# Rational Zeros

## Key Questions

See explanation...

#### Explanation:

The rational zeros theorem can be stated:

Given a polynomial in a single variable with integer coefficients:

${a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{0}$

with ${a}_{n} \ne 0$ and ${a}_{0} \ne 0$, any rational zeros of that polynomial are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term ${a}_{0}$ and $q$ a divisor of the coefficient ${a}_{n}$ of the leading term.

Interestingly, this also holds if we replace "integers" with the element of any integral domain. For example it works with Gaussian integers - that is numbers of the form $a + b i$ where $a , b \in \mathbb{Z}$ and $i$ is the imaginary unit.

• The Rational Zeros Theorem states: If $P \left(x\right)$ is a polynomial with integer coefficients and if $\frac{p}{q}$ is a zero of $P \left(x\right)$ , ($P \left(\frac{p}{q}\right) = 0$), then $p$ is a factor of the constant term of $P \left(x\right)$ and q is a factor of the leading coefficient of P(x) .

See explanation...

#### Explanation:

A polynomial in a variable $x$ is a sum of finitely many terms, each of which takes the form ${a}_{k} {x}^{k}$ for some constant ${a}_{k}$ and non-negative integer $k$.

So some examples of typical polynomials might be:

${x}^{2} + 3 x - 4$

$3 {x}^{3} - \frac{5}{2} {x}^{2} + 7$

A polynomial function is a function wholse values are defined by a polynomial. For example:

$f \left(x\right) = {x}^{2} + 3 x - 4$

$g \left(x\right) = 3 {x}^{3} - \frac{5}{2} {x}^{2} + 7$

A zero of a polynomial $f \left(x\right)$ is a value of $x$ such that $f \left(x\right) = 0$.

For example, $x = - 4$ is a zero of $f \left(x\right) = {x}^{2} + 3 x - 4$.

A rational zero is a zero that is also a rational number, that is, it is expressible in the form $\frac{p}{q}$ for some integers $p , q$ with $q \ne 0$.

For example:

$h \left(x\right) = 2 {x}^{2} + x - 1$

has two rational zeros, $x = \frac{1}{2}$ and $x = - 1$

Note that any integer is a rational number since it can be expressed as a fraction with denominator $1$.

• You can use the rational root theorem:

Given a polynomial of the form:

${a}_{0} {x}^{n} + {a}_{1} {x}^{n - 1} + \ldots + {a}_{n}$ with ${a}_{0} , \ldots , {a}_{n}$ integers,

all rational roots of the form $\frac{p}{q}$ written in lowest terms (i.e. with $p$ and $q$ having no common factor) will satisfy.

$p | {a}_{n}$ and $q | {a}_{0}$

That is $p$ is a divisor of the constant term and $q$ is a divisor of the coefficient of the highest order term.

This gives you a finite number of possible rational roots to try.

For example, the rational roots of

$6 {x}^{4} - 7 {x}^{3} + {x}^{2} - 7 x - 5 = 0$

must be of the form $\frac{p}{q}$ where $p$ is $\pm 1$ or $\pm 5$ and
$q$ is $1$, $2$, $3$ or $6$.

You can try substituting each of the possible combinations of $p$ and $q$ as $x = \frac{p}{q}$ into the polynomial to see if they work.

In fact the only rational roots it has are $- \frac{1}{2}$ and $\frac{5}{3}$.

Once you have found one root, you can divide the polynomial by the corresponding factor to simplify the problem.

• To find the zeroes of a function, $f \left(x\right)$, set $f \left(x\right)$ to zero and solve.
For polynomials, you will have to factor.

For example: Find the zeroes of the function $f \left(x\right) = {x}^{2} + 12 x + 32$

First, because it's a polynomial, factor it
$f \left(x\right) = \left(x + 8\right) \left(x + 4\right)$

Then, set it equal to zero
$0 = \left(x + 8\right) \left(x + 4\right)$

Set each factor equal to zero and the answer is $x = - 8$ and $x = - 4$.

*Note that if the quadratic cannot be factored using the two numbers that add to this and multiple to be this method, then use the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

to factor an equation in the form of $a {x}^{2} + b x + c$.