SOLUTION: Find a polynomial of degree 3 such that when divided by x^2-5x has a remainder of 6x-15 and when divided by x^2-5x+8 has a remainder of 2x-7. Please explain as well. Thank you.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find a polynomial of degree 3 such that when divided by x^2-5x has a remainder of 6x-15 and when divided by x^2-5x+8 has a remainder of 2x-7. Please explain as well. Thank you.      Log On


   

Question 1154891: Find a polynomial of degree 3 such that when divided by x^2-5x has a remainder of
6x-15 and when divided by x^2-5x+8 has a remainder of 2x-7.
Please explain as well.
Thank you.
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Find a polynomial of degree 3 such that when divided by x%5E2-5x has a remainder of 6x-15 and when divided by x%5E2-5x%2B8 has a remainder of 2x-7.


f%28x%29%2Fg%28x%29=q%28x%29%2Breminder
f%28x%29=q%28x%29%2Ag%28x%29%2Breminder

if given g%28x%29=+x%5E2-5x and reminder+=6x-15, we have
f%28x%29=q%28x%29%2A%28x%5E2-5x+%29%2B6x-15.......eq.1

and if given g%28x%29=+x%5E2-5x%2B8 and reminder+=2x-7, we have
f%28x%29=q%28x%29%2A%28x%5E2-5x%2B8%29%2B2x-7.......eq.2

from eq.1 and eq.2 we have

q%28x%29%2A%28x%5E2-5x+%29%2B6x-15=q%28x%29%2A%28x%5E2-5x%2B8%29%2B2x-7
q%28x%29%2A%28x%5E2-5x+%29-q%28x%29%2A%28x%5E2-5x%2B8%29=2x-6x%2B15-7.........solve for q%28x%29
q%28x%29%28x%5E2-5x+-%28x%5E2-5x%2B8%29%29=-4x%2B8
q%28x%29%28x%5E2-5x+-x%5E2%2B5x-8%29=-4x%2B8.......simplify
q%28x%29%28-8%29=-4x%2B8
q%28x%29=-4x%2F%28-8%29%2B8%2F%28-8%29
q%28x%29=x%2F2-1-> your quotient

now find f%28x%29
f%28x%29=q%28x%29%2Ag%28x%29%2Breminder
f%28x%29=%28x%2F2-1%29%2A%28x%5E2-5x+%29%2B6x-15.......eq.1
f%28x%29=x%5E3%2F2-5x%5E2%2F2-x%5E2+%2B5x%2B6x-15
f%28x%29=x%5E3%2F2+-+7x%5E2%2F2+%2B11x+-+15=> your polynomial of degree 3

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Let px+q be the quotient when the polynomial is divided by x²-5x leaving
a remainder of 6x-15, and
Let rx+s be the quotient when the polynomial is divided by x²-5x+8 leaving
a remainder of 2x-7,

We use the fact that

(divisor)(quotient) + remainder = polynomial of degree 3

   (x²-5x)(px+q) + (6x-15) = polynomial 
  (x²-5x+8)(rx+s) + (2x-7) = polynomial 

Thus we have the identity

   (x²-5x)(px+q) + (6x-15) = (x²-5x+8)(rx+s) + (2x-7) 

which must be true for all values of x. Lots of terms will become 0
if we substitute x=0 and x=5

If we substitute x=0 and solve for s, we get s = -1

   (x²-5x)(px+q) + (6x-15) = (x²-5x+8)(rx-1) + (2x-7)

If we substitute x=5, we get r = 1/2 = 0.5

Thus the polynomial is equal to

  (x²-5x+8)(0.5x-1) + (2x-7)

When we multiply that out, we get

  0.5x³-3.5x²+11x-15   <-- answer

--------------------------------------------

Checking:
                   0.5x- 1                            0.5x- 1 
x²-5x+0)0.5x³-3.5x²+11x-15         x²-5x+8)0.5x³-3.5x²+11x-15
        0.5x³-2.5x²+ 0x                    0.5x³-2.5x²+ 4x 
                -x²+11x-15                         -x²+ 7x-15
                -x²+ 5x+ 0                         -x²+ 5x- 8
                     6x-15                              2x- 7
                  
Edwin