Question 1183613
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The expression ax^3 + bx^2 - 5x + 2a is exactly divisible by x^2 - 3x - 4. 
Calculate the value of a and of b and factorise the expression completely.
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            Tutor @KMST provided very detailed long solution,  covering different possible options.

            I will try to give shorter solution in hope that it has its own charm.



<pre>
The trinomial  x^2 - 3x - 4  is factorable:  x^2 - 3x - 4 = (x-4)*(x+1).


It means that the given polynomial of the degree 3,  ax^3 + bx^2 - 5x + 2a,  is divisible 
by both binomials  (x-4)  and  (x+1).


Due to the Remainder theorem, it means that the values x= 4  and  x= -1  are the roots of that polynomial.


So, we substitute the values x= 4  and  x= -1 into the given polynomial, equate it to zero and
obtain two equations for the unknown coefficients  "a"  and  "b"


    a*4^3    + b*4^2 - 5*4    + 2a = 0      (1)

    a*(-1)^3 + b*1^2 - 5*(-1) + 2a = 0      (2)


Simplifying, you get


    64a + 16b - 20 + 2a = 0                 (1')

    -a  +   b +  5 + 2a = 0                 (2')


Simplifying further, you get

    66a + 16b = 20                          (1'')

      a +   b = -5                          (2'')
   

After solving the system, you get  a= 2,  b= -7.


Now the problem requires to find the third linear binomial, which is a third divisor to the given polynomial.


Use the Vieta's theorem:  the sum of the roots is equal to  {{{-b/a}}} = {{{-(-7)/2}}} = {{{7/2}}}.

    so, we write  4 + (-1) + t = {{{7/2}}}


where "t" is the third root, and we obtain from it

    t = {{{7/2 - 4 - (-1)}}} = {{{1/2}}}.


Thus the third root is  {{{1/2}}}, the associate binomial factor is  (x-1/2), and the required binomial decomposition is

    ax^3 + bx^2 - 5x + 2a = 2x^3 -7x^2 - 5x + 4 = {{{2*(x-1/2)*(x-4)*(x+1)}}} = (2x-1)*(x-4)*(x+1).
</pre>

Solved.