Question

PLEASE PLEASE HELP
Use synthetic division to find the quotient and remainder when

6x^6-x^4+7x^2+6 is divided by x-2

Quotient:______
Remainder:_______

Answer 1

Are you familiar with synthetic division?

Answer 2

You would take the zero of x-2, that is 2, and all the coefficients of the polynomial, and order them like this:

2 | 6 0 -1 0 7 6

Remember, 6x^6-x^4+7x^2+6 is really \(6x^6+0x^5+-1x^4+0x^3+7x^2+6\).

Answer 3

Now, the next step is, draw a line somewhat below the numbers, leaving some space between:

2 | 6 0 -1 0 7 6
|
----------------------- +

Then just drop the first coefficient (6) below the line:

2 | 6 0 -1 0 7 6
|
----------------------- +
6

Answer 4

Now, the actual work begins. Multiply 2 with the dropped 6 and put the result in the next column. Add:

2 | 6 0 -1 0 7 6
| 12
----------------------- +
6 12

Answer 5

Then, multiply 2 with the 12 below the line and put in the next column. Add:

2 | 6 0 -1 0 7 6
| 12 24
----------------------- +
6 12 23

Keep doing this:

2 | 6 0 -1 0 7 6
| 12 24 46 92 198
------------------------ +
6 12 23 46 99 204

Answer 6

okay understand so far so do we keep using synthetic division?

Answer 7

Awww, I see I still forgot one coefficient :( you have 0x, so we get:

2 | 6 0 -1 0 7 0 6
| 12 24 46 92 198 396
----------------------------- +
6 12 23 46 99 198 402

Answer 8

Now we are at the end.
The first six numbers below the -------- line are the new coefficients of our 5th degree polynomial. The last (402) is the remainder. So:

\[6x^6-x^4+7x^2+6=6x^5+12x^4+23x^3+46x^2+99x+198+\dfrac{402}{6x^6-x^4+7x^2+6}\]

Answer 9

The end result is:\[6x^5+12x^4+23x^3+46x^2+99^x+198+\dfrac{402}{6x^6-x^4+7x^2+6}\]

Answer 10

So the remainder is 402 and the quotient is 6x^5+12x^4+23x^3+46x^2+99x+98 ?? could it be more simplify

Answer 11

Yes, the remainder and qouotient are as you said.
It cannot be made simpler...

Answer 12

thank you!!

Answer 13

YW!