Question

. Find p(0),p(1) and p(2) for each of the following polynomials.
(1) p(x) = x^2-x+1
(2) p(y)=2 + y + 2y^2 - y^3
(3) p(z)=z^3
(4) p(t) = (t - 1) (t+1)
(5) P(x) = x^2 -3x+2​

Answer 1

Step-by-step explanation:

(i) p(x) = x² - x + 1

⇒ p(0) = (0)² - (0) + 1

= 0 - 0 + 1

= 1

⇒ p(1) = (1)² - (1) + 1

= 1 - 1 + 1

= 1

⇒ p(2) = (2)² - 2 + 1

= 4 - 2 + 1

= 3

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(ii) p(y) = 2 + y + 2y² - y³

= -y³ + 2y² + y + 2

⇒ p(0) = (-0)³ + 2(0)² + 0 + 2

= -0 + 0 + 0 + 2

= 2

⇒ p(1) = (-1)³ + 2(1)² + 1 + 2

= -1 + 2 + 1 + 2

= 4

⇒ p(2) = (-2)³ + 2(2)² + 2

= -8 + 8 + 2 + 2

= 4

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(iii) p(z) = z³

⇒ p(0) = (0)³

= 0

⇒ p(1) = (1)³

= 1

⇒ p(2) = (2)³

= 8

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(iv) p(t) = (t - 1)(t + 1)

⇒ p(0) = (0 - 1)(0 + 1)

= (-1)(1)

= -1

⇒ p(1) = (1 - 1)(1 + 1)

= (0)(2)

= 0

⇒ p(2) = (2 - 1)(2 + 1)

= (1)(3)

= 3

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(v) p(x) = x² - 3x + 2

⇒ p(0) = (0)² - 3(0) + 2

= 0 - 0 + 2

= 2

⇒ p(1) = (1)² - 3(1) + 2

= 1 - 3 + 2

= 0

⇒ p(2) = (2)² - 3(2) + 2

= 4 - 6 + 2

= 0

Answer 2

Step-by-step explanation:

1) 1 , 1 , 3

2) 2 , 2, 4

3) 0 , 1 , 8

4) -1 , 0 , 3

5) 2 , 0 , 0