# Factoring a trinomial using decomposition

Worksheet:Quadratic Expand and Factor Skill Sheet

# Review Let’s expand : We notice that gives us the ; and that gives us the constant .

To factor a trinomial of the form , we need two values such that and .

## Example: The factor pairs of 54 are (1,54); (2,27); (3,18); (6,9). Now the last pair has a difference of 3, which is what we need for the middle term.

Let and because and .

Then, # Factor , where ## Example 1:

First, let’s examine the expansion of a factored trinomial of this form: .

This time, notice that as expected, but . Rather, .

Instead of looking for two numbers that add to 17 and multiply to 10, we need to look for two numbers that add to 17 and multiply to 30.

## Example 2:

Factor: Look for two numbers, and , such that and .

The factor pairs of 40 are: (1,40); (2,20); (4,10); (5, 8)

To make 22, we need to use the pair (2,20).

We write a new line of work: (It doesn’t matter if you write the first or the first).

Now we look for a common factor in the first two terms: Perhaps suprisingly, (explained here), let’s look at the last two terms; Ha! is a common factor! Put together we have: Taking out as a common factor, we have: ## Example 3: Find two numbers, and such that and .

Factor pairs of 24 are (1,24); (2,12); (3,8); (4,6). Now so let and .

Decomposing the middle term we have: Grouping we have: Completing we have .

## Example 4: First, we notice that all three terms share a common factor 3. Let’s factor this out: Now we factor the reduced trinomial ; however the is kept present throughout. You might also try factoring without reducing to see how the result compares.

Now we look for and such that , and .

Factor pairs of 8 are (1, 8); (2, 4). To also satisfy , we need to use and .

Decomposing we have Grouping gives Completing we have # Try it out:

Correct fields show as green, incorrect as red.

Geogebra link to the old applet

Geogebra link to this applet, for full screen capability

Proof of the method

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