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algebra

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Can someone please explain these problems to me I have tried and cannot figure them out.


1)Use synthetic division to find p(-3) for p(x)=x^4-2x^3-4x+4

2)Find all the zeroes of the equation
x^4-6x^2-7x-6=0

  • algebra -

    1. X = -3.
    x+3 = 0.
    Using synthetic division, divide the given expression by x+3. You will get
    x^3 - 5x^2 + 15x - 49. + 151 Remainder.
    The fact that we got a remainder, means that x+3 is not a factor of the given
    expression and -3 is not a zero. Which
    means it will not give zero output when plugged into the Eq.

    2. All values of X that satisfy the given Eq is a "zero" of the Eq.

    x^4 - 6x^2 - 7x - 6 = 0.
    It was determined by trial and error that -2 satisfies the given Eq.
    X = -2.
    x+2 = 0
    Divide the given Eq by x+2 and get
    x^3 - 2x^2 - 2x -3. + 0 Remainder.
    (x+2)(x^3-2x^2-2x-3) = 0.
    3 satisfies the cubic polynomial:
    x = 3.
    x-3 = 0.
    Divide the cubic polynomial by x-3 and
    get x^2+x+1 + 0 Remainder.

    Our Eq 1s: (x+2)(x-3)(x^2+x+1) = 0.
    The 2nd degree polynomial cannot be
    factored and has no zeroes, because
    B^2 < 4AC. Therefore, the original Eq
    has only 2 zeroes: -2, and 3.
    OR
    (x,y)
    (-2,0)
    (3,0).

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