Name
September, 2013
Honors Math 3
Challenge Problem Set 1 page 1
In this course, a Challenge Problem Set is an extended homework assignment about an advanced or enrichment topic. Unless otherwise directed, students are encouraged to collaborate on these problem sets.
In this Challenge Problem Set, you will learn about an alternate calculation method for dividing a polynomial function by a linear function, called synthetic division . You are free to work with other people on the assignment. Your learning will be assessed by one or more problems on a quiz or test sometime during Chapter 1.
Synthetic division is a shorthand method of polynomial division in the special case of dividing by a linear factor of the form ( x – a ). It uses detached coefficients to cut down on the amount of writing compared to long division (the division method taught in class and in section 1.05).
Here is an example of synthetic division where
4 x 3  3 x 2  5 x  8
is divided by x  2
.
The value a = 2 is written at the far left. The coefficients of the polynomial being divided are listed across the top. The first step is to bring down the leading coefficient as shown below.
  a value

                      
2 4
 bring down

4
–3
5
–8
The calculation is done by repeated multiplication and addition. First multiply the 4 by the value a = 2, to get 8. Place the 8 under the next coefficient and add to get 5. Repeat this process:
5

2 = 10, 10 + 5 = 15, and so on. Keep repeating until you reach the end.
2 4
4
–3
8
5
5
10
15
–8
30
22
                 coefficien ts of quotient
remainder
The resulting list of numbers at the bottom represents the coefficients of the quotient and the remainder. Here, we can see that 4 x 3  3 x 2  5 x  8 divided by x  2 is 4 x 2  5 x  15 with a remainder of 22.
  


Name
September, 2013
Honors Math 3
Challenge Problem Set 1 page 2
Here is another example: x 4  x 3  3 x  2 divided by x  2
. Notice the zero coefficient for x
2 in the row of coefficients, and note that the a value is –2.
–2

1
1
1
–2
–1

0
2
2
3
–4
–1
–2
2
0
The result is x 4  x 3 x 3
 3 x  2
 x 2
.
 2 x  1
with a remainder of 0, which means that x + 2 is a factor of
1.
Divide 2 x
3
– 7 x
2
– 8 x – 34 by x – 5 using each of two methods: long division (as in section
1.05) and synthetic division (as taught on this sheet). Make sure your answers agree. Also analyze where the same numbers are present in both calculations, and compare their locations.
2.
Here is a synthetic division calculation. Write the long division calculation for the same division problem, making use of your observations from problem 1 so that you don’t have to redo the entire calculation.
–3
5
5
–17
–15
–32
83
96
179
–12
–537
–549
473
1647
2120
3.
Using synthetic division, divide x
3
– 4 x
2
+ 7 by 2 x – 3. Hint: Since the divisor is not of the form x – a , you will need to find a way to adapt the method to fit this situation.
4.
Using synthetic division, divide 2 x 4 – 7 x 3 – 10 x 2 + 53 x – 42 by x 2 – 5 x + 6. Hint: Since synthetic division is only for dividing by linear functions, you are going to need to do two synthetic divisions to divide by a quadratic.
5.
Use synthetic division to help you solve this problem: Find the values of a for which
(1 – a ) x
2
+ ( a + 1) x – 2 is divisible by ( x – a ).