College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson
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Transcript of College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson
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College AlgebraSixth EditionJames Stewart Lothar Redlin Saleem Watson
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Polynomial and Rational Functions3
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Dividing Polynomials3.3
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Long Division of Polynomials
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Long Division of Polynomials
Dividing polynomials is much like the
familiar process of dividing numbers.
• When we divide 38 by 7, the quotient is 5 and the remainder is 3.
• We write:
• To divide polynomials, we use long division, as follows.
38 35
7 7
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Division Algorithm
If P(x) and D(x) are polynomials, with
D(x) ≠ 0, then there exist unique polynomials
Q(x) and R(x), where R(x) is either 0 or of
degree less than the degree of D(x),
such that:
or P(x) = D(x) . Q(x) + R(x)
• The polynomials P(x) and D(x) are called the dividend and divisor, respectively.
• Q(x) is the quotient and R(x) is the remainder.
( ) ( )
( )( ) ( )
P x R xQ x
D x D x
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E.g. 1—Long Division of Polynomials
Divide 6x2 – 26x + 12 by x – 4.
• The dividend is 6x2 – 26x + 12 and the divisor is x – 4.
• We begin by arranging them as follows:
24 6 26 12x x x
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E.g. 1—Long Division of Polynomials
Next, we divide the leading term in the
dividend by the leading term in the divisor to
get the first term of the quotient:
6x2/x = 6x
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E.g. 1—Long Division of Polynomials
Then, we multiply the divisor by 6x and
subtract the result from the dividend.
2
2
6 26 16
4
6 2
2
2
4
12
xx
x
x
x
x x
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E.g. 1—Long Division of Polynomials
We repeat the process using the last
line –2x + 12 as the dividend.
2
2
2
6 26 12
6 24
2 12
2
84
64
2
xx
x
x x
x x
x
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E.g. 1—Long Division of Polynomials
The division process ends when the last
line is of lesser degree than the divisor.
• The last line then contains the remainder.
• The top line contains the quotient.
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E.g. 1—Long Division of Polynomials
The result of the division can be
interpreted in either of two ways:
•
•
26 26 12 46 2
4 4
x xx
x x
26 26 12 ( 4)(6 2) 4x x x x
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E.g. 2—Long Division of Polynomials
Let
P(x) = 8x4 + 6x2 – 3x + 1
and D(x) = 2x2 – x + 2
Find polynomials Q(x) and R(x)
such that:
P(x) = D(x) · Q(x) + R(x)
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E.g. 2—Long Division of Polynomials
We use long division after first inserting
the term 0x3 into the dividend to ensure that
the columns line up correctly.
• The process is complete at this point as –7x + 1 is of lesser degree than the divisor 2x2 – x + 2.
2
4 3 2
4
3
2 3
3 2
2
2
4 2
4
2 2 8 0 6 3 1
4 2
2 4
8
3
7
4 8
1
x
x x x
x x x x x
x
x x x
x
x x x
x
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E.g. 2—Long Division of Polynomials
From the long division, we see that:
Q(x) = 4x2 + 2x and R(x) = –7x + 1
• Thus,
8x4 + 6x2 – 3x + 1 = (2x2 – x + 2)(4x2 + 2x) + (–7x
+ 1)
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Synthetic Division
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Synthetic Division
Synthetic division is a quick method
of dividing polynomials.
• It can be used when the divisor is of the form x – c.
• In synthetic division, we write only the essential parts of the long division.
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Long Division vs. Synthetic Division
Compare the following long and synthetic
divisions, in which we divide 2x3 – 7x2 + 5
by x – 3.
• We’ll explain how to perform the synthetic division in Example 3.
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Long Division vs. Synthetic Division
2
3 2
3 2
2
2
2 3
3 2 7 0 5
2 6
0
3
3 5
3 9
4
x x
x x x x
x x
x x
x x
x
x
RemainderQuotient
3 2 7 0 5
6 3 9
2 1 43
Long Division Synthetic Division
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Synthetic Division
Note that, in synthetic division, we:
• Abbreviate 2x3 – 7x2 + 5 by writing only the coefficients: 2, –7, 0, and 5.
• Simply write 3 instead of x – 3.
(Writing 3 instead of –3 allows us to add instead of subtract. However, this changes the sign of all the numbers that appear in the boxes.)
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E.g. 3—Synthetic Division
Use synthetic division to divide
2x3 – 7x2 + 5 by x – 3.
