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ExtremeAnon

  • 2 years ago

NO ONE EVER HELPS ME ON HERE!!!!

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  1. ExtremeAnon
    • 2 years ago
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    Somebody please help me!! ALGEBRA 2 HELP PLEASE?!? Part 1: Create your own factorable trinomial. Part 2: Explain, in complete sentences, how the trinomial is factored. Part 3: Explain, in complete sentences, the process used to check the factors for accuracy.

  2. JahEmpress
    • 2 years ago
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    don't feel bad me too man

  3. Ahaanomegas
    • 2 years ago
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    Dude, this is not the place to complain. Be patient.

  4. Ahaanomegas
    • 2 years ago
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    Hint to Part 1: Multiply one-degree polynomials like (x-1)(x-2)(x+3).

  5. ExtremeAnon
    • 2 years ago
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    I don't need hints, i need help. I don't know how to do this, Lol.

  6. lgbasallote
    • 2 years ago
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    i think what you mean...is that you need *answers*..not help

  7. swissgirl
    • 2 years ago
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    hahah u can google this for the same price kiddo

  8. Ahaanomegas
    • 2 years ago
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    www.wolframalpha.com will do what you want. This place is not to solve your homework for you.

  9. swissgirl
    • 2 years ago
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    x^2+2x+1 can that be broken into two binomials? 2 years ago Report Abuse laney laney Best Answer - Chosen by Asker Yes it is. Because the leading coefficient is 1 (coefficient of x^2), you can simply factor this by finding two numbers that add up to 2 (the coefficient of x) and multiply to 1 (the constant). These two numbers are one (1+1=2 and 1X1=1). So if you were to factor this trinomial, you would get (x+1)(x+1) OR (x+1)^2. To check your answers, you would just multiple the two binomials and see if you end up with your original equation. If you do, then your answer is right. So to check: (x+1)^2 =(x+1)X(x+1) =x^2 + x + x + 1 =x^2 + 2x + 1 So yes, x^2+2x+1 is factorable and the two binomials are (x+1) and (x+1). Another way of writing this would be (x+1)^2 2 years ago

  10. swissgirl
    • 2 years ago
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    Ok this is a factorable trinomial x^2+2x+1

  11. swissgirl
    • 2 years ago
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    umms i got my answer from here: http://answers.yahoo.com/question/index?qid=20101008142201AAlC9JY

  12. ExtremeAnon
    • 2 years ago
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    Thanks:)

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