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Thread: Please help me with 4 quick polynomials

  1. #1
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    Please help me with 4 quick polynomials

    I just can't figure out these 4...Would be very happy if someone could help me out
    Solve the equations:
    1) (2x^3 -3x^2 -3x+2) / (x+1) < 0

    2) (x^3 -2x^2 -3x) > 0

    Solve the divisions
    3) (2x^3 +9x^2 -7x -4) : (x^2 +2x -3)

    4) x^4 : (x^2 - 4)
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  2. #2
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    Quote Originally Posted by pizdac
    I just can't figure out these 4...Would be very happy if someone could help me out
    Solve the equations:
    1) (2x^3 -3x^2 -3x+2) / (x+1) < 0
    Do the division

    $\displaystyle f(x)=\frac{2x^3-3x^2-3x+2}{x+1}=2x^2-5x+2$.

    Now as the coefficient of $\displaystyle x^2$ is positive $\displaystyle f(x)<0$ between the roots
    of the quadratic on the right (if these are real). The roots are $\displaystyle 2$ and $\displaystyle 1/2$. So:

    $\displaystyle f(x)=\frac{2x^3-3x^2-3x+2}{x+1}<0$, when $\displaystyle x \epsilon (1/2,2)$

    RonL
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  3. #3
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    Quote Originally Posted by pizdac
    2) (x^3 -2x^2 -3x) > 0
    The LHS of this inequality is a cubic, it has either three real roots, or but
    a single root. Also as $\displaystyle x \rightarrow \infty$ the cubic is positive.

    So if there is only a single root the cubic is $\displaystyle >0$ to the right
    of the root.

    If there are three roots it is $\displaystyle >0$ to the right of the greatest root,
    and again between the other two roots.

    (Sketch a cubic and you will see why this is so).

    As there is no constant term in the cubic we can factor it:

    $\displaystyle x(x^2 -2x -3)$,

    and so we can find the other roots using the quadratic formula.

    RonL
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    Problem 4
    $\displaystyle \frac{x^4}{x^2-4}=x^2+4+\frac{16}{x^2-4}$
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  5. #5
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    Problem 2)
    $\displaystyle x^3-2x^2-3x>0$
    Factor,
    $\displaystyle x(x^2-2x-3)$
    Once more,
    $\displaystyle x(x-3)(x+1)$.
    Now find the zero,
    $\displaystyle x=-1,0,3$
    Now observe each interval.
    $\displaystyle x<-1$
    $\displaystyle -1<x<0$
    $\displaystyle 0<x<3$
    $\displaystyle 3<x$
    Just pick points and see if the polynomial is positive or negative.
    Doing that we find that
    $\displaystyle x<-1$ This is Negative (Wrong)
    $\displaystyle -1<x<0$ This is Positive (Correct)
    $\displaystyle 0<x<3$ This is Negative (Wrong)
    $\displaystyle 3<x$ This is Positive (Correct)
    Answer:
    $\displaystyle -1<x<0 \mbox{ or } x>3$
    Or another way of writing $\displaystyle x\in (-1,0)\cup (3,\infty)$.
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    Thank you very much
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