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Algebra II: Polynomials

The Rational Zeros Theorem

Problems

Problems

Roots of a Polynomial

A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0 .

The Rational Zeros Theorem

The Rational Zeros Theorem states:

If P(x) is a polynomial with integer coefficients and if is a zero of P(x) ( P() = 0 ), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .

We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps:

  1. Arrange the polynomial in descending order
  2. Write down all the factors of the constant term. These are all the possible values of p .
  3. Write down all the factors of the leading coefficient. These are all the possible values of q .
  4. Write down all the possible values of . Remember that since factors can be negative, and - must both be included. Simplify each value and cross out any duplicates.
  5. Use synthetic division to determine the values of for which P() = 0 . These are all the rational roots of P(x) .


Example: Find all the rational zeros of P(x) = x 3 -9x + 9 + 2x 4 -19x 2 .

  1. P(x) = 2x 4 + x 3 -19x 2 - 9x + 9
  2. Factors of constant term: ±1 , ±3 , ±9 .
  3. Factors of leading coefficient: ±1 , ±2 .
  4. Possible values of : ± , ± , ± , ± , ± , ± . These can be simplified to: ±1 , ± , ±3 , ± , ±9 , ± .
  5. Use synthetic division:
Figure %: Synthetic Division
Thus, the rational roots of P(x) are x = - 3 , -1 , , and 3 .

We can often use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a . Next, we can use synthetic division to find one factor of the quotient. We can continue this process until the polynomial has been completely factored.


Example (as above): Factor P(x) = 2x 4 + x 3 -19x 2 - 9x + 9 .

As seen from the second synthetic division above, 2x 4 + x 3 -19x 2 -9x + 9÷x + 1 = 2x 3 - x 2 - 18x + 9 . Thus, P(x) = (x + 1)(2x 3 - x 2 - 18x + 9) . The second term can be divided synthetically by x + 3 to yield 2x 2 - 7x + 3 . Thus, P(x) = (x + 1)(x + 3)(2x 2 - 7x + 3) . The trinomial can then be factored into (x - 3)(2x - 1) . Thus, P(x) = (x + 1)(x + 3)(x - 3)(2x - 1) . We can see that this solution is correct because the four rational roots found above are zeros of our result.

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