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hawkfalcon

  • 2 years ago

factoring question

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  1. hawkfalcon
    • 2 years ago
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    \[5\cos^2x-5\sin^2x+cosx+sinx\]

  2. ILoveMath3006
    • 2 years ago
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    I think you should use the quadratic formula

  3. hawkfalcon
    • 2 years ago
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    I have to factor it :P

  4. ILoveMath3006
    • 2 years ago
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    oh sorry

  5. hawkfalcon
    • 2 years ago
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    Its okay:3 Anyone?

  6. hawkfalcon
    • 2 years ago
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    :(

  7. hawkfalcon
    • 2 years ago
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    *crosses fingers*

  8. jarmvel
    • 2 years ago
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    \[5 \cdot \cos^{2}x-5 \cdot \sin^{2}x +sinx+cosx=5\cdot(cosx+sinx)(cosx-sinx)+(sinx+cosx)\] and now \[5 \cdot \cos^{2}x-5 \cdot \sin^{2}x +sinx+cosx=5\cdot(sinx+cosx)(cosx-sinx +1)\]

  9. hawkfalcon
    • 2 years ago
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    o.O Okay

  10. satellite73
    • 2 years ago
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    if this is confusing, put \(a=\cos(x)\) and \(b=\sin(x)\) and factor \[5a^2-5b^2+a+b\]

  11. hawkfalcon
    • 2 years ago
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    That makes sense

  12. satellite73
    • 2 years ago
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    it has nothing whatsoever to do with trig, but the all those trig functions will tend to confuse

  13. satellite73
    • 2 years ago
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    \[5a^2-5b^2+a+b=5(a+b)(a-b)+a+b=(5(a-b)+1)(a+b)\]

  14. hawkfalcon
    • 2 years ago
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    :D figured it out! I can do "5a2−5b2+a+b" :P

  15. hawkfalcon
    • 2 years ago
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    Thanks :)

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