algebra review: factoring
TRANSCRIPT
Warmup
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Factoring Polynomials
Objective: Students will be able to solve quadratic functions by factoring using the grouping method.
* Question: What is a Quadratic Function?
Whatβs a Quadratic
A quadratic is a polynomial that is written in the form βππ₯2 + ππ₯ + π β where a, b, and c are all numbers.
Note that π, π = 0, but π β 0. Can you tell me why?
How to Factor A Quadratic by Grouping
Remember the standard form for a Quadratic: ππ₯2 + ππ₯ + π
First, find the product π β π (or ππ). Also, take note of the value of π (including the sign).
Next, find two numbers that multiply to get ππ and add into π or have a difference of π. It would be best to create a list of factors for ππ.
How to Factor A Quadratic by Grouping
Once you have identified the appropriate factors, you will replace βππ₯β with a sum/difference of those factors, also multiplied by βπ₯β
Next, factor completely by grouping (as we will see).
Practice Factoring: When π = π
Factor the following polynomial:
π₯2 + 5π₯ + 6
π = 1, π = 5, π = 6
Thus, the factors of ππ that
adds up to 5 are 2 and 3.
List of factors of ππ: β¦ 1,6 1 + 6 =
β¦ 2,3 2 + 3 =
Example 1 Continued
This mean π₯2 + 2π₯ + 3π₯ + 6
Group the first two and the second two terms π₯2 + 2π₯ + 3π₯ + 6
From the groups, factor out their GCF
π₯ π₯ + 2 + 3(π₯ + 2)
Notice that each group has a common term: βπ₯ + 2β
Factor that out from the group, and finally you have (π₯ + 2)(π₯ + 3)
Example
Factor the following polynomial:
π₯2 β 6π₯ + 8 Factors of ππ = 8
π = 1 ,π = β6 ,π = 8
The factors of 8 that add to make -6 are -2 and -4.
Thus π₯2 β 2π₯ β 4π₯ + 8
Next π₯ π₯ β 2 β 4(π₯ β 2)
Finally (π₯ β 2)(π₯ β 4)
Critical Question
How do we know if the factors of ππ have to add up to π or must have a difference of π.
How to remember the rules:
β¦ Add to π:
ππ₯2 + ππ₯ + π
ππ₯2 β ππ₯ + π
β¦ Difference of π:
ππ₯2 + ππ₯ β π
ππ₯2 β ππ₯ β π
Example
Factor the Polynomial completely: π₯2 + 3π₯ β 18
Add to π or Difference of π
Factors of ππ =
Then
π₯2 β 3π₯ + 6π₯ β 18 π₯ π₯ β 3 + 6(π₯ β 3)
(π₯ β 3)(π₯ + 6)
Day 2: Warmup
Factor the Following Polynomials
1. π₯2 + 9π₯ + 18
2. π₯2 β 12π₯ + 35
3. π₯2 + 5π₯ β 24
Day 2: What about when π > π?
The rules are the same, just a little more work.
Example:
2π₯2 + 7π₯ + 6 Factors of ππ = 12 that add to make π = 7 are
Next 2π₯2 + 4π₯ + 3π₯ + 6
2π₯ π₯ + 2 + 3(π₯ + 2)
(2π₯ + 3)(π₯ + 2)
ππ π
Example
Factor the Polynomial Completely 5π₯2 + 13π₯ β 6
Factors of ππ = β30 that have a difference of π = 13 are
5π₯2 + 15π₯ β 2π₯ β 6
5π₯ π₯ + 3 β 2 π₯ + 3
(π₯ + 3)(5π₯ β 2)
ππ π