Download 4-1B: Lecture Synthetic Division

Section 4 – 1B:
Synthetic Division
3x 3 + x 2 − 4 x − 13
= means divide
x−2
There are 2 ways to perform this division. Long Division or Synthetic Division
Long division requires you divide the binomial into the polynomial.
3x 3 + x 2 − 4 x − 13
= means divide x − 2 into 3x 3 + x 2 − 4 x − 13
x−2
Synthetic Division is a shortcut method for long division. It requires you divide the the
root of the binomial into the polynomial. If the denominator is x – a divide by a. If the
denominator is x + a divide by – a
Example 1
Example 2
3x 3 + x 2 − 4 x − 13
=
x−2
5x 3 + 3x 2 + x + 3
=
x +7
You could do "long division"
You could do "long division"
x − 2 3x 3 + x 2 − 4 x − 13
x + 7 5x 3 + 3x 2 + x + 3
or you could use use
synthetic division
and divide by 2
or you could use use
synthetic division
and divide by − 7
2| 3 1 −4
−7 | 5
)
− 13
)
↓
Section 4 – 1B Lecture
3 1 3
↓
Page 1
© 2015 Eitel
Synthetic Division
Example 1
Use Synthetic Division to divide
3x 3 + x 2 − 4x − 13
=
x−2
Step 1: If the denomitor is x − 2 divide by 2
Write the 2 and put the lines to the right of it and under it
Write the coefficient of each variable term but do not write the variable
Leave some space below these numbers and draw a line under the coefficients
2 | 3 1 − 4 − 13
Step 2: Bring the first coefficient (the 3) stright down below the line
2 | 3 1 − 4 − 13
↓
3
Step 3: Mutilply the 2 in the box by the 3 to get 6.
Put the 6 directly under the second coefficient (the 1) and above the line
2 | 3 1 − 4 − 13
↓ 6
3
Step 4: Combine the 1 and 6 to get 7.
Put the 7 below the line directly under the 1 and 6.
2 | 3 1 − 4 − 13
↓ 6
3 7
Step 5: Mutilply the 2 in the box by the 7 to get 14.
Put the 14 directly under third coefficient (the − 4) and above the line
2 | 3 1 − 4 − 13
↓ 6 14
3 7
Section 4 – 1B Lecture
Page 2
© 2015 Eitel
Step 6: Combine the − 4 and 14 to get 10.
Put the 10 below the line directly under the − 4 and 14
2 | 3 1 − 4 − 13
↓ 6 14
3 7 10
Step 7: Mutilply the 2 in the box by the 10 to get 20.
Put the 20 directly under forth coefficient (the − 13) and above the line
2 | 3 1 − 4 − 13
↓ 6 14
20
3 7 10
Step 7: Combine the − 13 and 20 to get 7.
Put the 7 below the line directly under the − 13 and 20
2 | 3 1 − 4 − 13
↓ 6 14
20
3 7 10
7
x2 x
#
Remainder
The original numerator in the problem had the highest powered term as x 3 . The polynomial answer
will start with a next lowest power x 2 and then have an x term next and then a constant term. The
coefficients of the first 2 terms are the first 2 numbers in the bottom row. The next number in the
bottom row is the constant term in the answer. The last number in the bottom row is the
remainder. We write the remainder as a fraction over the divisor x + 3
Answer:
3x2 + 7x + 10 +
7
x−2
NOTE: The example above rewrites each step with a new division graph. When you do the
problem all the steps are done on the same division graph. The entire problem with all the
steps included will look like the division below
Single Row Method
2| 3
1
Answer:
−4
− 13
↓
6
14
20
3
7
10
7
3x2 + 7x + 10 +
Section 4 – 1B Lecture
7
x−2
Page 3
© 2015 Eitel
Example 2
Use Synthetic Division to divide
x 3 + 2x 2 − 5x − 10
x +3
Step 1: If the denomitor is x + 3 divide by − 3
Write the − 3 and put the lines to the right of it and under it.
Write the coefficient of each variable term but do not write the variable.
Leave some space below these numbers and draw a line under the coefficients
−3 | 1 2 − 5 10
Step 2: Bring the first coefficient (the 1) stright down below the line
−3 |
1 2 − 5 − 10
↓
1
Step 3: Mutilply the − 3 in the box by the 1 to get − 3.
Put the − 3 directly under the second coefficient (the 2) and above the line
−3 |
1
2 − 5 − 10
↓ −3
1
Step 4: Combine the 2 and − 3 to get − 1.
Put the − 1 below the line directly under the 2 and − 3.
−3 |
1
2 − 5 − 10
↓ −3
1 −1
Step 5: Mutilply the − 3 in the box by the − 1 to get 3.
Put the 3 directly under third coefficient (the − 5) and above the line
−3 |
1
2 − 5 − 10
↓ −3
1 −1
3
Section 4 – 1B Lecture
Page 4
© 2015 Eitel
Step 6: Combine the − 5 and 3 to get − 2.
Put the − 2 below the line directly under the − 5 and 3
−3 | 1
2 − 5 − 10
↓ −3 3
1 −1 − 2
Step 7: Mutilply the − 3 in the box by the − 2 to get 6.
