LESSON X - Mathematics & Statistics
Pre-Class Problems 15 for Friday, October 13
These are the type of problems that you will be working on in class.
You can go to the solution for each problem by clicking on the problem number or letter.
Discussion on Long Division and Synthetic Division.
1. Use synthetic division to divide the following polynomials.
a. [pic]
b. [pic]
c. [pic]
d. [pic]
e. [pic]
f. [pic]
g. [pic] h. [pic]
i. [pic] j. [pic] k. [pic]
Discussion of the Remainder Theorem.
2. If [pic], then use the Remainder Theorem to find the following polynomial values.
a. [pic] b. [pic] c. [pic] d. [pic]
3. If [pic], then use the Remainder Theorem to find the following polynomial values.
a. [pic] b. [pic] c. [pic] d. [pic]
4. If [pic], then use the Remainder Theorem to determine if the given number is a zero (root) of the polynomial.
a. 3 b. [pic] c. 6 d. [pic]
5. If [pic], then use the Remainder Theorem to determine if the given number is a zero (root) of the polynomial.
a. 4 b. [pic] c. [pic] d. [pic]
Discussion of the Factor Theorem.
6. If [pic], then use the Factor Theorem to determine if the given binomial is a factor of the polynomial.
a. [pic] b. [pic] c. [pic] d. [pic]
7. Find a polynomial p that has the given zeros (roots) and multiplicity.
a. Degree 2 with zeros (roots) [pic] and [pic] each of multiplicity 1.
b. Degree 3 with zeros (roots) [pic], [pic], and 5 each of multiplicity 1.
c. Degree 3 with zeros (roots) 2 of multiplicity 1 and 4 of multiplicity 2.
d. Degree 4 with zeros (roots) [pic], 6, [pic], and [pic] each of multiplicity 1.
e. Degree 4 with zeros (roots) [pic] of multiplicity 2 and [pic] and [pic] each of multiplicity 1.
f. Degree 14 with zeros (roots) [pic] of multiplicity 5, [pic] of multiplicity 3, [pic] of multiplicity 4, and 6 of multiplicity 2.
g. Degree 4 with zeros (roots) [pic], [pic], [pic], and [pic] each of multiplicity 1.
h. Degree 8 with zeros (roots) [pic], 3, [pic], and [pic] each of multiplicity 2.
Problems available in the textbook: Page 325 … 23 – 82 and Examples 4 – 10 starting on page 303.
SOLUTIONS:
1a. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1b. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1c. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1d. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1e. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1f. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1g. [pic] Back to Problem 1.
[pic]
[pic] [pic]
Answer: [pic]
1h. [pic] Back to Problem 1.
[pic]
[pic] [pic]
Answer: [pic]
1i. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
NOTE: [pic]
This is the factorization of the difference of cubes [pic]. If you can remember that [pic] is one of the factors of [pic], then you can get the other factor of [pic] by synthetic division.
1j. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
1k. [pic] Back to Problem 1.
[pic] [pic]
Answer: [pic]
NOTE: [pic]
This is the factorization of the sum of cubes [pic]. If you can remember that [pic] is one of the factors of [pic], then you can get the other factor of [pic] by synthetic division.
2a. [pic] Back to Problem 2.
Find [pic].
[pic] [pic]
Answer: [pic]
2b. [pic] Back to Problem 2.
Find [pic].
[pic] [pic]
Answer: 751
2c. [pic] Back to Problem 2.
Find [pic].
[pic] [pic]
NOTE: In order to find [pic], I think it easier to do the evaluation of the function:
[pic]
[pic]
NOTE: [pic] Thus, [pic]
Answer: [pic]
2d. [pic] Back to Problem 2.
Find [pic].
[pic] [pic]
Answer: [pic]
3a. [pic] Back to Problem 3.
Find [pic].
[pic] [pic]
Answer: 2345
3b. [pic] Back to Problem 3.
Find [pic].
[pic] [pic]
Answer: 4366
3c. [pic] Back to Problem 3.
Find [pic].
[pic] [pic]
NOTE: In order to find [pic], I think it easier to do the evaluation of the function:
[pic]
[pic]
NOTE: [pic]
NOTE: [pic] Thus, [pic]
Answer: [pic]
3d. [pic] Back to Problem 3.
Find [pic].
[pic] [pic]
Answer: [pic]
4a. [pic] Back to Problem 4.
[pic] [pic]
[pic]
Answer: 3 is not a zero (root) of the polynomial
4b. [pic] Back to Problem 4.
[pic] [pic]
[pic]
Answer: [pic] is a zero (root) of the polynomial
4c. [pic] Back to Problem 4.
