Multiplying and Dividing Rational Expressions »

Numerator and denominator are linear functions

Example 1

Simplify the following rational expression:

$\frac{{2x + 4}}{{3x + 6}}$

Solution

1: Factor numerator: $2x + 4 = 2(x + 2)$

2: Factor denominator: $3x + 6 = 3(x + 2)$

3: Cancel common factors:

$\frac{{2x + 4}}{{3x + 6}} = \frac{{2\cancel{{(x + 2)}}}}{{3\cancel{{(x + 2)}}}} = \frac{2}{3}$

Example 2

Simplify the following rational expression:

$\frac{{4 - 2x}}{{3x - 6}}$

Solution

1: Factor numerator: $4 - 2x = 2(2 - x)$

2: Factor denominator: $3x - 6 = 3(x - 2)$

3: Cancel common factors:

$\frac{{4 - 2x}}{{3x - 6}} = \frac{{2(2 - x)}}{{3(x - 2)}}$

The factors 2 - x and x - 2 are almost the same, but not quite, so they can't be cancelled. Remember to switch the sign out front: 2 - x = -(x - 2)

$\frac{{4 - 2x}}{{3x - 6}} = \frac{{2(2 - x)}}{{3(x - 2)}} = \frac{{ - 2(x - 2)}}{{3(x - 2)}} = - \frac{2}{3}$

Exercise 1: Simplify the following expression

Numerator and denominator are quadratic trinomials

Factor a quadratic trinomial

To factor a quadratic trinomial we will use following formula

$a{x^2} + bx + c = a(x - {x_1}) + (x - {x_2})$ where ${x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Example 3 (IMPORTANT)

Factor the trinomial 2x² + 3x - 2

Solution 3

In this example a = 2, b = 3, c = -2. Plugging these numbers into the quadratic formula we get:

$${x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2z}}$$ $${x_{1,2}} = \frac{{ - 3 \pm \sqrt {{3^2} - 4 \cdot 2 \cdot ( - 2)} }}{{2 \cdot 2}}$$ $${x_{1,2}} = \frac{{ - 3 \pm 5}}{4}$$ $${x_1} = \frac{{ - 3 + 5}}{4}$$ $${x_2} = \frac{{ - 3 - 5}}{4}$$ $${x_1} = \frac{1}{2}$$ $${x_2} = - 2$$

We then have: $2{x^2} + 3x - 2 = 2 \cdot \left( {x - \frac{1}{2}} \right)\left( {x + 2} \right)$

Example 4

Simplify the following rational expression:

$\frac{{2{x^2} - x - 1}}{{ - {x^2} + 3x - 2}}$

Solution

Again, the first thing that we will do here is factor the numerator and denominator.

1: Factor numerator:

$2{x^2} - x - 1 = 2(x - 1)\left( {x + \frac{1}{2}} \right)$

2: Factor denominator:

$- {x^2} + 3x - 2 = - (x - 1)(x - 2)$

3: Cancel common factors:

$\frac{{2{x^2} - x - 1}}{{ - {x^2} + 3x - 2}} = \frac{{2\cancel{{(x - 1)}}\left( {x + \frac{1}{2}} \right)}}{{ - \cancel{{(x - 1)}}(x - 2)}} = \frac{{2\left( {x + \frac{1}{2}} \right)}}{{ - (x - 2)}} = \frac{{2x + 1}}{{ - x + 2}}$

Nothing else will cancel and so we have reduced this expression to lowest terms.

Exercise 2: Simplify the following expression