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stupidinmath

  • 3 years ago

how to factor .. (x-3)^2=36

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  1. stupidinmath
    • 3 years ago
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    (x-3)(x-3) x^2-3x-3x+9 x^2-6x+9=36 x^2-6x-27 right? then next?

  2. Yahoo!
    • 3 years ago
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    Sum = -6 Product = -27 So the No are -9 and 3

  3. Yahoo!
    • 3 years ago
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    so x^2-6x-27 = x^2 + 3x - 9x - 27 =0 Nw Factorize

  4. stupidinmath
    • 3 years ago
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    ?

  5. stupidinmath
    • 3 years ago
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    DONT GET IT :|

  6. heedcom
    • 3 years ago
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    by further factorising it comes to (x+3)(x-9)=0

  7. phi
    • 3 years ago
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    to find the factors of x^2-6x-27 first notice that the 27 is negative. that means you are looking for 2 numbers with different signs (either + and - or - and +) List the factors of 27 1,27 3,9 do any of the pairs give -6 , when the numbers have different signs? 1 and 27 could be -1 + 27 or +1-27 but neithert give -6 but +3-9 = -6 so we want +3 and -9 (x+3)(x-9) multiply out to check: x^2 -6x-27. it works

  8. anonymous
    • 3 years ago
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    you might try this \[(x-3)^2=36\] \[(x-3)^2-36=0\] then since \(a^2-b^2=(a+b)(a-b)\) you get \[(x-3+6)(x-3-6)\] and then add the numbers

  9. phi
    • 3 years ago
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    and not to confuse you, you could have started with (x-3)^2=36 take the square root of both sides x-3= ±6 and x= +6+3= 9 and x= -6+3= -3 this means the factors are (x-9)(x+3)

  10. anonymous
    • 3 years ago
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    if you are looking for the zeros, then there is no reason to factor, you find them in two steps, as @phi wrote above

  11. stupidinmath
    • 3 years ago
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    thanks guys!

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