Question

Question #733f9

Answer 1

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 `color(red)("linear")` equation is an algebraic equation of `color(blue)("degree one")`. So "thrid degree linear equation" is inconsistent.

If you mean third degree `color(red)"polynomial"` equations, i.e. equations
with a form `ax^3+bx^2+cx+d=0`, the method below is useful.

[Example] Solve `6x^3-7x^2-9x-2 =0`

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

If a third degree polynimial `color(brown)a` `x^3+bx^2+cx+` `color(blue)d` `=0` has rational roots,
they must be `x=+-("positive divisor of " color(blue)d)/("positive divisor of "color(brown)a)`.

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 `+-1,+-2,+-1/2,+-1/3,+-2/3` and `+-1/6`.

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

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

`f(x)=(x-2)(6x^2+5x+1)=(x-2)(2x+1)(3x+1)`
`->` The roots for `f(x)=0` are `x= 2, -1/2, -1/3`.

For more information, see the topics:
https://socratic.org/precalculus/real-zeros-of-polynomials/rational-zeros