• We begin by writing the appropriate coefficients to represent the divisor and the dividend.
3 2 7 0 5
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E.g. 3—Synthetic Division
We bring down the 2, multiply 3 · 2 = 6,
and write the result in the middle row.
Then, we add:
3 2 7 0 5
6
2 1
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E.g. 3—Synthetic Division
We repeat this process of
multiplying and then adding until
the table is complete.
3 2 7 0 5
6 3
2 1 3
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E.g. 3—Synthetic Division
• From the last line of the synthetic division we see that the quotient is 2x2 – x – 3 and the remainder is –4.
• Thus, 2x3 – 7x2 + 5 = (x – 3)(2x2 – x – 3) – 4.
3 2 7 0 5
6 3 9
2 1 3 4
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The Remainder and Factor Theorems
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Remainder Theorem
The Remainder Theorem shows how
synthetic division can be used to evaluate
polynomials easily.
If the polynomial P(x) is divided by x – c,
then the remainder is the value P(c).
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Remainder Theorem—Proof
If the divisor in the Division Algorithm is
of the form x – c for some real number c,
then the remainder must be a constant.
• This is because the degree of the remainder is less than the degree of the divisor.
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Remainder Theorem—Proof
If we call this constant r, then
P(x) = (x – c) · Q(x) + r
• Replacing x by c in this equation, we get: P(c) = (c – c) · Q(x) + r = 0 + r = r
• That is, P(c) is the remainder r.
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E.g. 4—Using the Remainder Theorem to Find the Value of a Polynomial
Let
P(x) = 3x5 + 5x4 – 4x3 + 7x + 3
(a) Find the quotient and remainder when P(x) is divided by x + 2.
(b) Use the Remainder Theorem to find P(–2).
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E.g. 4—Using Remainder Theorem
As x + 2 = x – (–2) , the synthetic division
for this problem takes the following form.
• The quotient is 3x4 – x3 – 2x2 + 4x – 1.• The remainder is 5.
Example (a)
2 3 5 4 0 7 3
6 2 4 8 2
3 1 2 4 1 5
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E.g. 4—Using Remainder Theorem
By the Remainder Theorem,
P(–2) is the remainder when P(x)
is divided by x – (–2) = x + 2.
• From part (a), the remainder is 5.
• Hence, P(–2) = 5.
Example (b)
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Factor Theorem
The Factor Theorem says that zeros
of polynomials correspond to factors.• We used this fact in Section 3.2 to graph
polynomials.
c is a zero of P if and only if x – c
is a factor of P(x).
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Factor Theorem—Proof
If P(x) factors as P(x) = (x – c) · Q(x),
then
P(c) = (c – c) · Q(c)
= 0 · Q(c)
= 0
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Factor Theorem—Proof
Conversely, if P(c) = 0, then, by
the Remainder Theorem,
P(x) = (x – c) · Q(x) + 0
= (x – c) · Q(x)
• So, x – c is a factor of P(x).
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E.g. 5—Factoring a Polynomial Using the Factor Theorem
Let
P(x) = x3 – 7x + 6
• Show that P(1) = 0.
• Use this fact to factor P(x) completely.
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Substituting, we see that:
P(1) = 13 – 7 · 1 + 6
= 0
• By the Factor Theorem, this means that x – 1 is a factor of P(x).
E.g. 5—Factoring a Polynomial Using the Factor Theorem
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Using synthetic or long division,
we see that:
P(x) = x3 – 7x + 6
= (x – 1)(x2 + x – 6)
= (x – 1)(x – 2)(x + 3)
E.g. 5—Factoring a Polynomial Using the Factor Theorem
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E.g. 6—Finding a Polynomial with Specified Zeros
Find a polynomial of degree 4 that has
zeros –3, 0, 1, and 5.
• By the Factor Theorem, x – (–3), x – 0, x – 1, and x – 5
must all be factors of the desired polynomial.
• So, let: P(x) = (x + 3)(x – 0) (x – 1)(x – 5)
= x4 – 3x3 – 13x2 + 15x
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E.g. 6—Finding a Polynomial with Specified Zeros
Since P(x) is of degree 4, it is a solution
of the problem.
• Any other solution of the problem must be a constant multiple of P(x).
• This is because only multiplication by a constant does not change the degree.
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The polynomial P of Example 6 is
graphed here.
• Note that the zeros of P correspond to the x-intercepts of the graph.
Finding a Polynomial with Specified Zeros