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§ 7.3. Factoring Trinomials whose Leading Coefficient is Not One. Factoring Trinomials.
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§7.3 Factoring Trinomials whose Leading Coefficient is Not One
Factoring Trinomials Now we will try to factor trinomials whose lead coefficient is not one. We will begin by making sure that there is not a common factor running through a trinomial that we might factor out. That should always be the first step in factoring – to look for a common factor to remove. Then we will try to factor the trinomials into a product of two binomials. The method we used in the previous section will work here as well. It’s true that we will have a longer list of choices here in this case though. So, get ready. We have more trials and more errors…and more satisfaction in finding the correct factorization. Blitzer, Introductory Algebra, 5e – Slide #2 Section 7.3
Factoring Trinomials A Strategy for Factoring ax2 + bx + c • Find two first terms whose product is ax2: (□x + )(□x + ) = ax2 + bx + c • Find two Last terms whose product is c. (□x + □ )(□x + □ ) = ax2 + bx + c Blitzer, Introductory Algebra, 5e – Slide #3 Section 7.3
Factoring Trinomials A Strategy for Factoring ax2 + bx + c • By trial and error, perform steps 1 and 2 until the sum of the Outside and Inside product is bx. (□x + □ )(□x + □ ) = ax2 + bx + c I O Sum of I + O Blitzer, Introductory Algebra, 5e – Slide #4 Section 7.3
Factoring Trinomials Factor 6x2 - 13x - 8 EXAMPLE Step 1. Find two First terms whose product is 6x2. 6x2 - 13x – 8 ? (3x )(2x ) ? (6x )(x ) Blitzer, Introductory Algebra, 5e – Slide #5 Section 7.3
Factoring Trinomials Factor 6x2 - 13x - 8 CONTINUED Step 2. Find two Last terms whose product is -8. 1(-8), -1(8), 2(-4), -2(4) Blitzer, Introductory Algebra, 5e – Slide #6 Section 7.3
Factoring Trinomials Factor 6x2 - 13x - 8 CONTINUED Required middle term. Step 3. Find various combinations of these factors. Blitzer, Introductory Algebra, 5e – Slide #7 Section 7.3
Factoring Trinomials Factor 6x2 - 13x - 8 CONTINUED Step 3 Continued. From our chart of possible combinations, the correct factorization is: (3x - 8)(2x +1) Check with FOIL (3x - 8)(2x + 1) = 6x2 + 3x - 16x – 8 = 6x2 – 13x - 8 Blitzer, Introductory Algebra, 5e – Slide #8 Section 7.3
Factoring Trinomials EXAMPLE Factor: SOLUTION The GCF of the three terms of the polynomial is 4y. Therefore, we begin by factoring out 4y. Then we factor the remaining trinomial. Factor out the GCF Begin factoring . Find two integers whose product is -18 and whose sum is 3. The integers are -3 and 6. Thus, Blitzer, Introductory Algebra, 5e – Slide #9 Section 7.3
Factoring Trinomials I O Sum of O + I If no such combinations exist, the polynomial is prime. Blitzer, Introductory Algebra, 5e – Slide #10 Section 7.3
Factoring Trinomials EXAMPLE Factor: SOLUTION 1) Find two First terms whose product is There is more than one choice for our two First terms. Those choices are cataloged below. ? ? 2) Find two Last terms whose product is 15. There is more than one choice for our two Last terms. Those choices are cataloged below. ? ? Blitzer, Introductory Algebra, 5e – Slide #11 Section 7.3
Factoring Trinomials CONTINUED 3) Try various combinations of these factors. The correct factorization of is the one in which the sum of the Outside and Inside products is equal to 19x. Here is a list of some of the possible factorizations. This is the required middle term. Blitzer, Introductory Algebra, 5e – Slide #12 Section 7.3
Factoring Trinomials CONTINUED Therefore, the factorization of is: (2x + 3)(3x + 5) . Determine which possible factorizations were not represented in the chart on the preceding page. Blitzer, Introductory Algebra, 5e – Slide #13 Section 7.3
Factoring Trinomials EXAMPLE Factor: SOLUTION The GCF of the three terms of the polynomial is . We begin by factoring out . 1) Find two First terms whose product is ? ? Blitzer, Introductory Algebra, 5e – Slide #14 Section 7.3
Factoring Trinomials CONTINUED 2) Find two Last terms whose product is 3. The only possible factorization is (-1)(-3) since the sum of the Outside and Inside products must be -17y, having a negative coefficient. 3) Try various combinations of these factors. The correct factorization of is the one in which the sum of the Outside and Inside products is equal to -17y. A list of the possible factorizations can be found on the next page. Blitzer, Introductory Algebra, 5e – Slide #15 Section 7.3
Factoring Trinomials CONTINUED This is the required middle term. The factorization of is (5y - 1)(2y - 3). Now we include the GCF in the complete factorization of the given polynomial. Thus, Blitzer, Introductory Algebra, 5e – Slide #16 Section 7.3
Factoring Trinomials There is a second method that is not as commonly used in practice for factoring in trinomials, but that may be worth your consideration. This is a method that uses both trial and error as well as grouping. We will simply call this method Grouping. Blitzer, Introductory Algebra, 5e – Slide #17 Section 7.3
Factoring Trinomials EXAMPLE Factor by grouping: SOLUTION The trinomial is of the form a = 1 b = 1 c = -12 1) Multiply the leading coefficient, a, and the constant, c. Using a = 1 and c = -12. Blitzer, Introductory Algebra, 5e – Slide #18 Section 7.3
Factoring Trinomials CONTINUED 2) Find the factors of ac whose sum is b. We want the factors of -12 whose sum is b, or 1. The factors of -12 whose sum is 1 are 4 and -3. 3) Rewrite the middle term, a, as a sum or difference using the factors from step 2: 4 and -3. Blitzer, Introductory Algebra, 5e – Slide #19 Section 7.3
Factoring Trinomials CONTINUED 4) Factor by grouping. Group terms Factor from each group Factor out a + 4, the common binomial factor Thus, Blitzer, Introductory Algebra, 5e – Slide #20 Section 7.3
Factoring Completely IMPORTANT TO NOTE 1) Always begin the process of factoring a polynomial by looking for a greatest common factor other than 1. If there is one, factor out the GCF first. 2) Once you have factored out the GCF, there will be no common factor within any of the binomial factors. Knowing this will help you to narrow down your list of possible factorizations. For example in factoring the trinomial 6x2 - 13x – 8, I would not even try the following as a possibility: (2x - 4)(3x +2) This would give a proper first term and a proper last term. However it is not in my list of possibilities. I know this won’t work since 2 and 4 share a common factor. Because if there is not a common factor in the given trinomial, there won’t be a common factor in one of its factors. Blitzer, Introductory Algebra, 5e – Slide #21 Section 7.3