This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

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A **factor **of a number is another number that divides into the original number evenly (i.e., without a remainder).

For example, \(4\) is a factor of \(12\) because it divides into \(12\) without a remainder, but \(5\) is not. Knowing that \(4\) is a factor, \(12\) can then be written in factored form:

\[ 12 = (4)(3) \]

\((4)(3)\) is one factored form of \(12\). \((2)(6)\) is another.

The same principle applies to polynomials - A factor of an algebraic expression is another algebraic expression that divides into the original one without remainder.

- \(x\) is a factor \(x^3\) because \(x^3 = (x)(x^2)\). It follows that \(x^2\) is also a factor of \(x^3\), and \((x)(x^2)\) is a factored form of \(x^3\)
- \(x+2\) is a factor of \(x^4 + 2x^3\) because \(x^4 + 2x^3 = (x^3)(x+2)\).

When you are asked to factor an expression, you need to rewrite it in a factored form like those above. Below, you will find a variety of factoring methods.

When factoring an expression, the **first type** of factoring you should try is common term factoring. To perform common term factoring,

- Find the greatest common factor of all terms in the expression
- Re-write each term as a product of the greatest common factor
- Factor out the greatest common factor.

As you get more comfortable using common tern factoring, you may skip Step 2.

`Example`

Factor the expression \(3x^3 + 6x^2\).

`Solution`

First, we need to identify the greatest common factor of \(3x^3\) and \(6x^2\). Starting with the numbers, the greatest common factor (GCF) of \(3\) and \(6\) is \(3\). Now looking at the \(x\) terms, the GCF of \(x^3\) and \(x^2\) is \(x^2\). Using the information, we can rewrite each term as a product of the GCF as follows:

\[ 3x^3 + 6x^2 = (3x^2)(x) + (3x^2)(2) \]

Now, we just need to factor the greatest common factor out:

\[ (3x^2)(x) + (3x^2)(2) = 3x^2(x + 2) \]

`Example`

See the video below for another example of common term factoring.

Another type of factoring you need to know is difference of squares. To perform difference of squares factoring, just apply the following formula:

\[ A^2 - B^2 = (A+B)(A-B) \]

This type of factoring should be applied if your expression consists of two terms separated by a minus sign. In many cases, the variable has an even power and the number is a perfect square, but this is not necessarily always the case.

`Example`

Factor \(x^2 - 4\).

`Solution`

First, rewrite the expressions so that it is in the form \((\hspace{0.25cm})^2 - (\hspace{0.25cm})^2\):

\[ x^2 - 4 = (x)^2 - (2)^2 \]

Now that it is in the required form, the formula above can be applied. In this context, \(A\) is \(x\) and \(B\) is \(2\). Applying the formula:

\[ (x)^2 - (2)^2 = (x+2)(x-2) \]

`Example`

See the video below for another examples of applying the difference of squares formula.

There are also formulas for factoring a sum and difference of cubes:

\[ A^3 - B^3 = (A-B)(A^2 + AB + B^2) \]

\[ A^3 + B^3 = (A+B)(A^2 - AB + B^2) \]

`Example`

Factor \( x^3 - 8 \).

`Solution`

Notice that there is a minus sign separating the terms in this expression, so the difference of cubes is the formula to apply. We start by rearranging the expression into the form \( (\hspace{.25cm})^3 - (\hspace{.25cm})^3 \):

\[ x^3 - 8 = (x)^3 - (2)^3 \]

Now that it is in the required form, apply the formula. In this case, \(A\) is \(x\) and \(B\) is \(2\)

\[ (x)^3 - (2)^3 = (x-2)(x^2 + (x)(2) + 2^2) \]

\[ = (x-2)(x^2 + 2x + 4) \]

`Example`

Factor \(8x^3 + 27\)

`Solution`

Again, start by rewriting into the form \( (\hspace{.25cm})^3 +(\hspace{.25cm})^3 \):

\[ 8x^3 + 27 = (2x)^3 + (3)^3 \]

Now we apply the sum of cubes formula where \(A\) is \(2x\) and \(B\) is \(3\)

\[ (2x)^3 + (3)^3 = (2x+3)((2x)^2 - (2x)(3) + 3^2) \]

\[ = (2x+3)(4x^2 - 6x + 9) \]

If you need to factor an expression with 4 terms (sometimes more) you may need to use factoring by grouping. The following are the steps you need to take in order to factor by grouping.

