A trinomial is an algebraic expression consisting of three terms. Most likely, you will begin to learn how to decompose quadratic trinomials, that is, written in the form x2 + bx + c. There are several tricks to learn that apply to different types of quadratic trinomials, but you will get better and faster just with practice. Polynomials of higher degree, with terms such as x3 or x4, are not always solvable by the same methods, but it is often possible to use simple decompositions or substitutions to transform them into problems that can be solved like any quadratic formula.
Steps
Method 1 of 3: Decompose x2 + bx + c
Step 1. Learn the FOIL technique
You may have already learned the FOIL method, ie "First, Outside, Inside, Last" or "First, outside, inside, last", to multiply expressions like (x + 2) (x + 4). It is useful to know how it works before we get to the breakdown:
- Multiply the terms First: (x+2)(x+4) = x2 + _
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Multiply the terms Outside: (x+2) (x +
Step 4.) = x2+ 4x + _
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Multiply the terms Inside: (x +
Step 2.)(x+4) = x2+ 4x + 2x + _
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Multiply the terms Last: (x +
Step 2.) (x
Step 4.) = x2+ 4x + 2x
Step 8.
- Simplify: x2+ 4x + 2x + 8 = x2+ 6x + 8
Step 2. Try to understand the factoring
When you multiply two binomials with the FOIL method, you arrive at a trinomial (an expression with three terms) in the form at x2 + b x + c, where a, b and c are any number. If you start from an equation in this form, you can break it down into two binomials.
- If the equation is not written in this order, move the terms. For example, rewrite 3x - 10 + x2 like x2 + 3x - 10.
- Since the highest exponent is 2 (x2), this type of expression is "quadratic".
Step 3. Write a space for the answer in FOIL form
For now, just write (_ _) (_ _) in the space where you can write the answer. We will complete it later.
Don't write + or - between the empty terms yet, as we don't know what they will be
Step 4. Fill in the first terms (First)
For simple exercises, where the first term of your trinomial is just x2, the terms in first (First) position will always be x And x. These are the factors of the term x2, since x for x = x2.
- Our example x2 + 3 x - 10 starts with x2, so we can write:
- (x _) (x _)
- We'll do some more complicated exercises in the next section, including trinomials starting with a term like 6x2 or -x2. For now, follow the example problem.
Step 5. Use the breakdown to guess the last (Last) terms
If you go back and reread the passage of the FOIL method, you will see that by multiplying the last terms (Last) together you will have the final term of the polynomial (the one without x). So, to do the decomposition, we need to find two numbers which, when multiplied, give the last term.
- In our example, x2 + 3 x - 10, the last term is -10.
- -10? Which two numbers multiplied together give -10?
- There are a few possibilities: -1 times 10, -10 times 1, -2 times 5, or -5 times 2. Write these pairs down somewhere to remember them.
- Don't change our answer yet. At the moment, we are at this point: (x _) (x _).
Step 6. Test which possibilities work with the external and internal multiplication (Outside and Inside) of the terms
We have narrowed down the last terms (Last) to a few possibilities. Go by trial and error to try every possibility, multiplying the external and internal terms (Outside and Inside) and comparing the result with our trinomial. Eg:
- Our original problem has an "x" term which is 3x, which is what we want to find with this proof.
- Try with -1 and 10: (x - 1) (x + 10). Outside + Inside = Outside + Inside = 10x - x = 9x. They are not good.
- Try 1 and -10: (x + 1) (x - 10). -10x + x = -9x. It's not true. In fact, once you try it with -1 and 10, you know that 1 and -10 will give just the opposite answer to the previous one: -9x instead of 9x.
- Try with -2 and 5: (x - 2) (x + 5). 5x - 2x = 3x. This matches the original polynomial, so this is the correct answer: (x - 2) (x + 5).
- In simple cases like this, when there is no number in front of the x, you can use a shortcut: just add the two factors together and put an "x" after it (-2 + 5 → 3x). This doesn't work with more complicated problems, though, so remember the "long way" described above.
Method 2 of 3: Decomposing More Complex Trinomes
Step 1. Use simple decomposition to ease more complicated problems
Suppose we want to simplify 3x2 + 9x - 30. Look for a common divisor for each of the three terms (the greatest common divisor, GCD). In this case, it is 3:
- 3x2 = (3) (x2)
- 9x = (3) (3x)
- -30 = (3)(-10)
- Therefore, 3x2 + 9 x - 30 = (3) (x2 + 3 x -10). We can decompose the trinomial again using the procedure in the previous section. Our final answer will be (3) (x - 2) (x + 5).
Step 2. Look for more complicated breakdowns
Sometimes, these may be variables or you may need to break it down a couple of times to find the simplest expression possible. Here are some examples:
- 2x2y + 14xy + 24y = (2y)(x2 + 7x + 12)
- x4 + 11x3 - 26x2 = (x2)(x2 + 11x - 26)
- -x2 + 6x - 9 = (-1)(x2 - 6x + 9)
- Don't forget to break it down further, using the procedure in Method 1. Check the result and find exercises similar to the examples at the bottom of this page.
Step 3. Solve problems with a number in front of the x2.
