SOLUTION: To Whom it May Concern, Could you help me with this question: Find the quadratic which has a remainder of -6 when divided by x - 1, a remainder of -4 when divided by x -3 and

Let the quadratic polynomial be Ax² + Bx + C


The remainder theorem says this:

If you substitute r into a polynomial, you will get the same answer as you will
get if you divide the polynomial by x-r and take the remainder.

>>Find the quadratic which has a remainder of -6 when divided by x - 1<<

So if r=1, then the remainder theorem says this:

If you substitute 1 into a polynomial, you will get the same answer as you will
get if you divide the polynomial by x-1 and take the remainder.

So let's substitute 1 into the quadratic polynomial and set it equal to the
remainder -6

A(1)² + B(1) + C = -6
       A + B + C = -6

>>a remainder of -4 when divided by x - 3<<

So if r=3, then the remainder theorem says this:

If you substitute 3 into a polynomial, you will get the same answer as you will
get if you divide the polynomial by x-3 and take the remainder.

So let's substitute 3 into the quadratic polynomial and set it equal to the
remainder -4

A(3)² + B(3) + C = -4
     9A + 3B + C = -4

>>and no remainder when divided by x + 1<<

Note that x + 1 is the same as x - (-1) 

So if r=-1, then the remainder theorem says this:

If you substitute -1 into a polynomial, you will get the same answer as you will
get if you divide the polynomial by x+1 and take the remainder.

So let's substitute -1 into the quadratic polynomial and set it equal to the
remainder 0.

A(-1)² + B(-1) + C = 0
         A - B + C = 9
 
So we have 3 equations in 3 unknowns:

 A +  B + C = -6
9A + 3B + C = -4
 A -  B + C =  0

Solve that and get A=1, B=-3, and C=-4

So Ax² + Bx + C becomes

    x² - 3x - 4  

Checking:

             x - 2
x - 1) x² - 3x - 4
       x² -  x
           -2x - 4
           -2x + 2
                -6   <--the remainder is -6 when we divide by x - 1.

 
             x - 0
x - 3) x² - 3x - 4
       x² - 3x
            0x - 4
            0x + 0
                -4   <--the remainder is -4 when we divide by x - 3.

             x - 4
x + 1) x² - 3x - 4
       x² +  x
           -4x - 4
           -4x - 4
                 0   <--the remainder is 0 (no remainder) when we divide by x - 1.

Edwin

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.    ANSWER


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