3. Factoring Trinomials
A trinomial is a 3 term polynomial. For example, 5x^{2} − 2x + 3 is a trinomial.
In many applications in mathematics, we need to solve an equation involving a trinomial. Factoring is an important part of this process. [See the related section: Solving Quadratic Equations.]
Example 1
Factor x^{2} − 5x − 6
Answer
Here, we are looking for an answer in the form:
x^{2} − 5x − 6 = (x ... )(x ... )
Note that the term we have at the beginning of the question is x^{2}, so we put x in each bracket. This gives us (x)(x) = x^{2}, which is what we need for that first term.
Now we need 2 numbers that multiply to give −6 and add to give −5. The possibilities are:
multiply to give | add to give | OK? | |
2 and −3 | 2 × −3 = −6 | 2 + −3 = −1 | No |
−2 and 3 | −2 × 3 = −6 | −2 + 3 = 1 | No |
6 and −1 | 6 × −1 = −6 | 6 + −1 = 5 | No |
−6 and 1 | −6 × 1 = −6 | −6 + 1 = −5 | OK |
So we have:
x^{2} − 5x − 6 = (x − 6)(x + 1)
This method of factoring can be tedious because we may need to try several combinations before we hit on the correct numbers (and letters).
After some practice, though, you can spot the most likely combination of letters and numbers. I've shown all the possibilities in the table to give you an idea of what can be involved.
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NOTE: Always check your answer by multiplying it out!
Example 2
Factor 2n^{2} − 13n − 7
Answer
Once again, we will do it the long way so you can see what is involved.
In this case, we need 2n^{2} so we start with:
2n^{2} − 13n − 7 = (2n ... )(n ... )
Now we need 2 numbers that multiply to give −7 and the sum of the inner and outer products must be −13. Don't forget the 2 in the first bracket!
Possibility 1: −7 and 1
Multiply to give −7
2n^{2} − 13n − 7 = (2n − 7)(n + 1)
Outer product plus inner product:
(2n)(1) + (−7)(n) = 2n − 7n = −5n
This is not the correct answer.
Possibility 2: 7 and −1
This would give us
Outer product plus inner product:
(2n)(−1) + (7)(n) = −2n + 7n = 5n
Possibility 3: −1 and 7
2n^{2} − 13n − 7 = (2n − 1)(n + 7)
Outer product plus inner product:
(2n)(7) + (−1)(n) = 14n − n = 13n
This is still not the correct answer.
Possibility 4: 1 and −7
2n^{2} − 13n − 7 = (2n + 1)(n − 7)
Outer product plus inner product:
(2n)(−7) + (1)(n) = −14n + n = −13n
This is the required answer, finally.
So
2n^{2} − 13n − 7 = (2n + 1)(n − 7)
I repeat, it won't always involve several steps like this. The more you do them, the easier and quicker they become (like most new skills).
Of course, after some practice, you will get a better sense of the numbers that will most likely work. It is unlikely that you will have to churn through all the possibilities before you find the right combination, like I have done above.
Now I'll show you a better method, one that reduces a lot of the guesswork.
Factoring by Grouping
This method requires the least amount of guessing and is recommended.
Example 3
Factor 6x^{2} + x − 12
Answer
If we were doing it the long way, we would need to consider all the factors of 6:
1 & 6;
−1 & −6;
2 & 3;
−2 & −3
and also many factors of −12:
1 & −12;
2 & −6;
3 & −4
and all the negatives of these.
We could spend a long time finding the correct combination of factors if we use the long method.
Factoring by Grouping Method
Using grouping method, first we find 2 things, the:
(a) Product of the outer 2 terms of the trinomial
(b) Inner number of the trinomial
Then, the only "guess and check" we need to do is to look for 2 numbers whose:
- Product is the result of (a) above
- Sum is the inner term (from (b) above)
So in our 6x^{2} + x − 12 example, we are looking for 2 terms whose:
(a) Product is 6x^{2} × −12 = −72x^{2} (multiply outer numbers)
and whose
(b) Sum is x (inner number).
We try some terms and easily get 9x and −8x.
