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The Mathematics 11 Competency Test Factoring Differences of Perfect Squares You need to know how to do just one more special factorization for the BCIT Mathematics 11 Competency Test. Notice that (u + v)(u – v) = (u + v)(u) + (u + v)(-v) = (u)(u + v) + (-v)(u + v) = (u)(u) + (u)(v) + (-v)(u) + (-v)(v) = u2 + uv – uv – v2 = u2 – v2 So, an expression of the form u2 – v2 (called a difference of two perfect squares) can always be factored into the form u 2 − v 2 = ( u + v )( u − v ) The formula in the box is a pattern. The symbols u and v may represent numbers, other algebraic symbols, or even algebraic expressions. It is the precise pattern which must be satisfied in each case. Example 1: Factor 4x2 – y2. solution: First, we check, and find that the two terms contain no monomial factors in common. Then, we note that this is a two-term expression, so the methods for factoring trinomials that were demonstrated in detail in the previous document in the series are not likely to apply. Whenever the expression to be factored contains just two terms, it is worth checking to see if the difference of squares pattern above can be used. You need to identify that the expression has two important features: (i) it is a difference of two terms – the second term must be subtracted from the first term. In this example, we do have a difference of two terms. (ii) it must be possible to write each of the two terms as a square of some expression. In this example, we see that 4x2 = (2x)2 and y2 = (y)2, so this condition is met. Matching the parts of this expression with the parts of the pattern in the box above, we get 4x2 – y2 = (2x)2 – (y)2 ⇔ u2 – v2 Thus, David W. Sabo (2003) Factoring Differences of Perfect Squares Page 1 of 3 and u2 = (2x)2 giving u = 2x v2 = (y)2 giving v = y Then 4x2 – y2 = u2 – v2 = (u + v)(u – v) = (2x + y)(2x – y) or, directly 4x2 – y2 = (2x + y)(2x – y) as the required factorization. It is each to check this result directly. (2x + y)(2x – y) = (2x + y)(2x) + (2x + y)(-y) = (2x)(2x + y) + (-y)(2x + y) = (2x)(2x) + (2x)(y) + (-y)(2x) + (-y)(-y) = 4x2 + 2xy – 2xy – y2 = 4x2 – y2 Thus our factorization checks. Example 2: Factor 9a4 – 36b2 . solution: Again, we check that there are no common monomial factors in the two terms (there are none). Then, recognizing the possibility that this is a difference of two perfect squares (since this expression obviously is the difference of precisely two terms), we note that: 9a4 = (3a2)2 and 36b2 = (6b)2 So, each of the two terms are perfect squares. So, we can apply the pattern formula given in the box above: and u2 = (3a2)2 or u = 3a2 v2 = (6b)2 or v = 6b to get 9a4 – 36b2 = (3a2 + 6b)(3a2 – 6b) You can easily verify by multiplication that this factorization is correct. David W. Sabo (2003) Factoring Differences of Perfect Squares Page 2 of 3 Example 3: Factor 9x2 + 4y2. solution: This expression consists of two terms, both of which are perfect squares: 9x2 + 4y2 = (3x)2 + (2y)2 However, it is a sum of two perfect squares rather than a difference of two perfect squares. We might be tempted to try ? 9 x 2 + 4 y 2 = (3 x + 2y )(3 x + 2y ) thinking that since a minus has become a plus on the left-hand side, perhaps the thing to do is to change the minus to a plus on the right-hand side. However, we should not simply assume this is the best we can do and move on, but rather, we must check to see if the assumption made here is correct. That is easy to do, as usual, by multiplication: (3x + 2y)(3x + 2y) = (3x + 2y)(3x) + (3x + 2y)(2y) = (3x)(3x + 2y) + (2y)(3x + 2y) = (3x)(3x) + (3x)(2y) + (2y)(3x) + (2y)(2y) = 9x2 + 6xy + 6xy + 4y2 = 9x2 + 12xy + 4y2 which is not the same as the original expression, 9x2 + 4y2. In fact, there is no way to factor the sum of two perfect squares (without introducing more advanced mathematical concepts that are well beyond the scope of the BCIT Mathematics 11 Competency Test). So, in answer to this example, we need to simply state that the given expression cannot be factored. David W. Sabo (2003) Factoring Differences of Perfect Squares Page 3 of 3