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Math Help - factor a cubic polynomial

  1. #1
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    factor a cubic polynomial

    How do I factor
    y^3- 18y^2+107y-216

    to equal (y-8)(y^2-10y+27)=0
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  2. #2
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    Re: factor a cubic polynomial

    you have P(y) = y^3- 18y^2+107y-216

    Then find a such that P(a) = 0 meaning y-a will be a factor, then divide P(y) by y-a

    Spoiler:
    Try y=8
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  3. #3
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    Re: factor a cubic polynomial

    Unfortunately, I'm not sure what you mean. I am familiar with factoring by grouping and MCD. Do these two methods not apply?
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    Re: factor a cubic polynomial

    Are you familiar with the Rational Roots Theorem? Because of the Rational Roots Theorem, the only possible roots of the given polynomial are factors of 216. One of those factors is 8, which turns out to be a root and the reason you can factor it as you describe.
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    Re: factor a cubic polynomial

    We need to use the factor theorem when faced with polynomials of order 3 or higher.

    It states that for a polynomial \displaystyle P(y) if \displaystyle P(a)=0 , then \displaystyle y-a is a linear factor of \displaystyle P(y).

    In your case \displaystyle P(y) = y^3- 18y^2+107y-216 substitute \displaystyle y=8 gives \displaystyle P(8) = 512-1152+856-216 = 0 therefore \displaystyle y-8 is a factor.

    So now all you need to do is find \displaystyle \frac{y^3- 18y^2+107y-216}{y-8} = \cdots

    and you will have your solution.
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    Re: factor a cubic polynomial

    We are given to factor:

    y^3-18y^2+107y-216

    Looking at the coefficient of the squared term, we find that if the coefficient of the linear term is 80, then we can factor the first 3 terms. We should also observe that 107 - 80 = 27 and 216/27 = 8, which is one of the factors of 80 we are using.

    y^3-18y^2+80y+27y-216

    y(y-8)(y-10)+27(y-8)

    (y-8)\left(y(y-10)+27\right)

    (y-8)\left(y^2-10y+27\right)
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    Re: factor a cubic polynomial

    Quote Originally Posted by icemanfan View Post
    Are you familiar with the Rational Roots Theorem? Because of the Rational Roots Theorem, the only possible roots of the given polynomial are factors of 216. One of those factors is 8, which turns out to be a root and the reason you can factor it as you describe.
    Quote Originally Posted by pickslides View Post
    We need to use the factor theorem when faced with polynomials of order 3 or higher.

    It states that for a polynomial \displaystyle P(y) if \displaystyle P(a)=0 , then \displaystyle y-a is a linear factor of \displaystyle P(y).

    In your case \displaystyle P(y) = y^3- 18y^2+107y-216 substitute \displaystyle y=8 gives \displaystyle P(8) = 512-1152+856-216 = 0 therefore \displaystyle y-8 is a factor.

    So now all you need to do is find \displaystyle \frac{y^3- 18y^2+107y-216}{y-8} = \cdots

    and you will have your solution.

    Ah, I see now that I needed to use the Integer Value Theorem. Now what you saying is making sense
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