Question

P (x) =x³ P(1) =(1) ³ P(0) =(0) ³=0 P(-2) =(-2) ³=-8​

Answer 1

Answer:

Given P(1)=1,P(2)=2,P(3)=3,P(4)=5

Let f(x)=(P(x)−x)

f(1)=P(1)−1=1−1=0

f(2)=P(2)−2=2−2=0

f(3)=P(3)−3=3−3=0

∴f(x)=0,x=1,2,3

⇒f(x)=a(x−1)(x−2)(x−3)

P(x)=a(x−1)(x−2)(x−3)+x

Put x=4

5=a(3)(2)(1)+4

⇒a=

6

1

∴P(x)=

6

1

(x−1)(x−2)(x−3)+x

∴P(6)=

6

1

(5)(4)(3)+6=16

Answer 2

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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