This pdf contains the following topics:- Greatest Common Factor and Factoring by Grouping Identify and factor out the Greatest Common Factor (GCF) Factoring by grouping Factoring Trinomials Factoring by Special Products & Integrated Review
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 6x2y is 2*3*x*x*y, the factored form of 9x2y3 is 3*3*x*x*y*y*y. The factors common to both monomials are 3*x*x*y, so the GCF is 3x2y. To factor out a GCF from a polynomial, we determine the GCF of the polynomial’s terms and “un- distribute” 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: 6x2y + 9x2y3 = 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 + 3y2 have no common Examples: 8y2 + 4 24x2 – 6x3 + 18x4 20a4b2 – 12a2b2 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: -2x3 + 4x2 – 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: y3 + 6y2 + 4y + 24 a2b2 + a2 – 3b2 – 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: x3 + 3x2 – 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 x2 + bx + c If we were to FOIL (x + 2)(x +3), we would see that (x + 2)(x + 3) = These are equivalent expressions, but x2 + 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 x2 + 5x + 6. In general the factored form of x2 + 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: x2 + 5x + 4 x2 + 9x + 18 2x2 + 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: x2 + 3x + 6 y2 – 4y + 5.
3. Factoring ax2 + bx + c Method 1-Trial and Check: Make a good guess for what ax2 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 ax2 + bx + c and rewrite as ax2 + px + qx + c. Now you can factor by grouping. 2x2 – 5x – 3 4y2 + 17y + 4 6a2 – a – 12 3x2 + 13x + 4 4x2 + 28xy + 49y2 14x2 – x – 3 Trick: 4a2 + 32a + 60 Helpful Hint—Sign Patterns ax2 + bx + c = (#x + #)(#x + #) ax2 – bx + c = (#x – #)(#x – #) ax2 + bx – c = (#x + #)(#x – #) ax2 – 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 a2 + 2ab + b2 = (a + b) 2 a2 – 2ab + b2 = (a – b) 2 Examples: x2 + 8x + 16 x2 – 14x + 49 4x3 – 32x2y + 64xy2 Factor the difference of two squares. Omit Example 5 a2 – b2 = (a + b)(a – b) Examples: x2 – 49 4y2 – 8112 – 3a2 Note: a2 + b2 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: a2 – b2 = (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: a2 + 2ab + b2 = (a + b) 2 a2 – 2ab + b2 = (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.
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