Mathematics for Technology I (Math 1131) Durham College, Mathematics
Free • 55 lessons
• 1 quizzes
• 10 week duration
• Numerical Computation

Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

• Measurements

An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

• Trigonometry with Right Triangles

Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

• Trigonometry with Oblique Triangles

This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

• Geometry

This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.

Mathematics for Technology I (Math 1131)

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 trial-and-error 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 intercepts (the roots). A whole unit will be dedicated to solving equations, including first and second degree equation. Otherwise, if it’s in general form, you can’t easily do that.

You’ll find out here that some quadratics whose b-term 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:

1. y = x² – 100   (easy)
2. y = 98a² – 450b²   (medium)
3. 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: