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)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