SOLUTION: 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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Question 1183613: 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.
Found 3 solutions by KMST, ikleyn, greenestamps:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
tells us
1) two of the factors we need to "factorise the expression completely" and
2) that the expression is zero for and for .
ONE OPTION:
Using the second finding we know that
--> --> and
--> -->
We can then solve the system of equations
or
to find
;
then re-write the original expression/polynomial as
,
and then find the other factor(s) we need to "factorise the expression completely" as
or as ,
depending on what is understood by factorise completely.
The missing factor(s) can be found by
1) dividing the original expression by , or by
2) dividing in two steps by and then by , or by
3) multiplying or as
and knowing that identical expressions means equal for every value of the variable, comparing to

to conclude that the the two ways to express the independent term mean
--> or -->

ANOTHER OPTION:
Multiply or as
and match the expressions of the coefficients in that product to the coefficients in
to find , , and either or
For example:

gives you
or and
which tells you
and substituting into you get
Answer by ikleyn(52925)   (Show Source): You can put this solution on YOUR website!
.
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.
~~~~~~~~~~~~~


            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. 


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   =  = .

    so, we write  4 + (-1) + t = 


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

    t =  = .


Thus the third root is  , 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 =  = (2x-1)*(x-4)*(x+1).

Solved.



Answer by greenestamps(13216)   (Show Source): You can put this solution on YOUR website!


And here is a third and very different way to solve the problem....

We need to have



Perform the multiplication on the left...



... and equate coefficients:

x^3: p=a [1]
x^2: q-3p=b [2]
x: -4p-3q=-5 [3]
constant: -4q=2a --> q=-0.5a [4]

We have 4 equations in 4 unknowns, which could have led to a messy solution. But things work out nicely....

Substitute [1] and [4] into [3]:
-4a+1.5a=-5
-2.5a=-5
a=2

Then
p=a=2
q=-0.5a = -1
b = q-3p=-1-6=-7

ANSWER: a=2; b=-7 (and p=2; q=-1)

CHECK:
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