The process for polynomial long division (like the process for numerical long
division) has been separated somewhat from its logical underpinnings for a more
efficient method to arrive at an answer. For particular types of polynomial long
division, we can even take this abstraction one step further. Synthetic Division is
a handy shortcut for polynomial long division problems in which we are dividing
by alinearpolynomial. This means that the highest power ofxwe are dividing by
needs to bex1_{. This limits the usefulness of Synthetic Division, but it will serve}

us well for certain purposes. Let’s examine where the coefficients in our answer come from when we divide by a linear polynomial:

2x3

x−5

2x4 _{−}_{6}_{x}3 _{−}_{23}_{x}2_{+ 16}_{x}_{−}_{5}

−2x4_{+ 10}_{x}3

4x3 _{−}_{23}_{x}2

Notice that the first coefficient in the answer is the same as the first coefficient in
the polynomial we’re dividing into. This is because we’re dividing by a polyno-
mial in the form1x1_{−}_{a}_{. This also makes the power of the first term in the answer}

one less than the power of the polynomial we are dividing into. Let’s see where the subsequent coefficients in the answer come from:

2x3 + 4x2
x−5 2x4 −6x3 −23x2+ 16x−5
−2x4_{+ 10}_{x}3
4x3 −23x2
−4x3 _{+ 20}_{x}2
−3x2_{+ 16}_{x}

The next coefficient in the answer (4) comes from the combination of the−6and the +10. The +10 came from multiplying the 2 in the answer by the 5 in the divisorx−5. The next coefficient in the answer will be −3, which comes from multiplying the 4 (in the answer) by the 5 (in the divisor) and combining it with the−23in the polynomial we’re dividing into:

2.7. SYNTHETIC DIVISION 121
2x3 _{+ 4}_{x}2 _{−}_{3}_{x}
x−5
2x4 _{−}_{6}_{x}3_{−}_{23}_{x}2_{+ 16}_{x}_{−}_{5}
−2x4_{+ 10}_{x}3
4x3_{−}_{23}_{x}2
−4x3 + 20x2
−3x2_{+ 16}_{x}
3x2−15x

The last part of our answer will come from multiplying the −3 in the answer times the 5 in the divisor (making −15) and combining this with the +16in the polynomial we’re dividing into:

2x3 _{+ 4}_{x}2 _{−}_{3}_{x}_{+ 1}
x−5 2x4 _{−}_{6}_{x}3_{−}_{23}_{x}2_{+ 16}_{x}_{−}_{5}
−2x4_{+ 10}_{x}3
4x3_{−}_{23}_{x}2
−4x3 + 20x2
−3x2_{+ 16}_{x}
3x2−15x
x−5
−x+ 5
0
1

The process of Synthetic Division uses these relationships as a shortcut to finding the answer. The set-up for a Synthetic Division problem is shown below:

This set-up allows us to complete the division problem2x

4 _{−}_{6}_{x}3_{−}_{23}_{x}2_{+ 16}_{x}_{−}_{5}

x−5 .

5 2 −6 −23 16 −5

The first coefficient in the answer is the same as the first coefficient in the polyno- mial we’re dividing into:

5 2 −6 −23 16 −5

↓

2

To get the next coefficient, we multiply the 2 by the 5 to get+10 and flll this in under the−6:

5 2 −6 −23 16 −5

↓ +10 2

Then,−6 + 10 = +4, which is the next coefficient in the answer:

5 2 −6 −23 16 −5

↓ +10

2 4

Then, we continue this process, multiplying the 4 by the 5 to get 20 and combin- ing this with the−23: −23 + 20 = −3:

5 2 −6 −23 16 −5

↓ +10 +20

2 4 −3

Next, multiply the−3by the 5 and combine the resulting−15with the 16:

5 2 −6 −23 16 −5

↓ +10 +20 −15

2.7. SYNTHETIC DIVISION 123 In the last step, multiply the 1 times the 5 and combine the result with the−5in the problem to get zero:

5 2 −6 −23 16 −5

↓ +10 +20 −15 5

2 4 −3 1 0

This last coefficient represents the remainder - in this case 0. The other numerals
in the answer represent the coefficients for the powers ofxin the answer. On the
far right is the remainder, then the constant (x0_{) term, then the linear (}_{x}1_{) term}

and so on:
5 2 −6 −23 16 −5
↓ +10 +20 −15 5
2x3 _{4}_{x}2 _{−}_{3}_{x}1 _{1}_{x}0 _{0}
2x4−6x3−23x2+ 16x−5
x−5 = 2x
3_{+ 4}_{x}2_{−}_{3}_{x}_{+ 1}

Let’s look at another example:

**Example**

Use Synthetic Division to divide:

