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Given a polynomial function f ( x ) , use the Rational Zero Theorem to find rational zeros.

  1. Determine all factors of the constant term and all factors of the leading coefficient.
  2. Determine all possible values of p q , where p is a factor of the constant term and q is a factor of the leading coefficient. Be sure to include both positive and negative candidates.
  3. Determine which possible zeros are actual zeros by evaluating each case of f ( p q ) .

Listing all possible rational zeros

List all possible rational zeros of f ( x ) = 2 x 4 5 x 3 + x 2 4.

The only possible rational zeros of f ( x ) are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.

The constant term is –4; the factors of –4 are p = ±1 , ±2 , ±4.

The leading coefficient is 2; the factors of 2 are q = ±1 , ±2.

If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.

p q = ± 1 1 , ± 1 2      p q = ± 2 1 , ± 2 2      p q = ± 4 1 , ± 4 2

Note that 2 2 = 1 and 4 2 = 2 , which have already been listed. So we can shorten our list.

p q = Factors of the last Factors of the first = ±1 , ±2 , ±4 , ± 1 2
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Using the rational zero theorem to find rational zeros

Use the Rational Zero Theorem to find the rational zeros of f ( x ) = 2 x 3 + x 2 4 x + 1.

The Rational Zero Theorem tells us that if p q is a zero of f ( x ) , then p is a factor of 1 and q is a factor of 2.

p q = factor of constant term factor of leading coefficient     = factor of 1 factor of 2

The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for p q are ±1 and ± 1 2 . These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for x in f ( x ) .

   f ( 1 ) = 2 ( 1 ) 3 + ( 1 ) 2 4 ( 1 ) + 1 = 4        f ( 1 ) = 2 ( 1 ) 3 + ( 1 ) 2 4 ( 1 ) + 1 = 0     f ( 1 2 ) = 2 ( 1 2 ) 3 + ( 1 2 ) 2 4 ( 1 2 ) + 1 = 3        f ( 1 2 ) = 2 ( 1 2 ) 3 + ( 1 2 ) 2 4 ( 1 2 ) + 1 = 1 2

Of those, −1, 1 2 ,  and  1 2 are not zeros of f ( x ) . 1 is the only rational zero of f ( x ) .

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Use the Rational Zero Theorem to find the rational zeros of f ( x ) = x 3 5 x 2 + 2 x + 1.

There are no rational zeros.

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Finding the zeros of polynomial functions

The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division    repeatedly to determine all of the zeros    of a polynomial function.

Given a polynomial function f , use synthetic division to find its zeros.

  1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
  2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
  3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
  4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

Finding the zeros of a polynomial function with repeated real zeros

Find the zeros of f ( x ) = 4 x 3 3 x 1.

The Rational Zero Theorem tells us that if p q is a zero of f ( x ) , then p is a factor of –1 and q is a factor of 4.

p q = factor of constant term factor of leading coefficient     = factor of –1 factor of 4

The factors of 1 are ±1 and the factors of 4 are ±1 , ±2 , and ±4. The possible values for p q are ±1 , ± 1 2 , and ± 1 4 . These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.

1 4 0 3 1 4 4 1    4  4   1     0

Dividing by ( x 1 ) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

( x 1 ) ( 4 x 2 + 4 x + 1 ) .

The quadratic is a perfect square. f ( x ) can be written as

( x 1 ) ( 2 x + 1 ) 2 .

We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

2 x + 1 = 0           x = 1 2

The zeros of the function are 1 and 1 2 with multiplicity 2.

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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