Put the 6 directly under forth coefficient (the − 10) and above the line
−3 | 1
2 − 5 − 10
↓ −3 3
1 −1 − 2
6
Step 7: Combine the − 10 and 6 to get − 4.
Put the − 4 below the line directly under the − 10 and 6
−3 | 1
2 − 5 − 10
↓ −3 3
1 −1 − 2
x2
Answer:
x
#
1x2 − 7x − 2 −
6
−4
remainder
4
x+3
The original numerator in the problem had the highest powered term as x 3 . The polynomial answer
will start with a next lowest power x 2 and then have an x term and then a constant term. The
coefficients of the 2 terms are the first 2 numbers in the bottom row. The next number in the bottom
row is the constant term in the answer. The last number in the bottom row is the remainder.
We write the remainder as a fraction over the divisor x + 3
Single Row Method
−3 | 1 2 − 5 − 10
↓ −3 3
1 −1 − 2
Answer:
1x2 − 7x − 2 −
Section 4 – 1B Lecture
6
−4
4
x+3
Page 5
© 2015 Eitel
Use Synthetic Division to find the answer to the following problems:
Example 1
Example 2
Example 3
3x 3 + x 2 − 4x − 13
=
x−2
x 3 + 2x 2 − 5x − 6
=
x +3
x 3 + 6x 2 + 5x − 12
=
x +4
2 | 3 1 − 4 − 13
↓ 6 14 20
−3 | 1 + 2 − 5 − 6
−4 | 1
3 7 10
↓ −3 3 6
1 −1 − 2 0
7
= 3x 2 + 7x + 10 +
5 − 12
↓ − 4 −8
1 2 −3
= x2 − x − 2
7
x−2
6
12
0
= x 2 + 2x − 3
Note: In examples 2 and 3 the remainder is zero. This means that the denominator
divides into the numerator evenly. When this happens the binomial in the denominator is a
factor of the original polynomial.
Synthetic Division with higher ordered polynomials.
The original numerator in the problems below have the highest powered term as x 4 . The polynomial
answer will start with the next lowest power x 3 and then have an x 2 and then an x term and then
a constant term. The coefficients of the 3 terms are the first 3 numbers in the bottom row. The next
number in the bottom row is the constant term in the answer. The last number in the bottom row
is the remainder. We write the remainder as a fraction over the divisor.
Example 4
Example 5
x 4 + 5x 3 + 8x 2 + 5x − 4
=
x+3
2x 4 − 9x 3 − 5x 2 + 3x − 15
=
x−5
−3 | 1 5
8 5 4
↓ − 3 −6 − 6 3
5 | 2 − 9 − 5 3 −15
↓ 10 5 0 15
1
2
2 −1
= x 3 + 2x 2 + 2x −1 +
Section 4 – 1B Lecture
7
2
1
0
3
0
= 2x 3 + x 2 + 3
7
x+3
Page 6
© 2015 Eitel
Some polynomials have missing powers of x. Each of the missing powers has a ZERO
coefficient. A zero must be included in the terms when you write the polynomial under the
division sign.
Example 6
Example 7
3x 3 + 17
=
x +2
4x 3 − 2x − 2
=
x +1
3x 3 + 17
4x 3 − 2x − 2
is really 3x 3 + 0x 2 + 0x + 17
is really 4x 3 + 0x 2 − 2x − 2
−1 | 4
−2 | 3
0 −2 −2
↓ −4
4 −4
0
17
↓ − 6 12 − 24
3 − 6 12 − 7
4 −2
2
0
= 3x 2 − 6x + 12 −
= 4x 2 − 4 x + 2
Section 4 – 1B Lecture
0
Page 7
7
x +2
© 2015 Eitel
The Remainder Theorem
When a Polynomial P(x) is divided by 1x − a using long division
the reminder of the division is P(a)
Note: This theorem can be used only when the coefficient of the x term is 1
Using Synthetic Division to find P(x)
To find P(x)
1. Use Synthetic Division to divide by x (no sign change).
2. The remainder of the Synthetic Division will be P(x)
Use the Synthetic Division to find P(c) for the given P(x)
Example 2
Example 2
P(x) = 4 x 4 − 5x 3 − 2x 2 − 3x − 4 find P(2)
P(x) = 5x 4 + 9x 3 + 2x 2 + 3x + 7 find P(−1)
2| 4 −5 −2 −3 −4
↓
8
6
8
10
4
3
4
5
6
−1 | 5
9
2 3 5
−5 −4 2 −5
5 4 −2 5 0
the remainder is 0
the remainder is 5
so P(2) = 6
so P(−1) = 0
Check:
P(2) = 4(2)4 − 5(2) 3 − 2(2)2 − 3(2) − 4
= 4(16) − 5(8) − 2(4) − 3(2) − 4
= 64 − 40 − 8 − 6 − 4
=6
Section 4 – 1B Lecture
Check:
P(−1) = 5(−1) 4 + 9(−1) 3 + 2(−1) 2 + 3(−1) + 5
= 5(1) + 9(−1) + 2(1) + 3(−1) + 7
=5−9 +2−3+5
=0
Page 8
© 2015 Eitel