Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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 hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.