**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 ifis a zero ofP(x)(P() = 0), thenpis a factor of the constant term ofP(x)andqis a factor of the leading coefficient ofP(x).

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

- Arrange the polynomial in descending order
- Write down all the factors of the constant term. These are all the possible values of
*p*. - Write down all the factors of the leading coefficient. These are all the possible values of
*q*. - 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.
- 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} -9*x* + 9 + 2*x*^{4} -19*x*^{2}.

*P*(*x*) = 2*x*^{4}+*x*^{3}-19*x*^{2}– 9*x*+ 9- Factors of constant term: ±1, ±3, ±9.
- Factors of leading coefficient: ±1, ±2.
- Possible values of : ±, ±, ±, ±, ±, ±. These can be simplified to: ±1, ±, ±3, ±, ±9, ±.
- 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*) = 2*x*^{4} + *x*^{3} -19*x*^{2} – 9*x* + 9.

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