Answered

Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

2. Find [tex]\( p(0), p(1) \)[/tex], and [tex]\( p(2) \)[/tex] for each of the following polynomials:

(i) [tex]\( p(x) = x^2 - x + 1 \)[/tex]

(ii) [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]

(iii) [tex]\( p(z) = z^3 \)[/tex]

(iv) [tex]\( p(t) = (t-1)(t+1) \)[/tex]

(v) [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]

Sagot :

GinnyAnswer
Certainly! Let's go through each of the given polynomials one by one and find [tex]\( p(0) \)[/tex], [tex]\( p(1) \)[/tex], and [tex]\( p(2) \)[/tex].

### (i) [tex]\( p(x) = x^2 - x + 1 \)[/tex]

1. [tex]\( p(0) \)[/tex]:
[tex]\[ p(0) = 0^2 - 0 + 1 = 1 \][/tex]

2. [tex]\( p(1) \)[/tex]:
[tex]\[ p(1) = 1^2 - 1 + 1 = 1 \][/tex]

3. [tex]\( p(2) \)[/tex]:
[tex]\[ p(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \][/tex]

So, the values are [tex]\( p(0) = 1 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].

### (ii) [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]

1. [tex]\( p(0) \)[/tex]:
[tex]\[ p(0) = 2 + 0 + 2 \cdot 0^2 - 0^3 = 2 \][/tex]

2. [tex]\( p(1) \)[/tex]:
[tex]\[ p(1) = 2 + 1 + 2 \cdot 1^2 - 1^3 = 2 + 1 + 2 - 1 = 4 \][/tex]

3. [tex]\( p(2) \)[/tex]:
[tex]\[ p(2) = 2 + 2 + 2 \cdot 2^2 - 2^3 = 2 + 2 + 8 - 8 = 4 \][/tex]

So, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 4 \)[/tex], and [tex]\( p(2) = 4 \)[/tex].

### (iii) [tex]\( p(z) = z^3 \)[/tex]

1. [tex]\( p(0) \)[/tex]:
[tex]\[ p(0) = 0^3 = 0 \][/tex]

2. [tex]\( p(1) \)[/tex]:
[tex]\[ p(1) = 1^3 = 1 \][/tex]

3. [tex]\( p(2) \)[/tex]:
[tex]\[ p(2) = 2^3 = 8 \][/tex]

So, the values are [tex]\( p(0) = 0 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 8 \)[/tex].

### (iv) [tex]\( p(t) = (t-1)(t+1) \)[/tex]

1. [tex]\( p(0) \)[/tex]:
[tex]\[ p(0) = (0-1)(0+1) = -1 \times 1 = -1 \][/tex]

2. [tex]\( p(1) \)[/tex]:
[tex]\[ p(1) = (1-1)(1+1) = 0 \times 2 = 0 \][/tex]

3. [tex]\( p(2) \)[/tex]:
[tex]\[ p(2) = (2-1)(2+1) = 1 \times 3 = 3 \][/tex]

So, the values are [tex]\( p(0) = -1 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].

### (v) [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]

1. [tex]\( p(0) \)[/tex]:
[tex]\[ p(0) = 0^2 - 3 \cdot 0 + 2 = 2 \][/tex]

2. [tex]\( p(1) \)[/tex]:
[tex]\[ p(1) = 1^2 - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0 \][/tex]

3. [tex]\( p(2) \)[/tex]:
[tex]\[ p(2) = 2^2 - 3 \cdot 2 + 2 = 4 - 6 + 2 = 0 \][/tex]

So, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 0 \)[/tex].

In summary, the calculated values are:

(i) [tex]\( p(x) = x^2 - x + 1 \)[/tex]:
[tex]\[ p(0) = 1, \quad p(1) = 1, \quad p(2) = 3 \][/tex]

(ii) [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 4, \quad p(2) = 4 \][/tex]

(iii) [tex]\( p(z) = z^3 \)[/tex]:
[tex]\[ p(0) = 0, \quad p(1) = 1, \quad p(2) = 8 \][/tex]

(iv) [tex]\( p(t) = (t-1)(t+1) \)[/tex]:
[tex]\[ p(0) = -1, \quad p(1) = 0, \quad p(2) = 3 \][/tex]

(v) [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 0, \quad p(2) = 0 \][/tex]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.