 34 lessons
 0 quizzes
 7 week duration

Unit 1: Linear Systems

Unit 2: Analytic Geometry

Unit 3: Geometric Properties

Unit 4: Quadratic Relations
Most relations that you have studied in mathematics have been linear. However, many nonlinear also exist in real life.

Unit 5: Quadratic Expressions

Unit 6: Quadratic Equations

Unit 7: Trigonometry
Factor a Difference of Squares
So far we’ve learned three factoring techniques. The first one, common factoring, is a technique that can be used for any polynomial. The other two were specific for quadratics, namely trialanderror and decomposition.
The whole purpose behind factoring any quadratic – if you haven’t discovered already – is convert it in such a format that enables you to solve for the x’s (the roots). Otherwise, if it’s in general form, you can’t easily do that.
You’ll find out here that some quadratics whose bterm is missing and whose a and c terms are being subtracted, can be factored by another technique known the difference of squares. Examples of quadratics that fit this pattern look like this:
 y = x² – 100 (easy)
 y = 98a² – 450b² (medium)
 y = (3x + 8)² – (x – 2)² (hard)
Also, as the name of the technique implies, both terms in the quadratic need to be perfect squares (squarootable).
Take equation (1) as an example. Taking the square root of x² yields x. In other words, √x² = x. Similarly, √100 = 10. In the second equation, after common factoring 2 from both terms, you get:
= 2(49a² – 225b²)
Notice now that √49a² = 7a and √225b² = 15b. Both terms within the parentheses are perfect squares; hence, this technique can be used.
In equation (3), the square root of (3x + 8)² is 3x + 8 and the square root of (x – 2)² is x – 2; therefore it fits the same mold. A summary of what’s required is written below.
Conditions required to factor a difference of squares:
 b term needs to be missing.
 The two remaining terms are being subtracted.
 The two remaining terms need to be a perfect square.
Let’s what a few examples: