Question

Find p(0),p(1) and p(2) for each of the following polynomial
(i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t -
(iii) p(x)= x
(iv) p(x) = (x - 1)(x + 1)​

Answer 1

ANSWER

(i) p(y)=y2−y+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

(ii) p(t)=2+t+2t2−t3

p(0)=2+(0)+2(0)2−t3=2+0+0−0=2

p(1)=2+(1)+2(2)2−(1)3=2+1+2−1=4

p(2)=2+(2)+2(2)2−(2)3=2+2+8−8=4

(iii) p(x)=x3

p(0)=(0)3=0

p(1)=(1)3=1

p(2)=(2)3=8

(iv) p(x)=(x−1)(x+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

Explanation:

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