- Group the terms in the expression into pairs. This may involve rearranging the order of the terms.
- Apply common term factoring to the each pair separately, factoring out the greatest common factor of each group.
- Apply common term factoring again to the resulting expression as a whole.

`Example`

Factor \(x^3-2x^2 + 6x - 12\).

`Solution`

Start by grouping the terms in the expression. Let's consider the first two terms as a group and the last two terms as a group.

\[ x^3-2x^2 + 6x - 12 = (x^3-2x^2) + (6x - 12) \]

Next, perform common term factoring on each of the groups:

\[ (x^3-2x^2) + (6x - 12) = x^2(x-2) + 6(x-2) \]

Now, we need to apply common term factoring one more time to the expression as a whole. Notice that \(x-2\) is present as a factor in both of the terms so that is what is factored out as the common term:

\[ x^2(x-2) + 6(x-2) = (x-2)(x^2 + 6) \]

`Example`

In this section we will learn how to factor trinomials in the form

\[ ax^2 + bx + c \]

**When a = 1:**

When \(a=1\), the trinomial is in the form

\[ x^2 + bx + c \]

For these cases, you need to find two numbers that satisfy the following rules:

- The numbers multiply to make \(c\)
- The sum of the numbers is \(b\)

When you have found those numbers, just drop them into the equation in the spaces below:

\[ x^2 + bx + c = (x + \underline{\hspace{.5cm}})(x + \underline{\hspace{.5cm}}) \]

`Example`

Factor \(x^2 - x - 6\).

`Solution`

To factor this expression, we need to find two numbers which

- Multiply to make \(-6\)
- Sum to \(-1\)

Start by listing all the whole numbers that multiply to make \(-6\). Once we've have them all, we choose the pair that add to make \(-1\) and those are the numbers we use in the formula.

Numbers multiplying to make \(-6\):

- \(6\) and \(-1\)
- \(-6\) and \(1\)
- \(3\) and \(-2\)
- \(-3\) and \(2\)

The pair of numbers that also sum to make \(-1\) are \(-3\) and \(2\), so those are the numbers we use in our factoring.

\[ x^2 -x -6 = (x-3)(x +2) \]

*When a \(\neq\) 1:*

When \(a \neq 1\), the method is different and involves factoring by grouping. To factor a trinomial \( ax^2 + bx + c \), you need numbers which

- Multiply to make \(a \times c\)
- Sum to make \(b\)

Suppose those numbers are \(M\) and \(N\). Remember, \(M\) and \(N\) sum to make \(b\), so rewrite the trinomial as follows:

\[ ax^2 + bx + c = ax^2 + Mx + Nx + c \]

and finish the problem using factoring by grouping.

`Example`

Factor \(2x^2 - x - 6\)

`Solution`

Following the steps outline above, we need to find the numbers that multiply to make \(-12\) and sum to \(-1\). Start of by listing the numbers that multiply to \(-12\):

- \(12\) and \(-1\)
- \(-12\) and \(1\)
- \(6\) and \(-2\)
- \(-6\) and \(2\)
- \(4\) and \(-3\)
- \(-4\) and \(3\)

The pair of numbers that also sum to make \(-1\) is \(-4\) and \(3\). Next,

\[ 2x^2 - x - 6 = 2x^2 -4x + 3x - 6 \]

Now we solve using factoring by grouping.

\[ 2x^2 -4x + 3x - 6 = 2x(x-2) + 3(x-2)\]

\[ = (x-2)(2x+3) \]

Special products are algebraic products that appear often. Being familiar with them and their factored form can make solving equations and factoring easier. Some common special products are:

- \(a(x+y) = ax + ay\) where \(a\) is a constant
- \((x+y)(x-y) = x^2-y^2\)
- \((x+y)^2 = x^2 + 2xy + y^2\)
- \((x-y)^2 = x^2 - 2xy - y^2\)
- \((x+a)(x+b) = x^2 + (a+b)x + ab\) where \(a, b\) are constants
- \((ax+b)(cx+d) = acx^2+(ad+bc)x+bd\) where \(a, b, c, d\) are constants
- \((x+y)^3 = x^3+3x^2y+3xy^2+y^3 = (x+y)(x^2-xy+y^2)\)
- \((x-y)^3 = x^3-3x^2y+3xy^2-y^3 = (x-y)(x^2+xy+y^2)\)

- Last Updated: Mar 4, 2024 10:18 AM
- URL: https://libraryguides.centennialcollege.ca/mathhelp
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