**Q: Solve for x: (x + 3)(x – 4) < 0**

A:

**Step 1: Find the zeroes**

Since the quadratic is already factored, this isn’t too tricky. If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 **or** x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line:

Now, you have to test each “interval” that is separated by the “zeroes”. There are **three intervals** to test.

Interval 1: The numbers to the left of -3 –> in interval notation this is (-infinity, -3)

Interval 2: The numbers between -3 and 4 –> in interval notation this is (-3, 4)

Interval 3: The numbers to the right of 4 –> in interval notation this is (4, infinity)

**Step 2: Test each interval**

Pick any number on interval 1 and test it into the original inequality. I’ll pick -5:

**(x + 3)(x – 4) < 0**

**(-5 + 3)(-5 – 4) < 0**

**(-2)(-9) < 0**

18 < 0 <— this is **false**, so numbers on this interval [interval 1] are **not part of the solution.**

Pick any number on interval 2 and test it into the original inequality. I’ll pick 0:

**(x + 3)(x – 4) < 0**

**(0 + 3)(0 – 4) < 0**

**(3)(-4) < 0**

-12 < 0 <— this is **true**, so numbers on this interval [interval 2] are a part of the** solution.**

Pick any number on interval 3 and test it into the original inequality. I’ll pick 5:

**(x + 3)(x – 4) < 0**

**(5 + 3)(5 – 4) < 0**

**(8)(1) < 0**

8 < 0 <— this is **false**, so numbers on this interval [interval 3] are **not part of the solution**.

SO: The only interval that “worked” was interval 2. Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4