1
Understanding the Process
Let’s factor this quadratic equation by grouping:
6x² + 13x + 6
To factor by grouping, we need to:
- Rewrite the middle term (bx) as a sum of two terms
- Create a four-term expression that can be grouped
2
Splitting the Middle Term
Our original equation:
6x2 + 13x + 6
We need to split the middle term 13x into two terms that:
- Add up to 13 (middle term coefficient)
- Multiply to give 36 (ac)
After splitting the middle term, we have:
6x² + 9x + 4x + 6
Group the first two and last two terms:
(6x² + 9x) + (4x + 6)
Factor out the GCF from each group:
3x(2x + 3) + 2(2x + 3)
We now have: 3x(2x + 3) + 2(2x + 3)
Just like the name implies, factoring by grouping means that you will group terms with common factors before factoring. Among various factoring techniques, factoring by grouping is particularly useful for trinomials and higher-degree polynomials where other methods might be cumbersome. This lesson delves deep into factoring by grouping, providing clear explanations and multiple examples to ensure a thorough understanding.
What is Factoring by Grouping?
Factoring by grouping involves organizing the terms of a polynomial into groups that share a common factor. Once grouped, each set of terms is factored individually, and if done correctly, a common binomial factor emerges, allowing the entire polynomial to be factored completely.
Key Steps:
1. Group Terms: Organize the polynomial into pairs (or groups) of terms that have common factors.
2. Factor Each Group: Factor out the greatest common factor (GCF) from each group.
3. Factor Out the Common Binomial: If the grouping is successful, a common binomial factor will appear, which can then be factored out.
More Examples Explaining Factoring by Grouping
Let’s explore several examples to illustrate the factoring by grouping process.
Example 1: Factoring a Simple Quadratic
Problem: Factor x2 + 5x + 6
Step 1: Analyze the Polynomial
- The polynomial has three terms, making it a trinomial.
- To use factoring by grouping, we need to express it as a four-term polynomial.
Step 2: Expand the Middle Term
- Find two numbers that multiply to 1 × 6 = 6 and add up to 5.
- 5x = 3x + 2x
Step 3: Rewrite the Polynomial
x2 + 5x + 6 = x2 + 3x + 2x + 6
Step 4: Group Terms
- Group the first two terms and the last two terms:
(x2 + 3x) + (2x + 6)
Step 5: Factor Each Group
- Factor out the GCF from each group:
x(x + 3) + 2(x + 3)
Step 6: Factor Out the Common Binomial
- Both groups contain (x + 3):
(x + 3)(x + 2)
Conclusion:
x2 + 5x + 6 = (x + 3)(x + 2)
Note: Grouping differently, such as pairing x2 with 2x and 3x with 6, yields the same result. Notice also that there is more than one way we can expand 5x. Therefore, different groupings are possible.
5x is also equal to 4x + x, 6x – x, 7x – 2x, 8x – 3x, and so forth…
However, not all groupings will work!
Example 2: A Slightly More Complex Quadratic
Problem: Factor x2 − 4x − 12
Step 1: Analyze the Polynomial
- The polynomial is a trinomial with coefficients and/or minus sign(s) that complicate simple factoring.
Step 2: Expand the Middle Term
Step 3: Rewrite the Polynomial: x2 − 4x − 12 = x2 − 6x + 2x − 12
Step 4: Group Terms: (x2 − 6x) + ( 2x − 12)
Step 5: Factor Each Group: x(x − 6) + 2(x − 6)
Step 6: Factor Out the Common Binomial: (x − 6)(x + 2)
Conclusion:
x2 − 4x − 12 = (x − 6)(x + 2)
Insight: Not all obvious groupings work, highlighting the importance of finding the correct combination.
Example 3: Factoring with Coefficients
Problem: Factor 3y2 + 14y + 8
Step 1: Analyze the Polynomial
- The trinomial has a leading coefficient other than 1, which requires careful grouping.
Step 2: Expand the Middle Term
- Multiply the leading coefficient by the constant: 3 × 8 = 24.
- Find two numbers that multiply to 24 and add to 14: 12 and 2.
- 14y = 12y + 2y
Step 3: Rewrite the Polynomial: 3y2 + 14y + 8 = 3y2 + 12y + 2y + 8
Step 4: Group Terms: (3y2 + 12y) + (2y + 8)
Step 5: Factor Each Group: 3y(y + 4) +2(y + 4)
Step 6: Factor Out the Common Binomial: (y + 4)(3y + 2)
Conclusion:
3y2 + 14y + 8 = (y + 4)(3y + 2)
Tip: When the leading coefficient is not 1, multiply it by the constant term to find suitable factors.
Example 4: A More Complex Quadratic
Problem: Factor 11x2 − 41x − 12
Step 1: Analyze the Polynomial
- The trinomial has large coefficients, making simple factoring challenging.
Step 2: Multiply the Leading Coefficient and Constant
Step 3: Find Two Numbers That Multiply to -132 and Add to -41
- Potential pairs:
…..
…..
-44 and 3→ Sum: −41
-46 and 5
-45 and 4
-40 + -1
-39 + -2
-38 + -3
-37 + -4
…..
….. - Correct Pair: −44 and 3 since this is the one that is equal to -132 when multiplied
Step 4: Rewrite the Polynomial: 11x2 − 41x − 12 = 11x2 − 44x + 3x − 12
Step 5: Group Terms: (11x2 − 44x) + (3x − 12)
Step 6: Factor Each Group: 11x(x − 4)+ 3(x − 4)
Step 7: Factor Out the Common Binomial: (x − 4)(11x + 3)
Conclusion:
11x2 − 41x − 12 = (x − 4)(11x + 3)
Insight: When multiple groupings are possible, utilizing the product of the leading coefficient and the constant term can expedite finding the correct factors.
Common Pitfalls and Tips when Factoring by Grouping
Incorrect Grouping: Not all groupings will lead to a common binomial factor. Ensure that the numbers chosen for expanding the middle term are correct.
Multiple Groupings: Sometimes, multiple groupings are possible. Test each to find the one that results in a common binomial factor.
Leading Coefficient Greater Than 1: When the leading coefficient isn’t 1, multiply it by the constant term to find suitable factors for the middle term.
Verification: Always multiply the factored binomials to verify the correctness of the factoring.
Advanced Example: Applying the Technique
Problem: Factor 12x3 + 8x2 − 18x − 12
Step 1: Analyze the Polynomial
- This is a four-term polynomial, making it suitable for factoring by grouping.
Step 2: Group Terms: (12x3 + 8x2)+(−18x − 12)
Step 3: Factor Each Group:
- First group: 12x3 + 8x2 = 4x2(3x + 2)
- Second group: −18x − 12 = −6(3x + 2)
Step 4: Factor Out the Common Binomial:
4x2(3x + 2) − 6(3x + 2) = (3x + 2)(4x2 − 6)
Step 5: Factor Further if Possible
- 4x2 − 6 = 2(2x2 − 3)
Final Factored Form:
12x3 + 8x2 − 18x − 12 = 2(3x + 2)(2x2 − 3)