Introduction
When factoring a polynomial, you want to change the sum and difference of many terms into the product of different polynomials.
For example, when asked to factor the number 12, you might write 2*2*3, because these are the smallest numbers that would multiply to make 12.
It is important that we be able to do this with polynomials as well. Breaking them down into the product of smaller pieces (factors) allows us to simplify problems.
When factoring a trinomial (a polynomial made up of 3 terms), you should first look for a GCF.
Then, we look to change the polynomial into the product of two binomials (polynomials made up of two terms), i.e. ( x + 3 )( x + 2)
To determine what the two binomials look like, we can see that the first piece of each binomial in each parentheses will simply be an “x”, since those are the only 2 things that multiply to x2.
To determine the two numbers we need in each parentheses, we need to determine the two number that would multiply to the “c” value, and also add to the “b” value.
Example
Given x2+ 5x + 6
(x + __) (x + __)
We know the beginning of each binomial will be just “x” we now look for two numbers that multiply to 6, but also add to 5. We have two choices of number pairs that multiply to 6: 1 · 6 and 2 · 3. We can see that the pair that also adds to 5 is 2 · 3. Therefore we can write:
(x + 2)(x + 3)
Check your answer by FOIL-ing or re-multiplying these two binomials together; ensure that you get the
original polynomial back again. This will confirm you have factored it correctly.
Here’s a video that shows some examples (for the first 3 minutes and 24 seconds):
Practice Problems: Now you try!
- x2 + 8x + 7
- x2 + 14x + 24
- x2– 4x – 21
- x2– x – 2
- x2 – 6x + 8
- 2x3– 18x2 + 40x
- (x + 7)(x + 1)
- (x + 2)(x + 12)
- (x – 7)(x + 3)
- (x – 2)(x + 1)
- (x – 2)(x – 4)
- 2x(x2– 9x + 20) took out a GCF first, then factored the polynomial inside the parentheses and got: 2x(x – 4 )(x – 5)
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