Polynomial Zeros
The zeros (or roots) of a polynomial P(x) are the values of x that make the polynomial equal to zero: $$ P(x) = 0 $$
A polynomial may have one, several, or no zeros at all.
Depending on the case, these zeros can be integers, rational numbers, real numbers, or complex numbers.
A Practical Example
Let’s take a look at the polynomial
$$ P(x) = x + 2 $$
To find its zeros, we set it equal to zero: P(x) = 0. That gives us the equation
$$ x + 2 = 0 $$
Using the principle of equality, we subtract 2 from both sides:
$$ x + 2 - 2 = 0 - 2 $$
$$ x = -2 $$
The polynomial evaluates to zero when x = -2.
Therefore, x = -2 is a zero of the polynomial.
Note. More specifically, it's an integer zero since -2 belongs to the set of integers.
How to Find the Zeros of a Polynomial
To find the zeros of a polynomial P(x), you set it equal to zero $$ P(x) = 0 $$ and solve the resulting equation.
This approach works for any polynomial.
However, for polynomials of degree higher than two, solving the equation directly can be time-consuming due to the complexity of the algebra involved.
Luckily, there are special techniques and algorithms for identifying integer and rational zeros.
Note. The algorithms below are much faster than solving the equation directly, but they’re not exhaustive: they only identify integer or rational zeros. So they won’t catch any irrational ones. Still, for most algebra problems, integer or rational zeros are all you need.
1] Method for Finding Integer Zeros
Integer zeros of a polynomial P(x) with integer coefficients can be found among the divisors of its constant term.
This method works only if the polynomial has integer coefficients, and it will only detect zeros that are integers - values of x belonging to the set of integers.
If the polynomial has rational or real zeros instead, this method won’t detect them.
Example
Consider this fourth-degree polynomial with integer coefficients:
$$ P(x):\ 3x^4 + 6x^2 + 5x - 4 $$
Let’s try to find its integer zeros without solving the full equation.
The constant term here is -4.
The divisors of -4 are:
$$ D_n = \{ 1, -1, 2, -2, 4, -4 \} $$
Now we test each of these values to see if any make the polynomial equal to zero:
$$ P(1) = 3(1)^4 + 6(1)^2 + 5(1) - 4 = 10 $$
$$ P(-1) = 3(-1)^4 + 6(-1)^2 + 5(-1) - 4 = 0 $$
We’ve found an integer zero:
x = -1
Note. Often, finding just one zero is enough - so it’s smart to start testing the smallest divisors first, since the calculations are simpler. Once a zero is found, you can stop there. Here’s the result of testing the remaining divisors for completeness: $$ P(2)=78 \\ P(-2)=58 \\ P(4)=880 \\ P(-4)=840 $$ No other integer zeros were found, but this doesn’t rule out the presence of rational or real ones.
2] Method for Finding Rational Zeros
Rational zeros of a polynomial P(x) with integer coefficients can be found by testing all fractions formed by taking a divisor of the constant term and dividing it by a divisor of the leading coefficient: $$ \frac{ \text{divisors of the constant term} }{ \text{divisors of the leading coefficient} } $$
This method works only for polynomials with integer coefficients, and it only finds zeros among the rational numbers.
If the polynomial has irrational or complex zeros, this method won’t find them.
Example
Consider this second-degree polynomial with integer coefficients:
$$ P(x):\ 8x^2 - 6x - 2 $$
The constant term is -2.
The leading coefficient (of the term 8x2) is 8.
First, list the divisors of -2 and 8:
$$ D(-2) = \{ 1, -1, 2, -2 \} $$
$$ D(8) = \{ 1, -1, 2, -2, 4, -4, 8, -8 \} $$
Now generate all possible fractions by placing each divisor of the constant term over each divisor of the leading coefficient:
$$ F = \{ \frac{1}{1}, \frac{1}{-1}, \frac{1}{2}, \frac{1}{-2}, \frac{1}{4}, \frac{1}{-4}, \frac{1}{8}, \frac{1}{-8}, \\ \ \ \ \ \ \ \frac{-1}{1}, \frac{-1}{-1}, \frac{-1}{2}, \frac{-1}{-2}, \frac{-1}{4}, \frac{-1}{-4}, \frac{-1}{8}, \frac{-1}{-8}, \\ \ \ \ \ \ \ \frac{2}{1}, \frac{2}{-1}, \frac{2}{2}, \frac{2}{-2}, \frac{2}{4}, \frac{2}{-4}, \frac{2}{8}, \frac{2}{-8}, \\ \ \ \ \ \ \ \frac{-2}{1}, \frac{-2}{-1}, \frac{-2}{2}, \frac{-2}{-2}, \frac{-2}{4}, \frac{-2}{-4}, \frac{-2}{8}, \frac{-2}{-8} \} $$
Simplify the fractions and remove duplicates:
$$ F = \{ \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{4}, \frac{1}{8}, -\frac{1}{8}, 1, -1, 2, -2 \} $$
If the polynomial has any rational zeros, they’ll be among these values.
Now test each one by plugging it into the polynomial:
$$ P\left( \frac{1}{2} \right) = -3 \\ P\left( -\frac{1}{2} \right) = 3 \\ P\left( \frac{1}{4} \right) = -3 \\ P\left( -\frac{1}{4} \right) = 0 $$
We’ve found a rational zero:
x = -1/4
Note. Since integers are a subset of rational numbers, this method is more comprehensive than the one that only finds integer zeros.
3] Method for Finding Real Zeros
If none of the previous methods work, you can turn to numerical techniques that approximate real solutions - values that may not be exact but are close enough for practical use.
One such method is the Newton-Raphson method, which provides an approximation of a real root of the polynomial.
However, this method requires knowledge of calculus, specifically derivatives.
And so on.
