# How to Solve Inequalities with Polynomial Fractions

In this post, let’s look at how to solve inequalities involving polynomial.

Let’s start off by looking at the steps needed to solve the polynomial inequalities:

Step 1: Determine the values that x cannot take (because denominator cannot be 0).

Step 2: Put all the non- zero terms on one side of the inequality, and make the other side of the inequality 0

Step 3: Combine everything on the non-zero side into a single fraction.

Step 4: Completely factorize the numerator and denominator.

Step 5: Determine the values that make numerator 0 or denominator 0.

Step 6: Put the numbers from step 5 into the number line. Put an open circle around the numbers that x cannot take (from step 1).

Step 7: Determine whether the polynomial fraction is positive or negative for each of the range.

Step 8: Determine the solution (remember to exclude values from step 1).

### Example: Solve the following inequality involving polynomial fractions:

Step 1: Determine the values that x cannot take (because denominator cannot be 0).

Since we only have 1 denominator here, x-3, it cannot be 0. Hence, x≠ 3.

Step 2: Put all the non- zero terms on one side of the inequality, and make the other side of the inequality 0.

Step 3: Combine everything on the non-zero side into a single fraction.

Step 4: Completely factorize the numerator and denominator.

Step 5: Determine the values that make numerator 0 or denominator 0.

Values that make numerator 0 are 5 and -3.

Values that make denominator 0 is 3.

Step 6: Put the numbers from step 5 into the number line. Put an open circle around the numbers that x cannot take (from step 1).

From step 5 we obtained these values of x: 5, -3 and 3.

We’ll put these numbers on the number line.

In addition, from step 1, x ≠ 3. Hence, we’ll put an open circle around 3 to remember that x cannot take this value.

Step 7: Determine whether the polynomial fraction is positive or negative for each of the intervals.

For interval A, where x is lesser than -3, we’ll substitute a value lesser than -3 into (-x+5)(x+3)/(x-3). For instance, we’ll substitute x=-4 in. (-(-4)+5)(-4+3)/(-4-3) = 9/7 > 0

For interval B, where x is between -3 and 3, we can substitute any value within this range into (-x+5)(x+3)/(x-3). For instance, we’ll substitute x = 0 in. (-0+5)(0+3)/(0-3)= -5 < 0.

For interval C, where x is between 3 and 5, we can substitute any value within this range into (-x+5)(x+3)/(x-3). For instance, we’ll substitute x = 4 in. (-4+5)(4+3)/(4-3)= 7 > 0.

For interval D, where x is more than 5, we’ll substitute a value more than 5 into (-x+5)(x+3)/(x-3). For instance, we’ll substitute x=6 in. (-(6)+5)(6+3)/(6-3) = -3 < 0

The number line now looks like this:

Step 8: Determine the solution (remember to exclude values from step 1).

Since we are interested in (-x+5)(x+3)/(x-3) ≥ 0, we are interested in the positive portion of (-x+5)(x+3)/(x-3) and the portions that gives (-x+5)(x+3)/(x-3) a value of 0.

From the number line, the solution for 12/(x-3)≥ x+1 is x≤-3 or 3<x≤5.

## Learn H2 A Level Math Inequalities

Here are the complete notes for solving inequalities for H2 Math:

## All the notes for H2 A Level Math

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