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Math Help - Algebra 2 question factoring polynomial

  1. #1
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    Algebra 2 question factoring polynomial

    hi can anyone help me solve these 2 problems
    1)Direction: solve each polynomial equation by factoring
    4x cubed + 7x squared - 5x = 6

    2)Direction:Identify the roots of each equation.state the mutiplicity of each root
    x to the fourth power + 2x cubed -11x squared - 12x +36=0

    and can you please explain to me how you would factor it if its factorable because im having a hard time factoring when x isnt to like the second power.hope this made sense..thanks
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  2. #2
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    Quote Originally Posted by miemie11 View Post
    ynomial equation by factoring
    4x cubed + 7x squared - 5x = 6
    4x^3 + 7x^2 - 5x = 6

    4x^3 + 7x^2 - 5x - 6 = 0

    There is no "nice" way to factor this thing (that I can see), so I'd use the rational root theorem to get a factor to start with then factor the rest in the "usual" way.

    Possible rational roots are \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, ....

    I'll cut to the chase. x - 1 is a factor. So do your division (synthetic or long) and you get that
    4x^3 + 7x^2 - 5x - 6 = (x - 1)(4x^2 + 11x + 6) = 0

    You can do the rest.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by miemie11 View Post
    2)Direction:Identify the roots of each equation.state the mutiplicity of each root
    x to the fourth power + 2x cubed -11x squared - 12x +36=0
    x^4 + 2x^3 - 11x^2 - 12x + 36 = 0

    Again this is a job for the rational root theorem.

    Possible rational roots are \pm (1, 2, 3, 4, 6, 9, 12, 18, 36), so 18 possibles in all.

    Starting off, I get that x - 2 is a factor. So
    x^4 + 2x^3 - 11x^2 - 12x + 36 = (x - 2)(x^3 + 4x^2 - 3x - 18) = 0

    Now let's apply the rational root theorem to the second factor. Possible rational roots of this are: \pm (1, 2, 3, 6, 9, 18) (Only 12 possibles this time!)

    Again I get that x - 2 is a factor. So
    x^4 + 2x^3 - 11x^2 - 12x + 36 = (x - 2)^2(x^2 + 6x + 9) = 0
    which I'm sure you can finish up on your own.

    -Dan
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    Quote Originally Posted by miemie11 View Post
    hi can anyone help me solve these 2 problems
    1)Direction: solve each polynomial equation by factoring
    4x cubed + 7x squared - 5x = 6

    2)Direction:Identify the roots of each equation.state the mutiplicity of each root
    x to the fourth power + 2x cubed -11x squared - 12x +36=0

    and can you please explain to me how you would factor it if its factorable because im having a hard time factoring when x isnt to like the second power.hope this made sense..thanks
    1) By trial and error you should be able to establish that x = 1 and x = -2 are roots of the equation 4x^3 + 7x^2 - 5x - 6 = 0. Therefore (x - 1) and (x + 2) are factors of 4x^3 + 7x^2 - 5x - 6.

    Divide these factors in to get the remaining linear factor and hence solve for the remaining root.

    Alternatively (x - 1)(x + 2) = x^2 + x - 2. Therefore 4x^3 + 7x^2 - 5x - 6 = (x^2 + x - 2)(ax + b) and it should be obvious what the value of a and b must be .......

    --------------------------------------------------------------------------

    2) Show that x = 2 and x = -3 are roots. In fact, they are repeated roots ......
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