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Posted by on Tuesday, January 13, 2009 at 12:32am.

I know how to solve if the leading coefficient is 1, but how would i solve it if its not 1.

Ex.

f(x) = 2x^3 + 7x^2 - 7x + 30

Can someone help me start that out? Then I'll use synthetic divison to solve for it.

  • Rational Zero Theorem - , Tuesday, January 13, 2009 at 1:05am

    If someone looks at this, please reply. I will for replies around 6:30am before school starts.

  • Rational Zero Theorem - , Tuesday, January 13, 2009 at 6:47am

    If the rational root is p/q, then p must be a divisor of 30 and q a divisor of 2.

  • Rational Zero Theorem - , Tuesday, January 13, 2009 at 7:26am

    Also, if there are many possible candidates like in this case, you should not try them all out. Instead, you should proceed as follows.

    Let's denote the polynomial by P(x).
    Suppose y is a possible rational root. But then we check if y is a root and see that P(y) is not equal to zero. You can then apply the Rational Roots Theorem to the polynomial:

    Q(x) = P(y + x)

    You don't have to actually replace x by y + x in the original polynomial and expand out all the powers. All you need to know to apply the Rational Roots Theorem are the coefficients of the highest power and the constant term.

    The coefficient of the highest power of x remains the same, the constant term is P(y).

    In this case, if you take y = 1, then
    P(y) = 32. So, the possible rational roots of Q(x) are of the form plus or minus 2^n for n ranging from minus 1 to 5. Therefore the possible rational roots of the original polynomial are of the form:

    1 plus or minus 2^n

    for -1<=n<=5

    You then check which of the rational root candidates on your original list are of this form. You are then left with:

    x = 1/2, 3/2, -1, 3, -3, 5, -15.

    If we check x = -1, we find:

    P(-1) = 42.

    This then generates the following candidate roots:

    x = -1 + p/q

    with p a divisor of 42 and q a divisor of 2. The remaining candidates are then:

    x = -3 and x = 5.

    None of them work, so there are no rational roots (provided I didn't make any mistakes :) )

  • Rational Zero Theorem - , Tuesday, January 13, 2009 at 7:53am

    Thanks alot!

  • Rational Zero Theorem - , Thursday, April 2, 2009 at 10:45pm

    f(x)=x^3+2x^2+4x+1

  • Rational Zero Theorem - , Sunday, October 11, 2009 at 12:50pm

    give more example by solving rational zero theorem.

  • Rational Zero Theorem - , Friday, October 28, 2011 at 11:21am

    x3-7x2+11x-5=0

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