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jchick

  • one year ago

solve and show all work for a2 – a – 20 and a2 – 5a – 20

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  1. myininaya
    • one year ago
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    what are you solving?

  2. jchick
    • one year ago
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    Factor each trinomial below. Please show your work and check your answer.

  3. myininaya
    • one year ago
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    ok for the first one we have a^2-a-20 a^2-1a-20 so first step is to ask yourself what two integers multiply to by -20 and add up to be -1?

  4. jchick
    • one year ago
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    What do you mean?

  5. myininaya
    • one year ago
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    \[ax^2+bx+c \\ \text{ when } a=1 \text{ and if the expression is "factorable" } \\ \text{ then you should be able to find } \\ \text{ two integers so that when you multiply them together you get } c \\ \text{ and when you add them together you get } b\] for example: \[x^2+5x+6 \\ \text{ the question here first to answer is: } \\ \text{ what two integers multiply to be 6 } \\ \text{ and add up to be 5}\] well 2(3)=6 and 2+3=5 so the factored form of \[x^2+5x+6 \text{ is } (x+2)(x+3)\]

  6. jchick
    • one year ago
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    a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5) a2 -5a+4a-2 -a a2 – 5a – 20 (a )(a ) 10*-2 -5*4 (a-5)(a+4) a2 + 12ab + 27b2 (a )(a ) 9*3 (a+3b)(a+9b) a2 +9ab+3ab+27b a2 +12ab+27b

  7. jchick
    • one year ago
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    Right?

  8. myininaya
    • one year ago
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    yes to the first one: a^2-a-20=(a-5)(a+4) since -5*4=-20 and -5+4=-1 no to the second one: a^2-5a-20 this does not equal (a-5)(a+4) -5(4)=-20 but -5+4 isn't -5

  9. myininaya
    • one year ago
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    and the third one looks good

  10. jchick
    • one year ago
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    a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5)

  11. myininaya
    • one year ago
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    so you got 2/3 correct

  12. myininaya
    • one year ago
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    you just need to fix the second one

  13. jchick
    • one year ago
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    Ok

  14. jchick
    • one year ago
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    I will try

  15. myininaya
    • one year ago
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    a^2-5a-20 are there two integers that multiply to be -20 and add up to be -5?

  16. myininaya
    • one year ago
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    possible pairings that multiply to be -20: 1(-20) -1(20) 2(-10) -2(10) 4(-5) -4(5)

  17. myininaya
    • one year ago
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    do any of those pairs also add up to -5?

  18. myininaya
    • one year ago
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    which equation is true if any: 1+(-20)=-5 -1+20=-5 2+(-10)=-5 -2+10=-5 4+(-5)=-5 -4+5=-5

  19. jchick
    • one year ago
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    None

  20. myininaya
    • one year ago
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    you are right

  21. jchick
    • one year ago
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    They are all false

  22. myininaya
    • one year ago
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    so that means that this is a prime trinomial

  23. myininaya
    • one year ago
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    means it is not "factorable"

  24. myininaya
    • one year ago
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    I put factorable in quotation marks because it only really means it isn't factorable over the integers

  25. jchick
    • one year ago
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    Ok

  26. jchick
    • one year ago
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    So I am not sure on the second one

  27. myininaya
    • one year ago
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    it isn't factorable

  28. myininaya
    • one year ago
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    it is prime

  29. jchick
    • one year ago
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    Ok thanks

  30. myininaya
    • one year ago
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    a^2-5a-20 does not factor

  31. myininaya
    • one year ago
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    unless you can factor over the reals?

  32. myininaya
    • one year ago
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    which I don't think that is the instructions

  33. jchick
    • one year ago
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    No but how would I write this just say it is prime?

  34. myininaya
    • one year ago
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    you can either say not factorable or prime

  35. myininaya
    • one year ago
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    which ever vocabulary your teacher prefers really but either saying is fine in general

  36. jchick
    • one year ago
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    Part 1: Factor each trinomial below. Please show your work and check your answer. For the first one: x^2 - 8 x + 15 x^2 - 5x - 3x + 15 x(x-5) -3(x-5) (x-5)(x-3) The check: x(x-3)-5(x-3) x^2 – 3x-5x+15 x^2-8x+15 Question 2: a2 – 5a – 20 Prime and cannot be factored. Question: 3 a^2 - a - 20 a^2 -5a+4a-20 a(a-5)+4(a-5) (a-5)(a+4) Question 4: a^2 + 12ab + 27b^2 a^2 + 9ab + 3ab + 27b^2 a(a+9b) + 3b(a+9b) (a+9b)(a+3b) Question 5: 2a2 + 30a + 100 2(a^2 + 15a + 50) the values that add to give 15a multiply to 50a^2 are 10a and 5a 2(a^2 + 5a + 10a + 50) 2[a(a+5) +10(a+5)] 2(a+5)(a+10) Question 6: Part 2: (5 points) It’s your turn to be a game show host! As you know, in the game of Math Time, the contestants are given an answer and they must come up with the question that corresponds to the given answer. Your task for this portion of the assignment is to create two different “answers” (and the questions that accompany them) that the host could use for the final round of Math Time. The questions and answers you create must be unique. Check out the example and hint below, if needed. 1 point for creating the two questions and answers. 2 points per question/answer for accuracy. Example: Question for Host: x2 + 6x + 5 is the product of these two binomials. Expected Question from Contestant: What is (x + 5)(x + 1)? HINT: Create the two binomials first and then use the distribution method to find their simplified product. The simplified product is the answer and the two binomial factors, the expected question. Hosts question 1: The product of these two binomials is x2+x-20 Answer: What is (x+5)(x-4) Hosts question 2: The product of these two binomials is x2+7xy+12y Answer: What is (x+3y)(x+4y)

  37. jchick
    • one year ago
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    how does this look?

  38. jchick
    • one year ago
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    @myininaya

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