These are correct since:
- (9x)(−8x) = −72x
- 9x + (−8x) = x
Now write the original expression replacing x with (9x − 8x), as follows:
6x^{2} + x − 12 = 6x^{2} + (9x − 8x) − 12
We now re-group the right-hand side:
6x^{2} + (9x − 8x) − 12 = (6x^{2} + 9x) − (8x + 12)
Now factor each of the bracketed terms:
(6x^{2} + 9x) − (8x + 12) = 3x(2x + 3) − 4(2x + 3)
On the right-hand side, we notice that each term in brackets is the same, so we can combine them as follows:
3x(2x + 3) − 4(2x + 3) = (3x − 4)(2x + 3)
[What just happened?
If we have 3xA − 4A, we can factor A out of each term, and write (3x − 4)A. This is how grouping method works. You always end up with brackets that have the same terms inside, and these can be factored out.]
So our answer is:
6x^{2} + x − 12 = (3x − 4)(2x + 3)
Always check your answer by multiplying it out!
NOTE: Of course, we may need to re-arrange our trinomial to get it into the correct form for grouping method to work. Normally this means we write our polynomial terms in decreasing powers of x.
Example 4
Let's return to Example 2 from above and do it again, but this time use grouping method.
Factor: 2n^{2} − 13n − 7
Answer
This time, the product of the outer terms is 2n^{2} × −7 = −14n^{2}.
The inner term is −13n.
So we are looking for 2 terms whose product is −14n^{2} and whose sum is −13n.
Those 2 terms are −14n and n.
(This step is nearly always easier to do with grouping method, compared to what we were doing at the top of the page.)
So we write:
2n^{2} − 13n − 7
= 2n^{2} − 14n + n − 7
= (2n^{2} − 14n) + (n − 7)
= 2n(n − 7) + (1)(n − 7)
= (2n + 1)(n − 7)
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Exercises
Factorise each of the following:
(1) 3n^{2} − 20n + 20 [Care with this one!!]
Answer
3n^{2}− 20n + 20
Cannot be further factored!!
When factoring, it's important to know when we cannot go any further. This example looks like we may be able to factor it, but we cannot.
(2) 3x^{2} + xy − 14y^{2}
Answer
Using grouping method, we have:
- Product of the outer 2 terms is (3x^{2})(−14y^{2}) = −42x^{2}y^{2}
- Inner term: xy
So we are looking for 2 terms whose product is −42x^{2}y^{2} and whose sum is xy.
A little bit of thinking gives us: 7xy and −6xy.
So we can factor our trinomial as follows:
3x^{2} + xy − 14y^{2}
= 3x^{2} + 7xy − 6xy − 14y^{2}
= (3x^{2} + 7xy) − (6xy + 14y^{2})
= x(3x + 7y) − 2y(3x + 7y)
= (x − 2y)(3x + 7y)
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(3) 4r^{2} + 11rs − 3s^{2}
Answer
Using grouping method, we have:
- Product of the outer 2 terms is: (4r^{2})(−3s^{2}) = −12r^{2}s^{2}
- Inner term: 11rs
So we are looking for 2 terms whose product is −12r^{2}s^{2}and whose sum is 11rs.
Those 2 terms will be 12rs and −rs.
So we can factor our trinomial as follows:
4r^{2} + 11rs − 3s^{2}
= 4r^{2} + 12rs − rs − 3s^{2}
= (4r^{2} + 12rs) − (rs + 3s^{2})
= 4r(r + 3s) − s(r + 3s)
= (4r − s)(r + 3s)
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(4) 6x^{4} − 13x^{3} + 5x^{2}
Answer
The first step with this example is to factor out x^{2} from each term:
6x^{4} − 13x^{3} + 5x^{2} = x^{2}(6x^{2} − 13x+ 5)
Using grouping method on the (6x^{2} − 13x + 5) part, we have:
- Product of the outer 2 terms is: (6x^{2})(5^{}) = 30x^{2}
- Inner term: −13x
So we are looking for 2 terms whose product is 30x^{2} and whose sum is −13x.
Those 2 terms are −10x and −3x.
So we can factor our trinomial as follows:
6x^{4} − 13x^{3} + 5x^{2}
= x^{2}(6x^{2} − 13x+ 5)
= x^{2}[6x^{2} − 10x − 3x + 5]
= x^{2}[(6x^{2} − 10x) − (3x − 5)]
= x^{2}[2x(3x − 5) − (1)(3x − 5)]
= x^{2}[(2x − 1)(3x − 5)]
= x^{2}(2x − 1)(3x − 5)
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