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2.6. Zeros of Polynomial Functions

A problem of key importance is finding the * zeros*, also known as

*, of a function. These are values of the variable x at which f(x)=0. In particular, it is highly useful to know how to find the zeros of a univariate polynomial. Finding the zeros of linear polynomials is completely trivial. In Section 2.4 we discussed three methods for finding the zeros of a quadratic function, namely trial and error, completing the square, and the quadratic formula. In this section we will discuss how to find rational zeros of polynomials of higher degree. First we state and prove a useful theorem.*

**roots**

**Theorem 2.6.1:** The linear polynomial x - a is a factor of a polynomial p(x) if and only if a is a zero of f, i.e. p(a)=0.

**Proof: **Clearly if x - a is a factor of p(x) then p(a) must be zero, for this factor becomes zero when x = a. On the other hand, suppose p(a) = 0. By the Factor Theorem from the previous section, there exist polynomials q(x) and r(x) with deg(r) < deg(x - a) = 1 such that p(x) = q(x)(x - a) + r(x). Now since deg(r)=0, we see that r(x) must be equal to a constant, call it r. Thus we have p(x) - q(x)(x - a) + r. But then we have p(a) = q(a)(a - a) + r = r = 0, whence r=0 and p(x) = q(x)(x - a). Thus we see that x-a is a factor of p(x). **QED**

According to Theorem 2.6.1, finding a root of a polynomial is equivalent to finding a linear factor. We use this fact to prove a very useful theorem pertaining to the rational roots of a polynomial.

**Theorem 2.6.2:** Let p(x) = c_{n}x^{n} + c_{n-1}x^{n-1} + ... + c_{2}x^{2} + c_{1}x + c_{0} be a polynomial of degree n with integer coefficients and leading coefficient c_{n} nonzero, and let a be a rational root of p(x). Then there exist integers j and k with j dividing c_{0} and k dividing c_{n} such that a = j / k.

**Proof: **According to Theorem 2.6.1, a is a zero of p(x) if and only if x-a is a linear factor of p(x). Now since a is rational, a must have the form j / k for some integers j and k with k nonzero. Thus x - a = x - j/k is a linear factor of p(x). But this implies that (k)(x - a) = (k)(x - j/k) = kx - j is also a linear factor of p(x). Let q(x) = p(x) / (kx - j). Then q(x) is a polynomial of degree n-1 with integer coefficients, i.e. we have q(x) = b_{n-1}x^{n-1} + ... + b_{2}x^{2} + b_{1}x + b_{0}. But then we see that p(x) = (b_{n-1}x^{n-1} + ... + b_{2}x^{2} + b_{1}x + b_{0})(kx - j) = c_{n}x^{n} + c_{n-1}x^{n-1} + ... + c_{2}x^{2} + c_{1}x + c_{0}, implying that kb_{n-1} = c_{n} and -jb_{0} = c_{0}. Thus we see that j divides c_{0} and k divides c_{n} as claimed. **QED**

Armed with Theorem 2.6.2, we now know how to find all rational zeros of an arbitrary polynomial with integer coefficients. We present a few examples below.

**Example 1: **Find the rational zeros of the polynomial p(x) = x^{3} + x^{2} - 4x - 4.

**Solution: **According to Theorem 2.6.2, the possible rational zeros of p(x) are ±1, ±2, and ±4. We test each of these zeros in turn by synthetic division until we have found one. The results are shown below:

Thus we see that 1 is not a zero of p(x) but -1 is. We also see that we have p(x) = (x + 1)(x^{2} - 4) = (x + 1)(x + 2)(x - 2), where we have found the other two zeros, namely -2 and 2, simply by factoring the quotient x^{2} - 4.

**Example 2: **Find the rational zeros of the polynomial p(x) = 2x^{3} - x^{2} - x - 3.

**Solution: **According to Theorem 2.6.2, the possible rational zeros of p(x) are ±1, ±3, ±1/2, and ±3/2. We test each of these zeros in turn by synthetic division until we have found one. The results are shown below:

Thus we see that 3/2 is the only rational zero of p(x). We also see that we have the factorization p(x) = (x - 3/2)(2x^{2} + 2x + 2) = (2x - 3)(x^{2} + x + 1). In fact, 3/2 is the only real zero of p(x) since the cofactor, x^{2} + x + 1, has no real zeros.

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