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| Question 89434:  Divide by using synthetic division
  (x^3 + 3x^2 -34x + 48) divided by (x - 3) Found 2 solutions by  stanbon, Edwin McCravy:
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! Divide by using synthetic division  (x^3 + 3x^2 -34x + 48) divided by (x - 3) -------------
 3)....1....3....-34....48
 ........1....6.....-16...|..0
 =======================
 Remainder = 0
 Quotient: x^2+6x-16
 =========================
 Cheers,
 stan H.
Answer by Edwin McCravy(20064)
      (Show Source): 
You can put this solution on YOUR website! 
Divide by using synthetic division
(x³ + 3x² - 34x + 48) divided by (x - 3)
Put a 1 before the first term.  It is already
in descending order.
1x³ + 3x² - 34x + 48
Erase the variables:
From (x - 3), change the sign of -3 to +3, or 3, and write this:
 3|1   3 -34  48
  |_____________
Bring down the 1
 3|1   3 -34  48
  |             
   1
Multiply the 1 at the bottom by the 3 on the far left, getting
3, and so  we put the 3 ABOVE AND TO THE RIGHT OF the 1 we just
brought down.  So we have this:
 3|1   3 -34  48
  |    3        
   1
Now we add the column with the 3 and the 3 and get 6. So we write
that 6 under the two 3's we just added,
 3|1   3 -34  48
  |    3              
   1   6 
Now we add the column with the 3 and the 3 and get 6. So we write
that 6 under the two 3's we just added,
Now we multiply the 6 by the 3 at the far left, getting 18, and write
that ABOVE AND TO THE RIGHT of the 6
 3|1   3 -34  48
  |    3  18          
   1   6 
Now we add the -34 and the 18 and get -16, so we write that under
their column:
 3|1   3 -34  48
  |    3  18          
   1   6 -16 
Then we multiply the -16 by the 3 at the far left, getting -48.
Then we write that ABOVE AND TO THE RIGHT of -16:
 3|1   3 -34  48
  |    3  18 -48      
   1   6 -16    
Only one more thing to do.  Combine the 48 and the -48 and get 0.
 3|1   3 -34  48
  |    3  18 -48      
   1   6 -16   0
Now we have to interpret the synthetic answer:
The last number, 0 is the remainder.
The other three numbers 
   1   6 -16
are a synthetic form of the quotient, remembering 
it will be a polynomial of one less degree than 
the original polynomial. So since the original 
polynomial has degree 3, it will have degree 2 
and so will begin with an x² term. So the
quotient i
   1   6 -16
   |   |  |
   1x²+6x-16
Then we write the answer and we would put the remainder
over the divisor:
 x² + 6x - 16 
Edwin
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