Question 1149382
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            I think that I know a SIMPLER way to solve the problem, comparing with what the two other tutors proposed.



<pre>
Since the polynomial does not give a remainder when divided by (x+1), it (the polynomial) can be written in the form


    p(x) = a*(x-t)*(x+1),   (1)


where "a" is the leading coefficient and t is another root of the polynomial.


Thus I need to find only two unknown values of "a" and "t.

For it, use two other given conditions.


The fact that  "the quadratic has a remainder of -6 when divided by x-1" means that  p(1) = -6 (the Remainder theorem)

    p(1) = a*(1-t)*(1+1) = -6,   or  a*(1-t)*2  = -6,   or  a*(1-t) = -3,   or

    a - at = -3.      (2)



The fact that  "the quadratic has a remainder of -4 when divided by x-3" means that  p(3) = -4 (the Remainder theorem, again)

    p(3) = a*(3-t)*(3+1) = -4,   or  a*(3-t)*4  = -4,   or  a*(3-t) = -1,   or

    3a - at = -1.      (3)


Now subtract equation (2) from equation (3).  You will get

     3a - a = -1 - (-3) = -1 + 3 = 2,  i.e.   2a = 2,   or  a = 1.


Next substitute the value of a= 1 into equation (2).  You will get

     1 - t = -3,  1 + 3 = t,   t = 4.


Hence, the polynomial under the question is


     p(x) = 1*(x-4)*(x+1) = x^2 - 3x - 4.    <U>ANSWER</U>


You may check it on your own that all conditions of the problem are satisfied.
</pre>

Solved.


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In this way,  I make you free from the necessity to solve 3x3-system of equations.



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