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anonymous

  • one year ago

will fan and medal~ polynomials and identities ~ Basically I have an assignment where I need to make my own polynomial identity

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  1. anonymous
    • one year ago
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    Create your own using the columns below. See what happens when different binomials or trinomials are combined. Square one factor from column A and add it to one factor from column B to develop your own identity. Column A: (x − y) (x + y) (y + x) (y - x) Column B: (x2 + 2xy + y2) (x2 − 2xy + y2) (ax + b) (cy + d)

  2. anonymous
    • one year ago
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    So do what they say. Square one factor from column A and add it to one factor from column B to develop your own identity. Column A: \( (x + y)^2 \) Column B: \( (x^2 + 2xy + y^2) \) add them \( (x + y)^2 + (x^2 + 2xy + y^2) = ?? \)

  3. anonymous
    • one year ago
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    @Nixy so I'm literally just adding them together? It's just \[(x+y)^{2} + (x^{2} + 2xy + y^{2})\] ?

  4. anonymous
    • one year ago
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    Yes. An identity is an equation that is always true. Once you solve by adding them together and put it on the other side of the = you will have an equation that is always true (identity)

  5. anonymous
    • one year ago
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    For example. \( \huge \frac{a}{2} = a × 0.5 \) is an identitiy and is always true

  6. anonymous
    • one year ago
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    so for the example you gave me, am I supposed to use the distributive property then?

  7. anonymous
    • one year ago
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    i mean, for the first example

  8. anonymous
    • one year ago
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    You need to expand this first (x+y)^2 and then add all like terms

  9. anonymous
    • one year ago
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    Expand \( \huge(x+y)^2 \) and then add all like terms

  10. anonymous
    • one year ago
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    so, expanding (x+y)^2 \[(x+y) \times (x+y)\] right?

  11. anonymous
    • one year ago
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    \( \huge (x+y)(x+y) = ???\) is correct

  12. anonymous
    • one year ago
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    Now times them using foil

  13. anonymous
    • one year ago
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    x^2 + xy^2 + y^2 ?

  14. anonymous
    • one year ago
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    So we have \( \huge x^2 + 2xy + y^2 \)

  15. anonymous
    • one year ago
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    Now we have \( \huge x^2 + 2xy + y^2 + x^2 + 2xy + y^2 \) combine all like terms now.

  16. anonymous
    • one year ago
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    Add all terms that can be added together.

  17. anonymous
    • one year ago
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    x^4 + 4xy + y^4 ?

  18. anonymous
    • one year ago
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    We should have \( \huge 2x^2+4xy+2y^2 \) You don't add the exponents x^2 + x^2 = 2x^2

  19. anonymous
    • one year ago
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    So we have an identity now and no matter what value we use for x or y is always true.. Below is our identity So we have an identity below now. \( \large (x+y)^{2} + (x^{2} + 2xy + y^{2}) = 2x^2+4xy+2y^2 \)

  20. anonymous
    • one year ago
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    Thank you so much for helping me, I'm actually understanding it a lot better now.

  21. anonymous
    • one year ago
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    You can combine them in all types of ways to make an identity

  22. anonymous
    • one year ago
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    You can subtract, divide, times, add and square them. They can get pretty complex or they can be simple. Any questions?

  23. anonymous
    • one year ago
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    I think I've got it, thank you c:

  24. anonymous
    • one year ago
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    YW, time for me to get to bed :-) Almost 12 AM here :-)

  25. anonymous
    • one year ago
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    Haha, same here. This is part of my last assignment, and I just wanted to get it done right! Thanks again for all your help. You're a lifesaver.

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