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calculusxy

  • one year ago

factoring quadratic expression...

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  1. calculusxy
    • one year ago
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    \[\huge 294a^2 - 54 \]

  2. calculusxy
    • one year ago
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    @Nnesha

  3. calculusxy
    • one year ago
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    @jim_thompson5910 @hartnn @Jhannybean

  4. calculusxy
    • one year ago
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    \[\large 294a^2 - 0a - 54\]

  5. anonymous
    • one year ago
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    larget common factor of \(294\) and \(24\) is?

  6. anonymous
    • one year ago
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    get rid of that stupid zero

  7. calculusxy
    • one year ago
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    okay sorry

  8. calculusxy
    • one year ago
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    don't you mean the gcf of 294 and 54 ?

  9. anonymous
    • one year ago
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    yes i get 6

  10. Jhannybean
    • one year ago
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    Hahahaha

  11. anonymous
    • one year ago
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    so start with \[6(49a^2-9)\] then factor \[49a^2-9\] as the difference of two squares

  12. Jhannybean
    • one year ago
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    \[a^2-b^2 = (a-b)(a+b)\]

  13. calculusxy
    • one year ago
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    (7a + 3)(7a - 3)

  14. Jhannybean
    • one year ago
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    or rather,.... \(ca^2-b^2 = (ca-b)(ca+b)\) :') \(c\) is the coefficient of one of the terms lol

  15. anonymous
    • one year ago
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    \[\heartsuit^2-\spadesuit^2=(\heartsuit+\spadesuit)(\heartsuit -\spadesuit)\][

  16. Jhannybean
    • one year ago
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    Oh you think you're fancy huh

  17. anonymous
    • one year ago
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    just showing off

  18. Jhannybean
    • one year ago
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    lol

  19. anonymous
    • one year ago
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    \[\xi^2 -\omega^2=(\xi+\omega)(\xi-\omega)\]

  20. Jhannybean
    • one year ago
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    You are correct @calculusxy , just don't forget to multiply by the 6

  21. calculusxy
    • one year ago
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    can i make sure the answer to another question?

  22. calculusxy
    • one year ago
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    \[\large 12x^2 + 12x + 3\] \[\huge Answer \] \[\large 3(2x+1)(2x+1)\]

  23. anonymous
    • one year ago
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    looks good to me you @Jhannybean ?

  24. calculusxy
    • one year ago
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    well i did the same question, but in a different way and got a different answer ( can u catch my mistake) 12x^2 + 12x + 3 12x^2 + 6x + 6x + 3 2x(2x+3) 3(2x + 3) (2x - 3)(2x + 3)

  25. Jhannybean
    • one year ago
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    \[\begin{align}12x^2+12x+3 &\rightarrow 4x^2+4x+1 \\&\rightarrow x^2+4x+4 \\&\rightarrow (x+2)(x+2) \\&\rightarrow \left(x+\frac{2}{4}\right)\left(x+\frac{2}{4}\right) \\&\rightarrow \left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right) \\&\rightarrow \color{red}{(2x+1)(2x+1)}\end{align} \]

  26. Jhannybean
    • one year ago
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    For this method i simplify the quadratic by putting it in standard form \(ax^2+bx+c\) typically where \(a=1\) find the factors divide the factors by the leading coefficient we used to simplify our equation If the simplification of the factors does not give you an integer value but a fractional one, multiply the variable with the denominator of the fraction. You'll have your factorization

  27. Nnesha
    • one year ago
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    if there is a GCF better to take it out |dw:1444172023392:dw|

  28. Nnesha
    • one year ago
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    |dw:1444172039740:dw|

  29. Nnesha
    • one year ago
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    \[12x^2+6x ---> 6x(2x+1)\] these are the mistakes :=)

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