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Home-Learning Review

Topic: Factoring x 2 + bx + c Essential Question: How can you write a trinomial as the product of two binomials?. Home-Learning Review. Quiz #10. Topic : Factoring x 2 + bx + c. FLASHBACK. Factors. Factors (either numbers or polynomials)

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Home-Learning Review

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  1. Topic: Factoring x2 + bx + cEssential Question: How can you write a trinomial as the product of two binomials?

  2. Home-Learning Review

  3. Quiz #10

  4. Topic: Factoringx2 + bx + c

  5. FLASHBACK

  6. Factors Factors (either numbers or polynomials) When an integer is written as a product of integers, each of the integers in the product is a factor of the original number. When a polynomial is written as a product of polynomials, each of the polynomials in the product is a factor of the original polynomial. Factoring – writing a polynomial as a product of polynomials.

  7. Greatest Common Factor Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved.

  8. Greatest Common Factor Example Find the GCF of each list of numbers. • A) 12 and 8 • B) 6, 8, and 46 12 = 8 = So the GCF is 6 = 8 = 46 = So the GCF is

  9. Greatest Common Factor Example Find the GCF of each list of terms. C) x3 and x7 D) 6x5 and 4x3 x3 = x ·x·x x7 = x ·x·x·x ·x·x·x So the GCF is x · x· x = x3 • 6x5 = 2 · 3 · x · x· x·x·x 4x3 = 2 · 2 ·x ·x·x So the GCF is 2·x ·x·x = 2x3

  10. Greatest Common Factor Example Find the GCF of the following list of terms. • E) a3b2, a2b5 and a4b7 a3b2 = a ·a·a· b· b a2b5 = a · a· b· b · b· b· b a4b7 = a · a· a· a· b· b · b· b· b· b· b • So the GCF is a · a· b· b = a2b2 Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.

  11. Lets make the connection! Ready Set Go

  12. Factoring Polynomials The first step in factoring a polynomial is to find the GCF of all its terms. Then we write the polynomial as a product by factoring out the GCF from all the terms. The remaining factors in each term will form a polynomial.

  13. Factoring out the GCF Example Factor out the GCF in each of the following polynomials. F) 6x3 – 9x2 + 12x =

  14. Factoring out the GCF Practice Factor out the GCF in each of the following polynomials. 3) 14x3y + 7x2y – 7xy =

  15. Factoring Trinomials of the Form x2 + bx + c

  16. Factoring Trinomials Recall by using the FOIL method that F O I L (x + 2)(x + 4) = x2 + 4x + 2x + 8 = x2 + 6x + 8 To factor x2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers. So we’ll be looking for 2 numbers whose product is c and whose sum is b. Note: there are fewer choices for the product, so that’s why we start there first.

  17. 3 x 10 3 + 10 = 13 √ Factoring Polynomials Example G) Factor the polynomial x2 + 13x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of 30 Sum of Factors = 13 1 x 30 1 + 30 = 31 x 2 x 15 2 + 15 = 17 x Note, there are other factors, but once we find a pair that works, we do not have to continue searching. • So x2 + 13x + 30 = (x + 3)(x + 10).

  18. x x -2 -4 1x + 8x = 9x 2x + 4x = 6x -1x - 8x = -9x -2x - 4x = -6x Factors of +8: 1 & 8 2 & 4 -1 & -8 -2 & -4

  19. F O I L Check your answer by using FOIL

  20. – 3 x – 8 – 3 + – 8 = – 11 √ Factoring Polynomials Practice: 4) Factor the polynomial x2 – 11x + 24. Since our two numbers must have a product of 24 and a sum of -11, the two numbers must both be negative. Negative factors of 24 Sum of Factors = –11 – 1 x – 24 – 1 + – 24= – 25 x – 2 x – 12 – 2 + – 12 = – 14 x So x2 – 11x + 24 = (x – 3)(x – 8).

  21. Factoring Polynomials Practice: 5) Factor the polynomial x2 – 2x – 35

  22. Prime Polynomials Example Since our two numbers must have a product of 10 and a sum of – 6, the two numbers will have to both be negative. Negative factors of 10 Sum of Factors – 1, – 10 – 11 – 2, – 5 – 7 Factor the polynomial x2 – 6x + 10. Since there is not a factor pair whose sum is – 6, x2 – 6x +10 is not factorable and we call it a prime polynomial.

  23. Check Your Result! You should always check your factoring results by multiplying the factored polynomial to verify that it is equal to the original polynomial. Many times you can detect computational errors or errors in the signs of your numbers by checking your results.