[pic] [pic]
[pic]
Answer: 6 is a zero (root) of the polynomial
4d. [pic] Back to Problem 4.
[pic] [pic]
[pic]
Answer: [pic] is not a zero (root) of the polynomial
5a. [pic] Back to Problem 5.
[pic] [pic]
[pic]
Answer: 4 is not a zero (root) of the polynomial
5b. [pic] Back to Problem 5.
[pic] [pic]
[pic]
Answer: [pic] is a zero (root) of the polynomial
5c. [pic] Back to Problem 5.
[pic] [pic]
[pic]
NOTE: In order to find [pic], I think it easier to do the evaluation of the function:
[pic]
[pic]
Answer: [pic] is a zero (root) of the polynomial
5d. [pic] Back to Problem 5.
[pic] [pic]
[pic]
Answer: [pic] is not a zero (root) of the polynomial
6a. [pic] Back to Problem 6.
[pic] is a factor of the polynomial f if and only if [pic].
[pic] [pic]
[pic]
Answer: [pic] is not a factor of the polynomial
6b. [pic] Back to Problem 6.
[pic] is a factor of the polynomial f if and only if [pic].
[pic] [pic]
[pic]
Answer: [pic] is a factor of the polynomial
6c. [pic] Back to Problem 6.
[pic] is a factor of the polynomial f if and only if [pic].
[pic] [pic]
[pic]
Answer: [pic] is not a factor of the polynomial
6d. [pic] Back to Problem 6.
[pic] is a factor of the polynomial f if and only if [pic].
[pic] [pic]
[pic]
Answer: [pic] is a factor of the polynomial
7a. Zero (Root) Multiplicity Back to Problem 7.
[pic] 1
[pic] 1
In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Since [pic] = [pic] = [pic], then [pic]
Answer: [pic], where a is any nonzero real
number
7b. Zero (Root) Multiplicity Back to Problem 7.
[pic] 1
[pic] 1
5 1
In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for 5 to be a zero (root) of multiplicity 1, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Using the special product formula [pic], we have that [pic] = [pic].
Thus, [pic] = [pic] =
[pic] = [pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
7c. Zero (Root) Multiplicity Back to Problem 7.
2 1
4 2
In order for 2 to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for 4 to be a zero (root) of multiplicity 2, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Using the special product formula [pic], we have that [pic].
Thus, [pic] = [pic] =
[pic] = [pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
7d. Zero (Root) Multiplicity Back to Problem 7.
[pic] 1
6 1
[pic] 1
[pic] 1
In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for 6 to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
We will use the special product formula [pic] for [pic]. Thus, [pic]. Since [pic], then [pic]. Thus, [pic].
Since [pic], then
[pic] = [pic] =
[pic] = [pic] =
[pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
7e. Zero (Root) Multiplicity Back to Problem 7.
[pic] 2
[pic] 1
[pic] 1
In order for [pic] to be a zero (root) of multiplicity 2, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Since [pic] = [pic], then using the special product formula [pic], we have that [pic] = [pic]. Using the special product formula [pic], we have that [pic]. Thus, [pic] = [pic]
= [pic].
Using the special product formula [pic], we have that [pic].
Thus, [pic] =
[pic] = [pic]
[pic] =
[pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
7f. Zero (Root) Multiplicity Back to Problem 7.
[pic] 5
[pic] 3
[pic] 4
6 2
In order for [pic] to be a zero (root) of multiplicity 5, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 3, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 4, [pic] must be a factor of p. In order for 6 to be a zero (root) of multiplicity 2, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Answer: [pic], where a is any nonzero real number
7g. Zero (Root) Multiplicity Back to Problem 7.
[pic] 1
[pic] 1
[pic] 1
[pic] 1
In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 1, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
We will use the special product formula [pic] for [pic]. Thus, [pic].
We will use the special product formula [pic] for [pic]. Thus, [pic]. Thus, [pic].
Thus, [pic] = [pic] =
[pic] = [pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
7h. Zero (Root) Multiplicity Back to Problem 7.
[pic] 2
3 2
[pic] 2
[pic] 2
In order for [pic] to be a zero (root) of multiplicity 2, [pic] must be a factor of p. In order for 3 to be a zero (root) of multiplicity 2, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 2, [pic] must be a factor of p. In order for [pic] to be a zero (root) of multiplicity 2, [pic] must be a factor of p.
Thus, [pic], where a is any nonzero real number.
Since [pic], then [pic] = [pic] =
[pic] = [pic] and [pic] =
[pic] = [pic] = [pic]
Thus, [pic] =
[pic] = [pic]
[pic] = [pic]
Thus, [pic]
Answer: [pic], where a is any nonzero real number
Example Find [pic].
