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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    i would factor it, then make a table hen graph iy

  2. UsukiDoll
    • one year ago
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    almost.... you need to factor more

  3. UsukiDoll
    • one year ago
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    ok... you have a perfect square in the form of \[(a^2-b^2) = (a+b)(a-b) \]

  4. UsukiDoll
    • one year ago
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    \[(x^2-4) \] is a perfect square.. what is the square root of x^2 and what is the square root of 4

  5. UsukiDoll
    • one year ago
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    no.

  6. Zale101
    • one year ago
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    |dw:1434948176809:dw|

  7. UsukiDoll
    • one year ago
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    a perfect square is the product of a rational number multiplied by itself. the product of a polynomial multiplied by itself.

  8. UsukiDoll
    • one year ago
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    A perfect square is a number that can be expressed as the product of two equal integers.

  9. UsukiDoll
    • one year ago
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    what are you doing?

  10. Zale101
    • one year ago
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    a is squared and so does x b is squared but 4 is the square of what number?

  11. UsukiDoll
    • one year ago
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    \[f(x) = x^3 + 4x^2 - x - 4\] factor by grouping which you did and that's fine you got to \[(x^2-4)(x+1) \] but you need to factor more..

  12. Zale101
    • one year ago
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    \(\Large (a^2-b^2)=(a+b)(a-b)\) \(\Large (x^2-4)=(x^2-2^2)\) clear?

  13. UsukiDoll
    • one year ago
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    something's wrong here.

  14. UsukiDoll
    • one year ago
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    factor my grouping (at least when I did it) had something else as a perfect square

  15. UsukiDoll
    • one year ago
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    \[f(x) = x^3 + 4x^2 - x - 4 \rightarrow x^2(x+4)-1(x+4)\]

  16. UsukiDoll
    • one year ago
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    \[(x+4)(x^2-1) \]

  17. UsukiDoll
    • one year ago
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    you shouldn't rearrange the terms.. -_- that's why you had a different answer.

  18. UsukiDoll
    • one year ago
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    second line is where everything got messed up x^3+x-4x^2-4 x(x^2+1)-4(x^2+1) you have an imaginary number case

  19. UsukiDoll
    • one year ago
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    Leave the function alone! *FIRST STEP* Second step use factor by grouping (two pairs!) first and second term has a common factor.. third and fourth term has a common factor

  20. UsukiDoll
    • one year ago
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    \[f(x) = x^3 + 4x^2 - x - 4 \] look at the first two terms.. what do they have in common?

  21. UsukiDoll
    • one year ago
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    ok .. we have four terms x^3 <- 1st term 4x^2 - 2nd term -x <- 3rd term -4 < - 4th term focus on the first and second only what do they have in common?

  22. UsukiDoll
    • one year ago
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    yes... how many x's though?

  23. UsukiDoll
    • one year ago
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    \[x^3-4x^2 => (x)(x)(x)-4(x)(x) \]

  24. UsukiDoll
    • one year ago
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    how many x's can I take out?

  25. UsukiDoll
    • one year ago
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    WHAT! I have three x's and I have 2 x's

  26. UsukiDoll
    • one year ago
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    -x-4 now what can I take out? =_=

  27. UsukiDoll
    • one year ago
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    what can I yank out to make sure I have (x+4) ?

  28. UsukiDoll
    • one year ago
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    I have -x-4 I want x+4 what do I need to yank out?

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