Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. 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].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.