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)
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3)....1....3....-34....48
........1....6.....-16...|..0
=======================
Remainder = 0
Quotient: x^2+6x-16
=========================
Cheers,
stan H.
Answer by Edwin McCravy(20056) (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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