Math, asked by sammanidanaiah816, 1 month ago

2. Find p(0),p(1) and p(2) for each of the following polynomials. p(x) = x2-x+1 (ii) p(v) = 2 + y + 2y2 - y3 (iii) p(z)=23 (iv) p(t) = (1 - 1) (t + 1) (v) p(x) = x2 – 3x+2 ​

Answers

Answered by JSP2008
5

i) p(x) = x²-x+1

=> P(0) = 0²-0+1

=> 0-0+1

=> 1

=>P(1) = 1²-1+1

=> 1 - 1 + 1

=> 1

=> P(2) = 2²-2+1

=> 4 - 2 + 1

=> 3

ii) p(y) = 2 + y + 2y² - y³

=>p(0) = 2+0+2(0)²-0³

=>2 + 0 + 0 - 0

=> 2

=>p(1) = 2 + 1 + 2(1)² - 1³

=> 3 + 2(1) - 1

=> 3 + 2 - 1

=> 4

=> p(2) = 2 + 2 + 2(2)² - 2³

=> 4 + 2(4) - 8

=> 4 + 8 - 8

=> 4

iii) p(z)=2³ – this question is incomplete.

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

=>p(0) = (1-1)(0+1)

=> (0)(1)

=> 0

=>p(1) = (1-1)(1+1)

=> (0)(2)

=> 0

=>p(2) = (1-1)(2-1)

=> (0)(1)

=> 1

v) p(x) = x²– 3x+2

=> p(0) = 0² - 3(0)+2

=> 0 - 0 + 2

=> 2

=>p(1)= 1² - 3(1) + 2

=> 1 - 3 + 2

=> 0

=>p(2)=2² - 3(2) + 2

=> 4 - 6 + 2

=> 0

HOPE IT HELPS!!!

Answered by sritharina3
2

Answer:

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Step-by-step explanation:

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