Question 928275
{{{6x^3 +5x^2 -4x}}}
 
STEP 1: Look for common factors and take out common factors if you find any:
{{{6x^3 +5x^2 -4x=x(6x^2 +5x -4)}}}
Always look for common factors, at the beginning and in further steps.
 
STEP 2: If the highest degree of your factors is 2,
factoring further )factoring the factors) may be possible, and even easy.
If {{{6x^2 +5x -4}}} can be factored further,
we will get to the factoring by looking for factors of {{{6*(-4)=-24}}} ,
the product of the first and last coefficients.
We need a pair of factors that multiply to yield {{{-24}}} and add up to {{{5}}} ,
the middle coefficient.
It turns out that {{{(-3)*8=-24}}} and {{{(-3)+8=5}}} ,
so {{{-3}}} and {{{8}}} are the factors we are looking for.
We re-write {{{+5x}}} as {{{-3x+8x}}} , and then "factor by grouping" :
{{{6x^2 +5x -4=6x^2-3x+8x-4=(6x^2-3x)+(8x-4)=3x(2x-1)+4(x-2)=(3x+4)(2x-1)}}} .
There we took {{{3x}}} as a common factor out of the group {{{6x^2-3x=3x(x-2)}}} ,
and we took common factor {{{4}}} out of group {{{8x-4=4(2x-1)}}} .
Next we took common factor {{{(2x-1)}}} out of the whole thing to get
{{{3x(2x-1)+4(2x-1)=(3x+4)(2x-1)}}} .
 
STEP #3: Put it all together:
{{{6x^3 +5x^2 -4x=x(6x^2 +5x -4)=highlight(x(3x+4)(x-2))}}} .
 
NOTE: If you cannot figure out how to factor a quadratic factor/polynomial,
like {{{6x^2 +5x -4}}} , there is a sort of cheat.
You could try making it equal to zero,
and solving the resulting quadratic equation by using the quadratic formula.
If there is no solution, the quadratic factor/polynomial cannot be factored.
If there is a solution
{{{6x^2 +5x -4=0}}}-->{{{x=(-5 +- sqrt(5^2-4*6*(-4) ))/(2*6)=(-5 +- sqrt(25+96))/12=(-5 +- sqrt(121))/12=(-5 +- 11)/12}}}-->{{{system(x=(-5-11)12=-16/12=-4/3,"or",x=(-5+11)12=6/12=1/2)}}}
The solutions to the equation, subtracted from {{{x}}} are factors.
In this case, {{{x-(-4/3)=x+4/3}}} and {{{x-1/2}}} are factors of {{{6x^2 +5x -4}}} .
Since the denominators make them look ugly, we multiply times the denominators to find the factors
{{{3(x+4/3)=3x+4}}} and {{{2(x-1/2)=2x-1}}} .