*Note: This guide assumed a general understanding of polynomials. For a review of this topic, click here.
Introduction
When factoring a polynomial, the objective is to change the sum and difference of many terms into the product of different, smaller polynomials. For example, if one were asked to factor the number 12, one might write 2*2*3, because these are the smaller 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 binomial (a polynomial made up of 2 terms), you should first look for a Greatest Common Factor.
Then, we look to change the polynomial into the product of two smaller binomials (polynomials made up of two terms), i.e. (x + 3)(x -- 3)
Putting it Together
In order to factor a binomial using the “Difference of 2 Squares” method, each of the two terms need to be perfect squares. That means the numbers have to be perfect squares which are the result of multiplying a number with itself, i.e. 1, 4, 9, 16, 25... (1 times 1 is 1; 2 times 2 is 4; 3 times 3 is 9 and so on). And the variables have to be perfect squares, which will be true as long as the variable’s exponent is an even number (2, 4, 6...). Finally, the two terms have to be separated by a minus sign (which is why we refer to it as taking the “difference” of 2 squares).
To determine what the two binomial factors should look like, we take the square root of each term (determine the expression that if you multiplied it by itself would give you the original term), put them in each parentheses twice, but in between the terms on one of them you’ll put a plus sign “+” and on the other you’ll put a minus sign “- ”
Examples
Given x2 – 9
We verify that “x2” is a square (x times x), 9 is a perfect square (3 times 3), and there is a minus sign in between
(x + 3)(x – 3)
so we write the square roots of each piece (x and 3) in each parentheses, but one has a “+” sign and one has a “-“ sign.
Here’s an example in which you have to take out a GCF first. Given: 45x2– 80
Notice that neither 45 nor 80 are perfect squares; neither number can be achieved by multiplying a number with itself.
However, each term (45x2 and 80) has a GCF in common. It’s 5. So let’s factor that out.
5(9x2 – 16)
Now that the expression inside the parentheses is the difference (minus sign) of two squares (3 times 3 is 9; 4 times 4 is 16), we can continue to factor the inside of this set of parentheses.
5(3x + 4)(3x -- 4)
Note: it doesn’t matter which order you write these binomials in the parentheses, whether it’s:
(3x + 4)(3x – 4) or (3x – 4)(3x + 4)
It is a good idea to check your answer by FOIL-ing or re-multiplying these expressions together to verify that you get the original polynomial back again, thereby confirming you factored it correctly.
Here’s a video that shows some examples:
Now you try!
- x2 -- 25
- 64 -- y2
- 4a2 -- 81b2
- 2w2 -- 98
- x2 + 100
- 121x5 -- x
- (x + 5)(x -- 5)
- (8 -- y)(8 + y)
- (2a + 9b)(2a -- 9b)
- 2(w2 -- 49) Take out a “2” as a GCF, then factor the difference of 2 squares 2(w + 7)(w -- 7)
- Not factorable. Remember, it needs to have a minus sign between the terms
- x(121x4 -- 1) Take out an “x” as a GCF, then factor the difference of 2 squares x(11x2 + 1)(11x2 -- 1)
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