The underlining in my answer below did not come out on the actual display of the answer. So sorry. It must be a shortcoming in the Wyzant editing. I think you can still make out what I am saying. MK ---------------------------------------

I will show you how to do part A. Once you know how to do synthetic division, you can do part B.

x4 − x3 + x2 − x + 2/ x-2

Write the "zero" that is represented by the divisor in the top right corner.

Since the divisor is (x - 2), the "zero" would be +2.

It is always the opposite sign in a simple binomial like (x - 2).

So write...

2 with a half box around it...

2 |

Then write the coefficients of the dividend to the right.

2 | 1 -1 1 -1 2

You must be certain that every descending power of x is represented in the list. For example, if there is not x^2 term, you would place a zero in the list to the right.

Next draw a line, leaving a blank space under the previous line.

2 | 1 -1 1 -1 2

Next, bring down the first coefficient (the coefficient of the highest degree term). In this problem, it is the 1 understood to be in front of the x^4. It is always the first coefficient in the list to the right of the divisor.

2 | 1 -1 1 -1 2

↓

1

Next, multiply the divisor by the number you brought down, and place the product under the next coefficient. (under the -1 in this problem.)

Like this...

2 | 1 -1 1 -1 2

↓

1

Next, multiply the divisor by the number you brought down, and place the product under the next coefficient. (under the -1 in this problem.)

Like this...

2 | 1 -1 1 -1 2

↓ 2

1

↓ 2

1

That is... the 2 | times the 1 at the bottom is equal to the 2 that we write under the -1.

Next, we add down the second column, that is the -1 plus the 2 we just wrote. We add those and get 1. Write that 1 under the 2 like this....

2 | 1 -1 1 -1 2

↓ 2

1 1

The procedure to remember here is "multiply and then add."

Repeat the procedure by multiplying the 2 | by the 1 that we just wrote, getting 2. Write that 2 below the next coefficient which was the 1. Like this...

2 | 1 -1 1 -1 2

↓ 2 2

1 1

Then add down again to get 3, and put the 3 underneath ...

2 | 1 -1 1 -1 2

↓ 2 2

1 1 3

Multiply and add again to get....

2 | 1 -1 1 -1 2

↓ 2 2 6

1 1 3 5

↓ 2 2 6

1 1 3 5

And one more time...

2 | 1 -1 1 -1 2

↓ 2 2 6 10

1 1 3 5 12

↓ 2 2 6 10

1 1 3 5 12

Now the sequence at the bottom represents the quotient.

1 1 3 5 12

Start at the right with the 12. It is the remainder. It goes on top of a fraction with the divisor as its denominator. It is the last term in the quotient. So the last term of this problem is 12 / (x-2)

The next number to the left is the 5. It is the constant term with no x. (you can think of it as the x^0 term.)

The the term is just 5.

Now we have 5 + 12/(x-2)

The next number to the left is the coefficient of the x term. (You can think of it as the x^1 term.)

It is 3x^1... which is simply 3x.

So far we have 3x + 5 + 12/(x-2)

The next number to the left is the coefficient of the x^2 term.

It is 1x^2, or simply x^2.

So far we have x^2 + 3x + 5 + 12/(x-2)

The next number to the left is the final number, and it is 1.

It is the coefficient of the x^3 term.

It is 1x^3, or simply x^3.

So the whole quotient is x^3 + x^2 + 3x + 5 + 12/(x-2)

The last term is the "remainder."

So 12/(x+2) is the remainder.

Note that we have now two "factors" of the original polynomial.

That is [x^3 + x^2 + 3x + 5 + 12/(x-2)] times (x-2) is equal to x4 − x3 + x2 − x + 2 (the original polynomial.

I will leave it to you to do part B. Use the same procedure.

Once you get used to synthetic division, you will find it very fast. You will use it to factor polynomials in order to find their "zeroes," which will be the x-intercepts on the graph.