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Math Help - Helppp!!!! factor and solve

  1. #1
    mathallychallenged
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    Exclamation Helppp!!!! factor and solve

    I have been sitting in my room for... 5 hours now, 5 hours!! Trying to do this algebra 2 sheet on quadratic equations and I am completley lost!!!



    I can do the easy one's like
    (t^2-9)=0
    (t-3)(t+3)=0
    t-3=0 t+3=0
    t=3 t=-3

    {3,-3}

    the steps i was taught were
    get zero on one side of equation
    factor completely
    set each to zero
    solve

    but I am stuck on these problems I can get them all set to zero and then that's it, i'm stuck after that
    ^2 = squared ^3 = cubed ^4=to the 4th power

    1) t^3-t=0

    2)3u^2+7u-6=0

    3)x^2-6x-72=0

    4)(s-1)^2(s-3)^2

    5)0.2x^2-1.1x+0.5=0

    6) (x^2-3)^2+(x^2-3)-2=0

    7) (x^2-3x+2)^3=0

    8) f(x)=(x+1)^4 +4(x+1)^3


    If anyone could help me that would be amazing!!!!!!! I need help a.s.a.p!!!!
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  2. #2
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    Quote Originally Posted by mathallychallenged View Post
    I have been sitting in my room for... 5 hours now, 5 hours!!
    If you were watching TV Simultaneously it does not count. But if that is true something to be proud of, most time I ever spend was 3 hours, trying to understand a proof to a theorem.

    I do the harder ones.

    7) (x^2-3x+2)^3=0
    (x^2-3x+2)(x^2-3x+2)(x^2-3x+2)=0
    x^2-3x+2=0
    (x-1)(x-2)=0
    8) f(x)=(x+1)^4 +4(x+1)^3
    (x+1)^3((x+1)+4)=0
    (x+1)(x+1)(x+1)(x+5)=0
    (x+1)=0
    (x+5)=0
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  3. #3
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    Hello, mathallychallenged!

    The steps i was taught were:
    . . get zero on one side of equation
    . . factor completely
    . . set each to zero
    . . solve . . . . . . . . Correct!

    In some of these, the factoring may be tricky.


    1)\;\;t^3-t\:=\:0
    Factor: . t(t^2 - 1)\:=\:\quad\Rightarrow\quad t(t-1)(t+1)\:=\:0

    Then we have: . \begin{Bmatrix}t \,= \,0 & \Rightarrow &t \,=\,0\\ t - 1 \:=\:0& \Rightarrow & t\,=\,1 \\ t + 1 \:=\:0 & \Rightarrow & t \,=\,\text{-}1\end{Bmatrix}



    2)\;\;3u^2+7u-6\:=\:0
    Factor: . (3u - 2)(u + 3)\:=\:0

    Then we have: . \begin{Bmatrix}3u - 2 \:=\:0 & \Rightarrow & u \,=\,\frac{2}{3} \\ u + 3 \:=\:0 & \Rightarrow & u = \text{-}3\end{Bmatrix}


    3)\;\;x^2-6x-72\:=\:0
    Factor: . (x - 12)(x + 6)\:=\:0 . . . etc.


    4)\;\;(s-1)^2(s-3)^2
    This is not an equation.
    I assume it's supposed to be: . (s-1)^2(s-3)^2\:=\:0

    It is already factored, right?

    So we have: . \begin{Bmatrix}(s-1)^2\:=\:0 & \Rightarrow & s-1 \:=\:0 & \Rightarrow & s\,=\,1 \\<br />
(s-3)^2\:=\:0 & \Rightarrow & s-3\:=\:0 & \Rightarrow & s = 3\end{Bmatrix}


    5)\;\;0.2x^2-1.1x+0.5\:=\:0
    Multiply by 10: . 2x^2 - 11x + 5\:=\:0

    Factor: . (x - 5)(2x - 1)\:=\:0 . . . etc.



    6)\;\;(x^2-3)^2+(x^2-3)-2\:=\:0

    Replace x^2 - 3 with u.

    The equation becomes: . u^2 + u - 2 \:=\:0
    . . which factors: . (u - 1)(u + 2)\:=\:0
    . . and we get: . \begin{Bmatrix} u = 1\\u=\text{-}2\end{Bmatrix}

    Since u \,=\,x^2 - 3
    . . we have: . \begin{Bmatrix}x^2 - 3 \:=\:1 & \Rightarrow & x^2 = 4 & \Rightarrow & x = \pm 2 \\<br />
x^3 - 3 \:=\:\text{-}2 & \Rightarrow & x^2 = 1 & \Rightarrow & x = \pm 1 \end{Bmatrix}



    7) (x^2-3x+2)^3\:=\:0

    It is already factored.

    Take the cube root of both sides: . x^2 - 3x + 2 \:=\:0

    Factor: . (x-1)(x-2) \:=\:0 . . . etc.



    8)\;\; f(x)\:=\:(x+1)^4 +4(x+1)^3 . Find the zeros.

    We have: . (x+1)^4 + 4(x+1)^3\:=\:0

    Factor: . (x+1)^3\,[(x+1) + 4] \:=\:0\quad\Rightarrow\quad(x+1)^3(x+5)\:=\:0

    And we have: . \begin{Bmatrix}(x+1)^3\:=\:0 & \Rightarrow & x + 1 \:=\:0 & \Rightarrow & x = \text{-}1 \\<br />
x + 5 \:=\:0 & & \Rightarrow & & x = \text{-}5\end{Bmatrix}

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