  24. Pair Practice: Page 483 (16, 22) Page 503 - 505 (10, 12, 14, 16, 20, 22, 24, 32, 42, 61, 62)

  25. Additional Practice Warm-Up: • 16p2 – 36p • m2 + 8m • x2 + 2x – 8

  26. Factoring Trinomials of the Form ax2 + bx + c Working with a value of ‘a’ that is higher than 1

  27. Factoring Trinomials Returning to the FOIL method, F O I L (3x + 2)(x + 4) = 3x2 + 12x + 2x + 8 = 3x2 + 14x + 8 To factor ax2 + bx + c into (#1·x + #2)(#3·x + #4), note that a is the product of the two first coefficients, c is the product of the two last coefficients and b is the sum of the products of the outside coefficients and inside coefficients. Note that b is the sum of 2 products, not just 2 numbers, as in the last section.

  28. Factoring Polynomials Example A) Factor the polynomial 25x2 + 20x + 4

  29. F O L I 5x(5x) + 5x(2) + 2(2) + 2(5x) Factoring Polynomials Example Continued Check the resulting factorization using the FOIL method. (5x + 2)(5x + 2) = = 25x2 + 10x + 10x + 4 = 25x2 + 20x + 4 So our final answer when asked to factor 25x2 + 20x + 4 will be (5x + 2)(5x + 2) or (5x + 2)2.

  30. Factoring Polynomials Example B) Factor the polynomial 21x2 – 41x + 10.

  31. F O L I 3x(7x) + 3x(-2) - 5(-2) - 5(7x) Factoring Polynomials Example Continued Check the resulting factorization using the FOIL method. (3x – 5)(7x – 2)= = 21x2 – 6x – 35x + 10 = 21x2 – 41x + 10 So our final answer when asked to factor 21x2 – 41x + 10 will be (3x – 5)(7x – 2).

  32. Factoring Trinomials (Method 2*) This time, the x2 term DOES have a coefficient (other than 1)! Factor. 3x2 + 14x + 8 Step 1: Multiply 3 • 8 = 24 (the leading coefficient & constant). 1 • 24 = 24 2 • 12 = 24 3 • 8 = 24 4 • 6 = 24 Step 2: List all pairs of numbers that multiply to equal that product, 24. Step 3: Which pair adds up to 14?

  33. 3 3 3 2 ( x + )( x + ) 12 2 ( x + )( x + ) 12 Factoring Trinomials (Method 2*) Factor. 3x2 + 14x + 8 Step 4: Write temporary factors with the two numbers. 2 ( x + )( x + ) 12 Step 5: Put the original leading coefficient (3) under both numbers. 4 Step 6: Reduce the fractions, if possible. 3 ( 3x + 2 )( x + 4 ) Step 7: Move denominators in front of x.

  34. Factoring Trinomials (Method 2*) Factor. 3x2 + 14x + 8 You should always check the factors by distributing, especially since this process has more than a couple of steps. ( 3x + 2 )( x + 4 ) = 3x • x + 3x • 4 + 2 • x + 2 • 4 = 3x2 + 14 x + 8 √ 3x2 + 14x + 8 = (3x + 2)(x + 4)

  35. Pair-Practice: Practice: 1) 3x2 + 13x + 4

  36. Pair-Practice: Practice: • 2x2+ x – 21 • 25x2+ 20x + 4 • 4) 3r2 – 18r + 27

  37. Factoring Polynomials Example C) Factor the polynomial 3x2 – 7x + 6. The only possible factors for 3 are 1 and 3, so we know that, if factorable, the polynomial will have to look like (3x )(x ) in factored form, so that the product of the first two terms in the binomials will be 3x2. Since the middle term is negative, possible factors of 6 must both be negative: {1,  6} or { 2,  3}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Continued.

  38. Factoring Polynomials Example Continued We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 7x. Now we have a problem, because we have exhausted all possible choices for the factors, but have not found a pair where the sum of the products of the outside terms and the inside terms is –7. So 3x2 – 7x + 6 is a prime polynomial and will not factor.

  39. Factoring Polynomials Example D) Factor the polynomial 6x2y2 – 2xy2 – 60y2. Remember that the larger the coefficient, the greater the probability of having multiple pairs of factors to check. So it is important that you attempt to factor out any common factors first. 6x2y2 – 2xy2 – 60y2 = 2y2(3x2 – x – 30) The only possible factors for 3 are 1 and 3, so we know that, if we can factor the polynomial further, it will have to look like 2y2(3x )(x ) in factored form. Continued.

  40. Factoring Polynomials Example Continued Since the product of the last two terms of the binomials will have to be –30, we know that they must be different signs. Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15}, {2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}. We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to –x. Answer: 2y2(3x - 10)(x + 3)

  41. F O I L 3x(x) + 3x(3) – 10(x) – 10(3) Factoring Polynomials Example Continued Check the resulting factorization using the FOIL method. (3x – 10)(x + 3) = = 3x2 + 9x – 10x – 30 = 3x2 – x – 30 So our final answer when asked to factor the polynomial 6x2y2 – 2xy2 – 60y2 will be 2y2(3x – 10)(x + 3).

  42. Home-Learning Assignment #3: Page 508-510 (1-5, 8, 14, 22, 28, 51)

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