Question 1183613
{{{x^2-3x-4=(x+1)(x-4)}}} tells us
1) two of the factors we need to "factorise the expression completely" and
2) that the expression is zero for {{{x=-1}}} and for {{{x=4}}} .

ONE OPTION:
Using the second finding we know that
{{{a(-1)^3+b(-1)^2-5(-1)+2a=0}}} --> {{{-a+b+5+2a=0}}} --> {{{a+b+5=0}}} and
{{{a(4)^3+b(4)^2-5(4)+2a=0}}} --> {{{64a+16b-20+2a=0}}} --> {{{66a+16b-20=0}}}
We can then solve the system of equations
{{{system(a+b+5=0,66a+16b-20=0)}}} or {{{system(a+b=-5,66a+16b=20)}}}
to find
{{{system(a=2,b=-7)}}} ;
then re-write the original expression/polynomial as
{{{2x^3-7x^2-5x+4}}} ,
and then find the other factor(s) we need to "factorise the expression completely" as
{{{(2x+c)(x+1)(x-4)}}} or as {{{2(x-r)(x+1)(x-4)}}} ,
depending on what is understood by factorise completely.
The missing factor(s) can be found by
1) dividing the original expression by {{{x^2-3x-4}}} , or by
2) dividing in two steps by {{{(x+1)}}} and then by {{{(x-4)}}} , or by
3) multiplying {{{(2x+c)(x^2-3x-4)}}} or as {{{2(x-r)(x^2-3x-4)}}}
and knowing that identical expressions means equal for every value of the variable, comparing to
{{{2x^3-7x^2-5x+4}}}
to conclude that the the two ways to express the independent term mean
{{{-4c=4}}} --> {{{c=-1}}} or {{{2r=4}}} --> {{{r=1/2}}}
 
ANOTHER OPTION:
Multiply {{{(ax+c)(x^2-3x-4)}}} or as {{{a(x-r)(x^2-3x-4)}}}
and match the expressions of the coefficients in that product to the coefficients in
{{{ax^3 + bx^2 - 5x + 2a}}} to find {{{a}}} , {{{b}}} , and either {{{c}}} or {{{r}}}
For example:
{{{(ax+c)(x^2-3x-4)}}}{{{"="}}}{{{ax^3-3ax^2-4ax+cx^2-3cx-4c}}}
{{{"="}}}{{{ax^3+(-3a+c)x^2+(-4a-3c)x-4c}}} gives you
{{{system(a=a,b=-3a+c,-5=-4a-3c,2a=-4c)}}} or {{{system(-5=-4a-3c,2a=-4c)}}} and {{{b=-3a+c}}}
which tells you
{{{system(a=2,c=-1)}}} and substituting into {{{b=-3a+c}}} you get {{{b=-7}}}