# Question #733f9

Sep 30, 2017

There is no third degree linear equation.
Three degree polynomial equations with at least one rational root(s) can be solved by factorization. See below.

#### Explanation:

A $\textcolor{red}{\text{linear}}$ equation is an algebraic equation of $\textcolor{b l u e}{\text{degree one}}$. So "thrid degree linear equation" is inconsistent.

If you mean third degree $\textcolor{red}{\text{polynomial}}$ equations, i.e. equations
with a form $a {x}^{3} + b {x}^{2} + c x + d = 0$, the method below is useful.

[Example] Solve $6 {x}^{3} - 7 {x}^{2} - 9 x - 2 = 0$

[Step 1] Find a rational zero on the equation. This step needs some perseverance.

If a third degree polynimial $\textcolor{b r o w n}{a}$${x}^{3} + b {x}^{2} + c x +$$\textcolor{b l u e}{d}$$= 0$ has rational roots,
they must be $x = \pm \left(\text{positive divisor of " color(blue)d)/("positive divisor of } \textcolor{b r o w n}{a}\right)$.

In this example, divisors of $a$ are $1 , 2 , 3 , 6$ and divisors of $b$ are $1 , 2$ .
Therefore the rational "canditates" for the equation are $\pm 1 , \pm 2 , \pm \frac{1}{2} , \pm \frac{1}{3} , \pm \frac{2}{3}$ and $\pm \frac{1}{6}$.

Let $f \left(x\right) = 6 {x}^{3} - 7 {x}^{2} - 9 x - 2$ and substitute these values to $x$
and you will find $f \left(2\right) = 0$.
This means that $x = 2$ is one of the roots and thus $f \left(x\right)$ is divisivle by $\left(x - 2\right)$.

[Step2] Factor the polynomial using the result of [Step 1].
In this case, $f \left(x\right) = 6 {x}^{3} - 7 {x}^{2} - 9 x - 2$ is divisible by $\left(x - 2\right)$.
$6 {x}^{3} - 7 {x}^{2} - 9 x - 2$ divided by $x - 2$ is $6 {x}^{2} + 5 x + 1$.

$f \left(x\right) = \left(x - 2\right) \left(6 {x}^{2} + 5 x + 1\right) = \left(x - 2\right) \left(2 x + 1\right) \left(3 x + 1\right)$
$\to$ The roots for $f \left(x\right) = 0$ are $x = 2 , - \frac{1}{2} , - \frac{1}{3}$.