# Mathematics I (Math 1132)

Durham College, Mathematics
Free
• 0 lessons
• 0 quizzes
• 10 week duration
• ##### 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.

• ##### Fractions, Percentage, Ratios and Proportion

Emphasis here is placed on understanding fractions, percent, and using ratios to compare quantities and set up proportions to solve problems.

• ##### Geometry

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

## Mathematics I (Math 1132)

### Simplify Expressions Through Factoring

The skills you’ll learn in this lesson will come in handy unexpectedly one day when you’re stuck trying to simplify what appears to be an impossible expression to reduce. Take a look at the three expressions below:

1. $\frac{ab+bc}{bc+bd}\phantom{\rule{0ex}{0ex}}$
2. $\frac{2{x}^{2}–5x–3}{4{x}^{2}–1}\phantom{\rule{0ex}{0ex}}$
3. $\frac{{x}^{2}–ax+2bx–2ab}{2{x}^{2}+ax–3{a}^{2}}\phantom{\rule{0ex}{0ex}}$

At first glance, you might be questioning how do I divide two polynomials when we’ve only learned how to divide polynomials with monomials? Factoring the numerator and denominator separately should always be considered as an option before devising other plans or giving up. By converting polynomials into factors, you open up the opportunity to cancel with similar factors in the other polynomial, as you’ll see below.

Question: Simplify each expression shown above.

 $\frac{ab+bc}{bc+bd}=\frac{b\left(a+c\right)}{b\left(c+d\right)}⇒\frac{\overline{)b}\left(a+c\right)}{\overline{)b}\left(c+d\right)}⇒\frac{\mathbf{a}\mathbf{+}\mathbf{c}}{\mathbf{c}\mathbf{+}\mathbf{d}}\phantom{\rule{0ex}{0ex}}$ ✔ Technique: Common factor both the numerator and denominator. $\frac{2{x}^{2}–5x–3}{4{x}^{2}–1}=\frac{\left(2x+1\right)\left(x-3\right)}{\left(2x–1\right)\left(2x+1\right)}⇒\frac{\overline{)\left(2x+1\right)}\left(x-3\right)}{\left(2x–1\right)\overline{)\left(2x+1\right)}}=\frac{\mathbf{x}\mathbf{–}\mathbf{3}}{\mathbf{2}\mathbf{x}\mathbf{–}\mathbf{1}}\phantom{\rule{0ex}{0ex}}$ ✔ Technique: The numerator is factored by decomposition; the denominator is a difference of squares. $\frac{{x}^{2}–ax+2bx–2ab}{2{x}^{2}+ax–3{a}^{2}}=\phantom{\rule{0ex}{0ex}}\frac{x\left(x–a\right)+2b\left(x–a\right)}{x\left(2x+3a\right)–a\left(2x+3a\right)}=\phantom{\rule{0ex}{0ex}}\frac{\left(x+2b\right)\left(x–a\right)}{\left(x–a\right)\left(2x+3a\right)}=\frac{\left(x+2b\right)\overline{)\left(x–a\right)}}{\overline{)\left(x–a\right)}\left(2x+3a\right)}=\frac{\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{a}}\phantom{\rule{0ex}{0ex}}$ ✔ Technique: Factor the numerator by grouping; the denominator is factored by decomposition.