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Thread: Find the real-number solutions of the equation

  1. #1

    Find the real-number solutions of the equation

    I"m confused as to how to begin to solve these (polynomials). Please help.

    x^3 +7x^2 +4x +28 = 0

    I don't know how to type these problems out on the computer, so I hope you understand. Here's where I started and then got lost.

    x(x^2 + 7x + 4....... that s as far as I got. Please help it makes no sense to me!

  2. #2
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    Re: Find the real-number solutions of the equation

    Quote Originally Posted by dtbrunson
    I"m confused as to how to begin to solve these (polynomials). Please help.

    x^3 +7x^2 +4x +28 = 0

    I don't know how to type these problems out on the computer, so I hope you understand. Here's where I started and then got lost.

    x(x^2 + 7x + 4....... that s as far as I got. Please help it makes no sense to me!
    Hi dtbrunson,

    You didn't notice you could factor this expression by grouping, did you?
    Well, you can.

    [tex]x^3+7x^2+4x+28=0[/tex]

    Group the first two terms and factor out [tex]x^2}[/tex]
    Then, group the last two terms and factor out a 4.

    [tex]x^2(x+7)+4(x+7)=0[/tex]

    Now you have: [tex](x^2+4)(x+7)=0[/tex]

    Looks like 1 real and 2 imaginary solutions.
    "There are 10 types of people in the world today - those who understand
    binary and those who don't"

  3. #3

    Re: Find the real-number solutions of the equation

    Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
    (x^2 + 4) (x + 7)
    x(x+4) (x+7)?.......

  4. #4

    Re: Find the real-number solutions of the equation

    Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
    (x^2 + 4) (x + 7)
    x(x+4) (x+7)?.......

  5. #5
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    Re: Find the real-number solutions of the equation

    Quote Originally Posted by dtbrunson
    Thanks but what do you mean when you say 1 real and 2 imaginary? Does it go further than that?
    (x^2 + 4) (x + 7)
    x(x+4) (x+7)?.......
    That's factored incorrectly. Start here:

    [tex](x^2+4)(x+7)=0[/tex]

    Use the 'zero product property' to set each factor to zero and solve.

    [tex]x^2+4=0[/tex]

    [tex]x^2=-4[/tex]

    [tex]x=\pm 2i[/tex]

    These are your two imaginary zeros.

    [tex]x+7=0[/tex]

    [tex]x=-7[/tex]

    This is your one real zero.
    "There are 10 types of people in the world today - those who understand
    binary and those who don't"

  6. #6
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    Re: Find the real-number solutions of the equation

    Quote Originally Posted by dtbrunson
    (x^2 + 4)
    x(x+4)
    NO. x(x + 4) = x^2 + 4x, NOT x^2 + 4.

    And if you don't understand what masters just told you, then you need help ftom your teacher...
    I'm just an imagination of your figment !

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