FACTORING GENERAL TRINOMIALS IN THREE DIFFERENT CASES | Q1 Week 1−2 Part 4
31This is a Module based lesson for MATH 8 with the topic FACTORING GENERAL TRINOMIALS IN THREE DIFFERENT CASES (Easy case - Leading coefficient is 1, Common Factor Case and Difficult Case - Leading coefficient other than one and has no common factor) for Quarter 1 Week 1−2 : Part 4
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Learning Goals and Targets
Factor trinomials with a leading coefficient of 1.
Factor trinomials with a common factor.
Factor trinomials with a leading coefficient other than one.
Video Flow:
0:00 Sneak Peek Preview
1:12 Introduction
1:49 Learning Goals and Targets
2:13 Terminologies Explained
2:58 Strategy steps in solving x2+bx+c
4:03 Example #1
7:34 Example #2
10:50 Steps in Solving ax2+bx+c with common factor
11:14 Example #1
14:14 Example #2
15:55 Steps in Factoring Trinomials with leading coefficient other than 1
17:43 Example #1
23:45 Closing Message and references
You may also watch connected videos:
Factoring polynomials with common Monomials: https://youtu.be/kcc-hOlp6cM
Difference of Two Squares: https://youtu.be/XvdX-iYaEzU
Perfect Square Trinomials: https://youtu.be/aI_9wzkm_f4
A polynomial with three terms is called a trinomial. Trinomials often (but not always!) have the form x2 + bx + c. At first glance, it may seem difficult to factor trinomials, but you can take advantage of some interesting mathematical patterns that I am going to share to you to factor even the most difficult-looking trinomials. So, how do you get from 6x2 + 2x – 20 to (2x + 4)(3x −5)? Let’s look.
The general form of a quadratic trinomial is written as ax2 + bx + c where a, b, and c are constants. The “easy” case happens when the value of aa is equal to +1 or −1, that is a=1 or a=−1. You don’t need to write the coefficient of 1 before the {x^2} term because it is understood.
The basic strategy to factor this type of trinomial is to find two numbers (factor pair) which when multiplied, gives the constant number c. More so, their sum (when added together) should equal to constant b, the coefficient of the x term.
In this trinomial, we need to identify the other relevant constants. Observe that b = 7 and c = 10. Next, find two numbers (factor pair) that when multiplied equals the constant value of c = 10, and when added equals the constant value of b = 7. Because the product of two numbers must be positive, the two numbers should be both positive, or both negative. To find the pair, you can perform several trial and error to find the correct combination. Here are some possible combinations. The only combination of numbers that can satisfy the two given requirements is the third option. The one with the green checkmark. We can finally express the binomial factors of this trinomial by writing down a pair of parenthesis with an x as the leading term. Since the correct combination of numbers are 5 and 2, our final answer should be (x+5)(x+2)
Not all trinomials look like x2 + 5x + 6, where the coefficient in front of the x2 term is 1. In these cases, your first step should be to look for common factors for the three terms.
The general form of trinomials with a leading coefficient of a is ax2 + bx + c. Sometimes the factor of a can be factored as you saw above; this happens when a can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an x2 term, instead of an ax2 term.
However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.
References:
https://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U12_L2_T1_text_final.html
https://www.chilimath.com/lessons/intermediate-algebra/factoring-trinomial-easy-case/
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