Question #733f9

1 Answer
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 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)ax^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