# Factoring expressions with 3 terms into single brackets. Single bracket factorization.

In this article I will show you how to factorise an expression containing three terms into a single bracket. All you need to do is look for the greatest common divisor of the 3 numbers in your expression, and look for any letters common to all 3 terms in the expression. Put these letters and numbers outside of the bracket, and then work out what you need inside the bracket to give the original expression once the bracket is expanded.

**Question 1**

Factorize 12a + 14ab + 8ac

The greatest common divisor of 12, 14 and 8 is 2. Put the 2 outside the bracket.

Also there is a letter a in each of the three terms. Put the an outside the bracket too.

Therefore, 2a is the factor outside the bracket.

To give 12a, you must multiply 2a by 6.

To give 14ab, you must multiply 2a by 7b.

To give 8ac, you must multiply 2a by 4c.

So 6, 7b and 4c go inside the bracket.

Therefore:

2a(6 + 7b + 4c)

**Question 2**

Factorize 15mn + 9n + 18mnp

The greatest common divisor of 15, 9 and 18 is 3. Put the 3 outside the bracket.

Also there is a letter n in each of the three terms. Put the n outside the bracket too.

Therefore, 3n is the factor outside the bracket.

To give 15mn, you must multiply 3n by 5m.

To give 9n, you must multiply 3n by 3.

To give 18mnp, you must multiply 3n by 6mp.

So 5m, 3 and 6mp go inside the bracket.

Therefore:

3n(5m + 3 + 6mp)

**Question 3**

Factorize 16x² + 20xy + 4x

The greatest common divisor of 16, 20 and 4 is 4. Put the 4 outside the bracket.

Also there is a letter x in each of the three terms. Put the x outside the bracket too.

Therefore, 4x is the factor outside the bracket.

To give 16x², you must multiply 4x by 4x.

To give 20xy, you must multiply 4x by 5y.

To give 4x, you must multiply 4x by 1.

So 4x, 5y and 1 go inside the bracket.

Therefore:

4x(4x + 5y + 1)

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