Can there be more than one correct factorization of a polynomial? There can be depending on the sign: -2x 3 + 4x 2 6x can factor to either

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1 MTH95 Day 9 Sections 5.5 & 5.6 Section 5.5: Greatest Common Factor and Factoring by Grouping Review: The difference between factors and terms Identify and factor out the Greatest Common Factor (GCF) Factoring is un-multiplying, 6 = 2 * 3, so 2 and 3 are factors of 6, and 2*3 is the factored form of 6. The factored form of 6x 2 y is 2*3*x*x*y, the factored form of 9x 2 y 3 is 3*3*x*x*y*y*y. The factors common to both monomials are 3*x*x*y, so the GCF is 3x 2 y. To factor out a GCF from a polynomial, we determine the GCF of the polynomial s terms and undistribute the GCF. Our goal is the have an expression equivalent to the original polynomial, but an expression that is a product of a monomial (the GCF) and a simpler polynomial. Example: 6x 2 y + 9x 2 y 3 = Note that if we were to multiply our result, we would have exactly what we started with so this is an equivalent expression but written in a different format. Note also that 2 + 3y 2 have no common factors. Examples: 8y x 2 6x x 4 20a 4 b 2 12a 2 b 2 Note: for the first example, the 1 is not optional! Without having the 1 as a placeholder, we would not have an equivalent expression the second term of the polynomial would be lost. Can there be more than one correct factorization of a polynomial? There can be depending on the sign: -2x 3 + 4x 2 6x can factor to either Factoring by grouping Start with: 5(x + y) + 2x(x + y) There are two terms: 5(x + y) is a term and 2x(x + y) is a term. Do the two terms have a common factor? Factored form: Be sure you understand this before going on. Factoring by grouping is a technique to try when there is no common factor for a polynomial and the polynomial has 4 terms.

2 First, group the terms into two pairs, then factor each pair and check whether a common factor appears between the pairs. Examples: y 3 + 6y 2 + 4y + 24 a 2 b 2 + a 2 3b 2 3 xy + 5x y 5 Note: Order of the factors does not matter for the final answer. Try rearranging the terms, do all pairings work? An example that does not factor: x 3 + 3x 2 5x 5 = There is no common factor so this polynomial is a prime polynomial. And it DOES NOT factor to Section 5.6 Factoring Trinomials: Factoring x 2 + bx + c If we were to FOIL (x + 2)(x +3), we would see that (x + 2)(x + 3) = These are equivalent expressions, but x 2 + 5x + 6 is the sum of 3 terms, while (x + 2)(x + 3) is the product of two factors. So (x + 2)(x + 3) is the factored form of x 2 + 5x + 6. In general the factored form of x 2 + bx + c will be (x + some number)(x + another number). How do we find those unknown numbers? We need to look for two numbers whose product equals c and whose sum equals b. Sometimes you will be able to just guess what the two numbers must be, but let s practice using a systematic approach. List all possible factors, check the sums to find the correct pair. Examples: x 2 + 5x + 4 x 2 + 9x x 2 + 6x + 4 Note: Always check to see if you have a GCF first! It can make a difficult problem much simpler. Some trinomials are prime and cannot be factored: x 2 + 3x + 6 y 2 4y + 5.

3 Factoring ax 2 + bx + c Method 1-Trial and Check: Make a good guess for what ax 2 would factor as, and then a good guess on how c will factor. Then check by FOILing to see if it worked and gave you the correct middle term. Method 2-Grouping: Find two numbers whose product is a * c and whose sum is b. Let s call those two numbers p and q. So p * q = a * c and p + q = b. Take ax 2 + bx + c and rewrite as ax 2 + px + qx + c. Now you can factor by grouping. 2x 2 5x 3 4y y + 4 6a 2 a 12 3x x + 4 4x xy + 49y 2 14x 2 x 3 Trick: 4a a + 60 Helpful Hint Sign Patterns ax 2 + bx + c = (#x + #)(#x + #) ax 2 bx + c = (#x #)(#x #) ax 2 + bx c = (#x + #)(#x #) ax 2 bx c = (#x + #)(#x #)

4 Section 5.7: Factoring by Special Products & Integrated Review This will be a short section to cover but you will also need to review factoring strategies from the Integrated Review. Factoring a perfect square trinomial a 2 + 2ab + b 2 = (a + b) 2 a 2 2ab + b 2 = (a b) 2 Examples: x 2 + 8x + 16 x 2 14x x 3 32x 2 y + 64xy 2 Factor the difference of two squares. Omit Example 5 a 2 b 2 = (a + b)(a b) Examples: x y a 2 Note: a 2 + b 2 will be a prime polynomial Omit Objective 3: Factor the sum or difference of two cubes. Omit Examples 6 9. Factoring a Polynomial Step 1. Are there any common factors? If so, factor the greatest common factor. Step 2. How many terms are in the polynomial? a. If there are two terms, decide if one of the following formulas may be used: I. Difference of two squares: a 2 b 2 = (a + b)(a b) II. III. (Omit the difference and sum of two cubes.) b. If there are three terms, try one of the following: I. Perfect square trinomial: a 2 + 2ab + b 2 = (a + b) 2 a 2 2ab + b 2 = (a b) 2 II. If not a perfect square, try Trial and Check or Grouping (ac method) c. If there are four terms, try factoring by grouping. Step 3. See whether any of the factors in the factored polynomial can be factored further. Examples:

5 12x 2 22x 20 7x 2 63 x 6x 2 6x 12 5x 2 2x 3 4x x x x + 4xy + 16y