Polynomial and Synthetic Division

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Polynomial and Synthetic Division

• Pre-Calculus
• Mrs.Volynskaya

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Polynomial Division

• Polynomial Division is very similar to long
  division.
• Example

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Polynomial Division
Subtract!!
Subtract!!
Subtract!!
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Polynomial Division

• Example
• Notice that there is no x term. However, we need
  to include it when we divide.

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Polynomial Division
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Try This

• Example
• Answer

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What does it mean if a number divides evenly into
another??
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Practice Problems

• 2.3 Page 140 8,14

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Now lets look at another method to divide

• Why???
• Sometimes it is easier

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Synthetic Division

• Synthetic Division is a shortcut for polynomial
  division that only works when dividing by a
  linear factor (x b).
• It involves the coefficients of the dividend, and
  the zero of the divisor.

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• (SUSPENSE IS BUILDING)

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Example

• Divide
• Step 1
• Write the coefficients of the dividend in a
  upside-down division symbol.

1
5
6
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Example

• Step 2
• Take the zero of the divisor, and write it on the
  left.
• The divisor is x 1, so the zero is 1.

1
5
6
1
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Example

• Step 3
• Carry down the first coefficient.

1
5
6
1
1
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Example

• Step 4
• Multiply the zero by this number. Write the
  product under the next coefficient.

1
5
6
1
1
1
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Example

• Step 5
• Add.

1
5
6
1
1
1
6
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Example

• Step etc.
• Repeat as necessary

1
5
6
1
1
6
12
1
6
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Example

• The numbers at the bottom represent the
  coefficients of the answer. The new polynomial
  will be one degree less than the original.

1
5
6
1
1
6
12
1
6
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Synthetic Division

• The pattern for synthetic division of a cubic
  polynomial is summarized as follows. (The pattern
  for higher-degree polynomials is similar.)

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Synthetic Division

• This algorithm for synthetic division works only
  for divisors of the form x k.
• Remember that x k x (k).

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Using Synthetic Division

• Use synthetic division to divide x4 10x2 2x
  4 by x 3.
• Solution
• You should set up the array as follows. Note that
  a zero is included for the missing x3-term in the
  dividend.

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Example Solution
contd

• Then, use the synthetic division pattern by
  adding terms in columns and multiplying the
  results by 3.
• So, you have
• 
  .

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Try These

• Examples
• (x4 x3 11x2 5x 30) ? (x 2)
• (x4 1) ? (x 1)Dont forget to include the
  missing terms!
• Answers
• x3 3x2 5x 15
• x3 x2 x 1

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Practice Problems

• 2.3 Page 140-141 20, 24, 50, 58

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Application of Long Division

• To begin, suppose you are given the graph of
• f (x) 6x3 19x2 16x 4.

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Long Division of Polynomials

• Notice that a zero of f occurs at x 2.
• Because x 2 is a zero of f,you know that (x
  2) isa factor of f (x). This means thatthere
  exists a second-degree polynomial q (x) such
  that
• f (x) (x 2) ? q(x).
• To find q(x), you can uselong division.

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Example - Long Division of Polynomials

• Divide 6x3 19x2 16x 4 by x 2, and use the
  result to factor the polynomial completely.

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Example 1 Solution
Multiply 6x2(x 2).
Subtract.
Multiply 7x(x 2).
Subtract.
Multiply 2(x 2).
Subtract.
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Example Solution
contd

• From this division, you can conclude that
• 6x3 19x2 16x 4 (x 2)(6x2 7x 2)
• and by factoring the quadratic 6x2 7x 2, you
  have
• 6x3 19x2 16x 4 (x 2)(2x 1)(3x
  2).

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Example 1 Solution
contd

• Note that this factorization agrees with the
  graph shown in Figure 2.28 in that the three
  x-intercepts occur atx 2, x , and x .

Figure 2.28
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Long Division of Polynomials

• In Example 1, x 2 is a factor of the
  polynomial6x3 19x2 16x 4, and the long
  division process produces a remainder of zero.
  Often, long division will produce a nonzero
  remainder.
• For instance, if you divide x2 3x 5 by x 1,
  you obtain the following.

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Long Division of Polynomials

• In fractional form, you can write this result as
  follows.
• This implies that
• x2 3x 5 (x 1)(x 2) 3
• which illustrates the following theorem, called
  the Division Algorithm.

Multiply each side by (x 1).
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Long Division of Polynomials
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The Remainder and Factor Theorems

• The remainder obtained in the synthetic division
  process has an important interpretation, as
  described in the Remainder Theorem.
• The Remainder Theorem tells you that synthetic
  division can be used to evaluate a polynomial
  function. That is, to evaluate a polynomial
  function f (x) when x k, divide f (x) by x k.
  The remainder will bef (k).

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Example Using the Remainder Theorem

• Use the Remainder Theorem to evaluate the
  following function at x 2.
• f (x) 3x3 8x2 5x 7
• Solution
• Using synthetic division, you obtain the
  following.

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Example Solution
contd

• Because the remainder is r 9, you can conclude
  that
• f (2) 9.
• This means that (2, 9) is a point on the graph
  of f. You can check this by substituting x 2
  in the original function.
• Check
• f (2) 3(2)3 8(2)2 5(2) 7
• 3(8) 8(4) 10 7
• 9

r f (k)
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The Remainder and Factor Theorems

• Another important theorem is the Factor Theorem,
  stated below.
• This theorem states that you can test to see
  whether a polynomial has (x k) as a factor by
  evaluating the polynomial at x k.
• If the result is 0, (x k) is a factor.

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Example Factoring a Polynomial Repeated
Division

• Show that (x 2) and (x 3) are factors of
• f (x) 2x4 7x3 4x2 27x 18.
• Then find the remaining factors of f (x).
• SolutionUsing synthetic division with the
  factor (x 2), you obtain the following.

0 remainder, so f (2) 0 and (x 2) is a factor.
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Example Solution
contd

• Take the result of this division and perform
  synthetic division again using the factor (x
  3).
• Because the resulting quadratic expression
  factors as
• 2x2 5x 3 (2x 3)(x 1)
• the complete factorization of f (x) is
• f (x) (x 2)(x 3)(2x 3)(x 1).

0 remainder, so f (3) 0 and (x 3) is a
factor.
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The Remainder and Factor Theorems

• For instance, if you find that x k divides
  evenly into f (x) (with no remainder), try
  sketching the graph of f.
• You should find that (k, 0) is an x-intercept of
  the graph.