note that this will give you TWO answers!! x = (-b + √(b2-4ac)) / 2a AND x = (-b - √(b2-4ac)) / 2a