Factorisable expressions

Example with comments

Question

Solve

\frac{3 x - 7}{x - 1} > 1 - x.

Solution

Step 0. Do not cross multiply!

When working with rational equations like

\frac{3 x - 7}{x - 1} = 1 - x,

it is very common to "cross-multiply".

That will not work any more when dealing with inequalities as the sign of the inequality changes under multiplication depending on whether we are multiplying by a positive or negative term.

Our denominator ${x - 1}$ is either positive or negative, depending on ${x}$ so applying the multiplication step to an inequality will lead to a wrong result.

Step 1. Move terms by addition or subtraction

Since cross multiplication is not valid, we will use addition or subtraction to move terms around such that one side of the inequality is zero.

For our example, this gives us

\begin{gather*} \frac{3 x - 7}{x - 1} > 1 - x \\ \frac{3 x - 7}{x - 1} - 1 + x > 0 \end{gather*}

Step 2. Combine into a single fraction

We then combine the terms into a single fraction

\begin{gather*} \frac{3 x - 7}{x - 1} - 1 + x > 0 \\ \frac{3 x - 7 + (- 1 + x)(x - 1)}{x - 1} > 0 \\ \frac{x^2 + x - 6}{x - 1} > 0 \end{gather*}

Step 3. Factorise everything

We proceed with factorization

\begin{align*} \frac{x^2 + x - 6}{x - 1} &> 0 \\ \frac{(x + 3)(x - 2)}{x - 1} &> 0 \\ \end{align*}

Step 4. The number line approach

The factors

{(x + 3),}

{(x - 2)}

and

{(x - 1)}

have corresponding roots

{- 3, 2}

and

{1}

. Let's plot them out on a number line

number_line_regions

We notice that the number line is now broken up into 4 regions which we name ${A,B,C}$ and ${D.}$

Now consider region ${A}$ on the right. That happens when ${x}$ is bigger than ${2.}$ Under such a case, our fraction ${\frac{(x + 3)(x - 2)}{x - 1}}$ is positive because, individually ${(x + 3),}$ ${(x - 2)}$ and ${(x - 1)}$ will be positive when ${x > 2.}$

Now consider region ${B,}$ where ${1 < x < 2.}$ Now ${(x-2)}$ will be negative while ${(x+3)}$ and ${(x-1)}$ will still be positive so our overall fraction ${\frac{(x+3)(x-2)}{x-1}}$ will be negative.

Repeating this analysis over regions ${C}$ and ${D}$ gives us the following picture:

number_line_regions_positive_and_negative

An alternative method to the analysis above will be a "test point" method, where we substitute test values (e.g.

{x=3}

for region

{A}

and

{x=1.5}

for region

{B}

) into the fraction to check if the result is positive or negative.

Step 5. The final answer

We are now ready to solve our inequality. We want

\frac{(x+3)(x-2)}{x-1} > 0

so that means the answers are located in the positive regions

{C}

and

{A.}

This gives us our final answer

- 3 < x < 1 \quad \textrm{ or } \quad x > 2 \quad \blacksquare

Practice

Question

Solve the following inequality

\frac{2 x - 2}{x + 1} > 1 - x

Answer

- 3 < x < -1 \quad \textrm{ or } \quad x > 1

Solution

\begin{gather*} \frac{2 x - 2}{x + 1} > 1 - x \\ \frac{2 x - 2}{x + 1} - 1 + x > 0 \\ \frac{2 x - 2 + (- 1 + x)(x + 1)}{x + 1} > 0 \\ \frac{x^2 + 2 x - 3}{x + 1} > 0 \\ \frac{(x + 3)(x - 1)}{x + 1} > 0 \end{gather*}

- 3 < x < -1 \quad \textrm{ or } \quad x > 1 \quad \blacksquare

Extensions

What if our quadratic is not factorisable?

Maybe the quadratic still has roots, but they happen to be irrational so we are unable to obtain nice factors.

In that case we can use our quadratic formula ${x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$ to find the roots and resume with our number line approach.

Next technique

The next technique will tackle the case if our quadratic is not factorizable because it has no real roots