When factoring expressions, there are a couple of special cases, or shortcuts, that can be very helpful. The trick is being able to recognize when these cases come up!

**The Difference of Perfect Squares**

This first special case isn’t that difficult to recognize.

Difference means subtraction and a perfect square is a number that you can take a square root of without getting a decimal answer. For example:

25 is a perfect square because \(\sqrt {25} = 5\).

144 is a perfect square because \(\sqrt {144} = 12\).

Even \({x^2}\) is perfect square because \(\sqrt {{x^2}} = x\).

So, if we had \(64{x^2}\), this would count as a perfect square as well because \(\sqrt {64{x^2}} = 8x\).

To fully understand this special case, we need to grasp the fact that if we had \({x^2} = 49\), x would equal 7, but it would ALSO equal -7 because \({( - 7)^2} = 49\).

Let’s see how we can put all these concepts together to get to our shortcut.

The general form for the difference of perfect squares is:

\({a^2} - {b^2} = (a + b)(a - b)\)

**Example 1:**

Factor the following. \({x^2} - 9\)

The square root of \({x^2}\) is **x** and the square root of 9 is **3**. This splits and factors into:

(x + 3)(x – 3)

**Example 2:**

Factor the following. \(25{x^2} - 4\)

The square root of \(25{x^2}\) is **5x** and the square root of 4 is** 2**. This splits and factors into:

(5x + 2)(5x – 2)

**Example 3:**

Factor the following. \({x^2} + 16\)

Don’t get fooled! This isn’t our special case because it’s not subtraction. This is *prime, *which means it cannot be factored.

**Trinomial Squares**

This special case shortcut can be helpful but is a bit tougher to recognize. It still has to do with perfect squares. The first coefficient and the last coefficient will be perfect squares. The middle term will be two times the product of those coefficients’ square roots. It will make more sense with an example.

The general form for a trinomial square is:

\({a^2} + 2ab + {b^2} = {(a + b)^2}\) or \({a^2} - 2ab + {b^2} = {(a - b)^2}\)

**Example 1:**

Factor the following. \({x^2} - 6x + 9\)

The square root of \({x^2}\) is **x** and the square root of 9 is **3**. The middle term matches our special case because \(2(x \bullet 3) = 6x\). This factors into:

\({(x - 3)^2}\)

**Example 2:**

Factor the following. \(4{x^2} + 36x + 81\)

The square root of \(4{x^2}\) is **2x** and the square root of 18 is **9**. The middle term matches our special case because \(2(2x \bullet 9) = 36x\). This factors into:

\({(2x + 9)^2}\)

**Example 3:**

Factor the following. \(486{x^2} + 864x + 384\)

This one is different. The first and last terms aren’t perfect squares! Let’s find a factor that will divide equally into each term. We can factor out a 6.

\(6(81{x^2} + 144x + 64)\)

Well, that’s better. The square root of \(81{x^2}\) is **9x** and the square root of 64 is **8**. The middle term matches our special case because \(2(9x \bullet 8) = 144x\). This factors into:

\(6{(9x + 8)^2}\)

Below you can **download** some **free** math worksheets and practice.

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**