[pic]
NOTE: [pic]
[pic]
[pic]
The expression [pic] is called the divisor in the division. The function [pic] is called the divisor function.
The expression [pic] is called the dividend in the division. The function [pic] is called the dividend function.
The expression [pic] is called the quotient in the division. The function [pic] is called the quotient function.
The expression [pic] is called the remainder in the division. The function [pic] is called the remainder function.
We have that
[pic]
Multiplying both sides of this equation by [pic], we have that
[pic] =
[pic]
Let a and b be polynomials. Then [pic]. The degree of the remainder polynomial r is less than the degree of divisor polynomial b, written deg r < deg b.
Multiplying both sides of the equation [pic] by [pic], we have that [pic].
Example Find [pic].
[pic]
NOTE: [pic]
[pic]
[pic]
[pic]
The quotient function is [pic] and the remainder function is [pic]. We have that
[pic].
Example Find [pic].
[pic]
The quotient function is [pic] and the remainder function is [pic]. We have that
[pic].
NOTE: In the example above, we had that
[pic]. Thus,
[pic] =
[pic] =
[pic]
Example Find [pic].
[pic]
The quotient function is [pic] and the remainder function is [pic]. We have that
[pic].
Consider the following.
[pic] [pic]
What do the numbers in the third row represent?
[pic] [pic]
Thus, the quotient function is [pic] and the remainder function is [pic]. These are the same answers that we obtained above using long division.
This process is called synthetic division. Synthetic division can only be used to divide a polynomial by another polynomial of degree one with a leading coefficient of one. Thus, you can NOT use synthetic division to find
[pic]. However, we can do the following division.
Example Use synthetic division to find [pic].
[pic] [pic]
Thus, the quotient function is [pic] and the
remainder function is [pic]. These are the same answers that we obtained above using long division.
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Example If [pic], then find [pic].
[pic] =
[pic] =
[pic]
This calculation would have been faster (and easier) using the fact that
[pic]
that we obtained in the example above. Thus, [pic], where [pic].
Thus, [pic].
This result can be explained by the following theorem.
Theorem (The Remainder Theorem) Let p be a polynomial. If [pic] is divided by [pic], then the remainder is [pic].
Proof If [pic] is divided by [pic], then [pic]. Thus, [pic].
Example If [pic], then find [pic].
Using synthetic division to find [pic], we have that
[pic] [pic]
Thus, the remainder is [pic]. Thus, [pic] = [pic].
Example If [pic], then find [pic].
Using synthetic division to find [pic], we have that
[pic] [pic]
Thus, the remainder is 41. Thus, [pic].
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Theorem (The Factor Theorem) Let p be a polynomial. The expression [pic] is a factor of [pic] if and only if [pic].
Proof [pic] Suppose that [pic] is a factor of [pic]. Then the remainder upon division by [pic] must be zero. By the Remainder Theorem, [pic].
[pic] Suppose that [pic]. By the Remainder Theorem, we have that
[pic]. Thus, [pic]. Thus, [pic] is a factor of [pic].
Example Show that [pic] is a factor of [pic].
We will use the Factor Theorem and show that [pic]. We will use the Remainder Theorem and synthetic division to find [pic].
[pic] [pic]
Thus, [pic]. Thus, by the Factor Theorem, [pic] is a factor of
[pic].
NOTE: The third row in the synthetic division gives us the coefficients of the other factor starting with [pic]. Thus, the other factor is [pic].
Thus, we have that [pic] = [pic].
Example Show that [pic] is not a factor of [pic].
We will use the Factor Theorem and show that [pic]. We will use the Remainder Theorem and synthetic division to find [pic].
[pic] [pic]
Thus, [pic]. Thus, by the Factor Theorem, [pic] is not a factor of
[pic].
Example Find the value(s) of c so that [pic] is a factor of
[pic].
By the Factor Theorem, [pic] is a factor of the polynomial f if and only if [pic].
[pic]
Thus, [pic].
Using the Remainder Theorem and synthetic division to find [pic], we have
[pic] [pic]
By the Remainder Theorem, [pic]. Thus, [pic]
[pic].
Answer: [pic]
Example Find the value(s) of c so that [pic] is a factor of
[pic].
By the Factor Theorem, [pic] is a factor of the polynomial g if and only if [pic].
[pic]
Thus, [pic].
Using the Remainder Theorem and synthetic division to find [pic], we have
[pic] [pic]
By the Remainder Theorem, [pic]. Thus, [pic]
[pic].
Answer: 8
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