Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

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

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

(iii) [tex]p(x)=x^3[/tex]

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

Sagot :

GinnyAnswer
Sure! Let's evaluate the given polynomials at specified points:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Now summarizing all the evaluated results:

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

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

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

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

These are the desired function values for each polynomial at [tex]\( 0 \)[/tex], [tex]\( 1 \)[/tex], and [tex]\( 2 \)[/tex].
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.