Rationals

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RRT

Let $f(x)$ be a polynomial with rational coefficients.

\[ f(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_2 x^2 + b_1 x + b_0 \]

where $b_k \in \mathbb {Q}$: $b_k = \frac {n_k}{d_k}$

Then we could get a common denominator of all of the coefficients and factor that out front

\[ f(x) = \frac {1}{d}(c_n x^n + c_{n-1} x^{n-1} + \cdots + c_2 x^2 + c_1 x + c_0) \]

These are the same polynomial function, so they have the same roots and factorization.

So, when we are investigating zeros and roots and factors, we can always assume that a polynomial with rational coefficients just has integer coefficents.

Rational Roots

Let $f(x)$ be a polynomial with integer coefficients.

\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \]

Suppose $f(x)$ has a rational root: $\frac {N}{D}$ in reduced form. That is $N$ and $D$ do not share any prime factors.

That means

\[ f \left ( \frac {N}{D} \right ) = a_n \left ( \frac {N}{D} \right )^n + a_{n-1} \left ( \frac {N}{D} \right )^{n-1} + \cdots + a_2 \left ( \frac {N}{D} \right )^2 + a_1 \left ( \frac {N}{D} \right ) + a_0 = 0 \]

Let’s factor out $\frac {1}{D^n}$

\[ \frac {1}{D^n} (a_n N^n + a_{n-1} N^{n-1} D + \cdots + a_2 N^2 D^{n-2}+ a_1 N D^{n-1} + a_0 D^n) = 0 \]

That tells us

\[ a_n N^n + a_{n-1} N^{n-1} D + \cdots + a_2 N^2 D^{n-2}+ a_1 N D^{n-1} + a_0 D^n = 0 \]

\[ a_n N^n = -(a_{n-1} N^{n-1} D + \cdots + a_2 N^2 D^{n-2}+ a_1 N D^{n-1} + a_0 D^n) \]

$D$ is a factor of the right side, so $D$ must be a factor of the left side, But $D$ is not a factor of $N$, because the original rational root was in reduced form. Therefore $D$ must be a factor of $a_n$.

Similarly,

\[ -(a_n N^n + a_{n-1} N^{n-1} D + \cdots + a_2 N^2 D^{n-2}+ a_1 N D^{n-1}) = a_0 D^n \]

$N$ is a factor of the left side, so $N$ must be a factor of the right side, But $N$ is not a factor of $D$, because the original rational root was in reduced form. Therefore $N$ must be a factor of $a_0$.

This is very helpful when factoring polynomials.

Rational Root Theorem

Let $ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 $ be a polynomial with integer coefficients.

Let $\frac {N}{D}$ be a rational root of $f(x)$.

Then $D$ is a factor of $a_n$, the leading coefficient and $N$ is a factor of $a_0$, the constant term.

Roots and Factoring

Let $T(h) = 2 h^2 - 9h - 5$

If $T(h)$ has a rational root, $\frac {N}{D}$, then

$N$ is a factor of $5$: $-5$, $-1$, $1$, or $5$
$D$ is a factor of $2$: $-2$, $-1$, $1$, or $2$

Possibilities are

\[ \frac {5}{2}, \frac {-5}{2}, \frac {5}{1}, \frac {-5}{1}, \frac {-1}{-1}, \frac {-1}{1}, \frac {1}{2}, \frac {-1}{2} \]

$T$ is a quadratic, so there are only two roots. If they are rational roots, then they are among these.

We can try each one.

$T\left ( \frac {5}{2} \right ) = -15$
$T\left ( \frac {-5}{2} \right ) = 30$
$T(5) = 0$ : a root
$T(-5) = 90$
$T(1) = -12$
$T(-1) = 6$
$T\left ( \frac {1}{2} \right ) = -9$
$T\left ( \frac {-1}{2} \right ) = 0$ : a root

We have two roots, which gives us two factors.

$T(h) = 2\left (h-\frac {1}{2} \right )(h-5)$

We could also multiply the $2$ inside the first factor.

$T(h) = (2h - 1)(h-5)$

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more examples can be found by following this link
More Examples of Zeros

2025-05-18 01:43:13