3x3_{+ 5}_{x}2 _{−}_{9}_{x}_{+ 9}

x+ 3

Since Synthetic Division is set up to divide by x−a, if we’re dividing by x+ 3

we’ll need to use a−3in the Synthetic Division:

−3 3 5 −9 9 ↓ 3 Then,3∗ −3 = −9: −3 3 5 −9 9 ↓ −9 3 −4

Next,−4∗ −3 = +12:

−3 3 5 −9 9

↓ −9 12 3 −4 3

And this example also has a zero remainder:

−3 3 5 −9 9

↓ −9 12 −9

3 −4 3 0

The answer here is3x2_{−}_{4}_{x}_{+ 3}_{:}

3x3+ 5x2−9x+ 9

x+ 3 = 3x

2_{−}_{4}_{x}_{+ 3}

and

3x3 _{+ 5}_{x}2_{−}_{9}_{x}_{+ 9 = (}_{x}_{+ 3)(3}_{x}2_{−}_{4}_{x}_{+ 3)}

Let’s look at an example that is a little bit different.

**Example**

Use Synthetic Division to divide:

6x4_{+}_{x}3_{+ 9}_{x}2_{+}_{x}_{−}_{2}
2x+ 1

Synthetic Division is set up to handle problems in which we are dividing by

1x−a. Clearly, this is not the case in this example, however we can work around this. Another way of looking at setting up the synthetic division is that we use the number that is the solution tox−a= 0. When we divided byx−5, we used

+5. When we were dividing byx+ 3, we used−3. So, if we’re going to divide by

2x+ 1, we’ll use−1

2 in the Synthetic Division:

−1

2 6 1 9 1 −2

↓

2.7. SYNTHETIC DIVISION 125 Then, we proceed as usual:

−1

2 6 1 9 1 −2

↓ −3 1 −5 2 6 −2 10 −4 0

Notice that, again, we have a zero remainder. Also, notice that each coefficient in our answer has a common factor of 2, which was the coefficient of the x in

2x+ 1, which we orginally were going to divide by. What we’ve done here is not
division by2x+ 1, but division byx+ 1_{2}.

So, in the end, our work tells us that:

6x4+x3+ 9x2+x−2
x+ 1_{2} = 6x
3_{−}_{2}_{x}2_{+ 10}_{x}_{−}_{4}
and
6x4+x3+ 9x2+x−2 =
x+ 1
2
(6x3−2x2+ 10x−4)

Notice that if we factor out the common factor of 2 from our answer, we can
multiply it back into thex+1_{2} and get an answer for our original problem:

6x4+x3+ 9x2+x−2 = x+1 2 2(3x3−x2+ 5x−2) = (2x+ 1)(3x3−x2+ 5x−2)

This means that:

6x4+x3+ 9x2+x−2 2x+ 1 = 3x

3_{−}

Another thing to understand about Synthetic Division is that if there is a missing power ofx, then you should include a zero as the coefficient of that power.

**Example**

Use Synthetic Division to divide: x3 + 4x−6

x−2

Since there is nox2 _{term in the polynomial we’re divding into, we’ll enter a zero}

as the coefficient for that term:

2 1 0 4 −6

↓

1

And then proceed as usual:

2 1 0 4 −6

↓ 2 4 16 1 2 8 10

2.7. SYNTHETIC DIVISION 127

**Exercises 2.7**

Use synthetic division to find the quotient in each problem.

1) x
3_{−}_{8}_{x}2_{+ 5}_{x}_{+ 50}
x−5 2)
x3+ 5x2−x−14
x+ 2
3) x
3_{+ 12}_{x}2_{+ 34}_{x}_{−}_{7}
x+ 7 4)
x3_{−}_{10}_{x}2_{+ 23}_{x}_{−}_{6}
x−3
5) x
4_{−}_{15}_{x}2_{+ 10}_{x}_{+ 24}
x+ 4 6)
x4_{−}_{3}_{x}3_{+ 4}_{x}2_{−}_{36}
x−3
7) x
4_{−}_{2}_{x}3_{−}_{x}_{+ 10}
x−2 8)
x4_{−}_{16}_{x}2_{−}_{5}_{x}_{−}_{24}
x+ 4
9) 2x
4_{−}_{x}3_{+ 2}_{x}_{−}_{1}
2x−1 10)
3x4_{+}_{x}3_{−}_{3}_{x}_{+ 1}
3x+ 1
11) 3x
4_{−}_{8}_{x}3_{+ 9}_{x}2_{−}_{2}_{x}_{−}_{2}
3x+ 1 12)
6x4_{−}_{7}_{x}3_{+ 5}_{x}2_{−}_{17}_{x}_{+ 10}
3x−2
13) 2x
3_{+ 7}_{x}2_{+ 6}_{x}_{−}_{5}
2x−1 14)
3x4−x3−21x2−11x+ 6
3x−1