Some trinomials cannot be simplified to factors. Learn to solve problems like 3x2 + 10x + 8, then practice on your own with the example problems at the bottom of the page:
- Set up the solution like this: (_ _)(_ _)
- Our first terms (First) will each have an x and multiply together to give 3x2. There is only one possible option here: (3x _) (x _).
- List the divisors of 8. The possible choices are 8 x 1 or 2 x 4.
- Try them out using the terms outside and inside (Outside and Inside). Note that the order of the factors is important, as the outer term is multiplied by 3x instead of x. Try all possible combinations until you get an Outside + Inside which gives 10x (from the original problem):
- (3x + 1) (x + 8) → 24x + x = 25x no
- (3x + 8) (x + 1) → 3x + 8x = 11x no
- (3x + 2) (x + 4) → 12x + 2x = 14x no
- (3x + 4) (x + 2) → 6x + 4x = 10x Yes It is the correct decomposition.
Step 4. Use substitution for higher degree trinomials
The math book might surprise you with a high exponent polynomial, such as x4, even after simplifying the problem. Try substituting a new variable so that you end up with an exercise you can solve. Eg:
- x5+ 13x3+ 36x
- = (x) (x4+ 13x2+36)
- Let's use a new variable. Suppose y = x2 and replace:
- (x) (y2+ 13y + 36)
- = (x) (y + 9) (y + 4). Now let's go back to the starting variable.
- = (x) (x2+9) (x2+4)
- = (x) (x ± 3) (x ± 2)
Method 3 of 3: Breakdown of Special Cases
Step 1. Check with prime numbers
Check if the constant in the first or third term of the trinomial is a prime number. A prime number is only divisible by itself and 1 only, so there are only a couple of possible factors.
- For example, in the trinomial x2 + 6x + 5, 5 is a prime number, so the binomial must be of the form (_ 5) (_ 1).
- In problem 3x2 + 10x + 8, 3 is a prime number, so the binomial must be of the form (3x _) (x _).
- For the 3x problem2 + 4x + 1, 3 and 1 are prime numbers, so the only possible solution is (3x + 1) (x + 1). (You should still multiply to check the work done, as some expressions just can't be factored - for example, 3x2 + 100x + 1 cannot be broken down into factors.)
Step 2. Check to see if the trinomial is a perfect square
A perfect square trinomial can be decomposed into two identical binomials and the factor is usually written (x + 1)2 instead of (x + 1) (x + 1). Here are some squares that often show up in problems:
- x2+ 2x + 1 = (x + 1)2 and x2-2x + 1 = (x-1)2
- x2+ 4x + 4 = (x + 2)2 and x2-4x + 4 = (x-2)2
- x2+ 6x + 9 = (x + 3)2 and x2-6x + 9 = (x-3)2
- A perfect square trinomial in the x-form2 + b x + c always has the terms a and c which are positive perfect squares (e.g. 1, 4, 9, 16 or 25) and a term b (positive or negative) which equals 2 (√a * √c).
Step 3. Check if there is no solution
Not all trinomials can be taken into account. If you are stuck on a trinomial (ax2 + bx + c), use the quadratic formula to find the answer. If the only answers are the square root of a negative number, there is no real solution, so there are no factors.
For non-quadratic trinomials, use Eisenstein's criterion, described in the Tips section
Example problems with Answers
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Find answers to deceptive problems with decompositions.
We have already simplified them into easier problems, so try to solve them using the steps seen in method 1, then check the result here:
- (2y) (x2 + 7x + 12) = (x + 3) (x + 4)
- (x2) (x2 + 11x - 26) = (x + 13) (x-2)
- (-1) (x2 - 6x + 9) = (x-3) (x-3) = (x-3)2
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Try more difficult decomposition problems.
These problems have a common factor in each term that must first be collected. Highlight the space after the equal signs to see the answer so you can check the work:
- 3 x 3 + 3 x 2 -6 x = (3x) (x + 2) (x-1) ← highlights the space to see the answer
- -5x3y2+ 30x2y2-25y2x = (-5xy ^ 2) (x-5) (x-1)
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Practice with difficult problems.
These problems cannot be broken down into easier equations, so you need to come up with an answer in the form of (x + _) (_ x + _) by trial and error:
- 2x2+ 3x-5 = (2x + 5) (x-1) ← highlight to see the answer
- 9 x 2 + 6 x + 1 = (3x + 1) (3x + 1) = (3x + 1)2 (Hint: You may need to try more than one pair of factors for 9 x.)
Advice
- If you can't figure out how to decompose a quadratic trinomial (ax2 + bx + c), you can always use the quadratic formula to find x.
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While not mandatory, you can use Eisenstein's criteria to quickly determine if a polynomial is irreducible and cannot be factored. These criteria work for any polynomial, but are especially good for trinomials. If there is a prime number p which is a factor of the last two terms and satisfies the following conditions, then the polynomial is irreducible:
- The constant term (for a trinomial in the form ax2 + bx + c, this is c) is a multiple of p, but not of p2.
- The initial term (which here is a) is not a multiple of p.
- For example, it allows you to quickly determine that 14x ^ 9 + 45x ^ 4 + 51 is irreducible, since 45 and 51, but not 14, are divisible by the prime number 3 and 51 is not divisible by 9.
