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5.5 Factoring

5.5 Factoring. Algebra 2 Mrs. Spitz Fall 2006. Objective:. Factor polynomials. Assignment. pg. 233 #5-54. Example 1. Try factoring 12y 2 + 15y First, write each term as the product of its prime factors. 12y 2 = 2  2  3  y  y 15y = 3  5  y The GCF of 12y 2 + 15y is 3y.

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5.5 Factoring

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  1. 5.5 Factoring Algebra 2 Mrs. Spitz Fall 2006

  2. Objective: • Factor polynomials Assignment • pg. 233 #5-54

  3. Example 1 • Try factoring 12y2 + 15y First, write each term as the product of its prime factors. 12y2 = 2  2  3  y  y 15y = 3  5  y The GCF of 12y2 + 15y is 3y

  4. Example 2: Factor 24a2b – 18ab2. Draw a geometric representation of the expression. 24a2b – 18ab2 = (2  2  2  3  a  a  b) – (2  3  3  a  b  b) = (6ab  4a) – (6ab  3b) = 6ab(4a – 3b) The rectangle is 6ab units wide and (4a – 3b) units long. Area = 6ab(4a – 3b) or 24a2b – 18ab2 6ab 4a – 3b

  5. Ex. 3: Factor 3s2t + 4st2 + st3 3s2t = 3  s  s  t 4st2 = 2  2  s  t  t st3 = s  t  t  t = (st  3s) + (st  4t) + (st  t2) = st(3s + 4t + t2) Sometimes just factoring out the GCF does not result in a completely factored polynomial. In those cases, rearranging and grouping the terms can be helpful.

  6. Ex. 4: Factor a2 – 2ab + a – 2b a2 – 2ab + a – 2b = (a2 – 2ab) + (a – 2b) = a(a – 2b) + 1(a – 2b) What is common now? = (a – 2b)(a + 1) Look at the terms in pairs and factor out the GCF of each. (a – 2b) is the GCF of the two new terms. a – 2b

  7. Binomials • In the last lesson, you learned how to multiply binomials using the FOIL method. Let’s see if we can find a way that the process can be reversed to factor a trinomial into its binomial factors. (x + r)(x + s) = x  x + x  s + r  x + r  s = x2 + xs + rx + rs = x2 + (s + r)x + rs Look at the coefficient of the second term r + s. It is the sum of the two constant terms in the original binomials. The constant term in the product is the product of the constants from the original binomial. Now that we know where this coefficient and constant come from, factoring a trinomial like this will be easier.

  8. Ex. 5: Factor x2 – 2x - 35 What are the factors of -35 that subtract to give you -2 • The two numbers are 5 and -7. So, x2 – 2x – 35 = (x +5)(x – 7) Check using FOIL

  9. Ex. 6: Factor x2 – 49 What are the factors of -49 whose middle terms drop out • The two numbers are 7 and -7. So, x2 – 49 = (x + 7)(x – 7) Check using FOIL

  10. Notice anything? • The two terms of the original polynomial are perfect squares. This is a special case called the “difference of two squares.” The difference of two squares is simple to factor. • Rule: For real numbers a and b, a2 – b2 = (a + b)(a – b)

  11. Ex. 7: What if there’s a coefficient in front of the first squared term? Yes, that is harder. Factor 2x2 – 11x – 21 The product of the coefficient and the constant term is 2  -21 or -42. So, the two coefficients of x must have a sum of -11 and a product of -42. They must be what? -11 1 x -42 = -42 2 x -21 = -19 3 x -14 = -11 6 x -7 = -1 Simplify, but keep this form, then write out in binomial form. -14 3 (2x + 3)(x - 7) Check using FOIL -42

  12. Example 8: What about perfect square trinomials? Factor 16r2 – 24r + 9 If you have a perfect square in the front and in the back, and there is a term in the middle, check for perfect square trinomials. Factors very easily into square root 1st term and last term. If the first sign is negative, then both binomial terms are negative. If the first sign is positive, then both binomial terms will be positive. Factors to (4r – 3)(4r – 3) or (4r – 3)2

  13. x3 + 64 = okay now you have cubes. It would be a good idea to memorize the easy ones. 1  1  1 = 1 2  2  2 = 8 3  3  3 = 27 4  4  4 = 64 5  5  5 = 125 6  6  6 = 216 7  7  7 = 343 8  8  8 = 512 9  9  9 = 729 10  10  10 =1000 Get the picture? Ex. 9: Factor x3 + 64

  14. Ex. 9: Factor x3 + 64 x3 + 64 = x3 + (4)3 = (x + 4)(x2 – 4  x + 42) = (x + 4)(x2 – 4x + 16) You guessed it, there is a pattern: a3 + b3 = (a + b)(a2 – 2ab + b2)

  15. Steps to factor a polynomial • Find the GCF of the terms and factor it out. • Check for special products. • If there are two terms, look for a difference of squares, difference of two cubes, or sum of two cubes. • If there are three terms, look for a perfect square trinomial. • Try other factoring methods • If there are three terms, try the trinomial pattern. • If there aare four or more terms, try grouping.

  16. Ex. 10: Factor 3x3 +6x2 – 3x – 6. 3x3 + 6x2 – 3x – 6 = 3(x3 + 2x2 – x – 2) = 3[x2(x + 2) – 1(x + 2)] = 3(x + 2)(x2 – 1) = 3(x + 2)(x + 1)(x – 1)

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