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Question Number 97919 by Lekhraj last updated on 10/Jun/20

Commented by MJS last updated on 10/Jun/20

p(x)=ax^5 +bx^4 +cx^3 +dx^2 +ex+f insert the given pairs (x, p(x)) and solve for the constants ⇒ p(x)=−(x^5 /(40))+((11x^4 )/(24))−((25x^3 )/8)+((241x^2 )/(24))−((287x)/(20))+8 p(7)=8

p(x)=ax5+bx4+cx3+dx2+ex+finsertthegivenpairs(x,p(x))andsolvefortheconstants⇒p(x)=−x540+11x424−25x38+241x224−287x20+8p(7)=8

Commented by mr W last updated on 10/Jun/20

answer correct sir. i got the same using an other method.

answercorrectsir.igotthesameusinganothermethod.

Answered by mr W last updated on 10/Jun/20

an attempt without solving eqn. and even without using a calculator: p(1)=1 ⇒p(x)=a(x)(x−1)+1 p(2)=a(2)(2−1)+1=1 ⇒a(2)=0 ⇒a(x)=b(x)(x−2) ⇒p(x)=b(x)(x−2)(x−1)+1 p(3)=b(3)(3−2)(3−1)+1=2 ⇒b(3)=(1/2) ⇒b(x)=c(x)(x−3)+(1/2) ⇒p(x)=[c(x)(x−3)+(1/2)](x−2)(x−1)+1 ⇒p(4)=[c(4)(4−3)+(1/2)](4−2)(4−1)+1=3 ⇒c(4)=−(1/6) ⇒c(x)=d(x)(x−4)−(1/6) ⇒p(x)={[d(x)(x−4)−(1/6)](x−3)+(1/2)}(x−2)(x−1)+1 p(5)={[d(5)(5−4)−(1/6)](5−3)+(1/2)}(5−2)(5−1)+1=5 ⇒d(5)=(1/(12)) ⇒d(x)=e(x)(x−5)+(1/(12)) ⇒p(x)={[{e(x)(x−5)+(1/(12))}(x−4)−(1/6)](x−3)+(1/2)}(x−2)(x−1)+1 p(6)={[{e(6)(6−5)+(1/(12))}(6−4)−(1/6)](6−3)+(1/2)}(6−2)(6−1)+1=8 ⇒e(6)=−(1/(40)) ⇒e(x)=−(1/(40))=constant, since p(x) is quintic. ⇒p(x)={{{−(1/(40))(x−5)+(1/(12))}(x−4)−(1/6)}(x−3)+(1/2)}(x−2)(x−1)+1 ⇒p(7)={{{−(1/(40))(7−5)+(1/(12))}(7−4)−(1/6)}(7−3)+(1/2)}(7−2)(7−1)+1 ={{{−(1/(20))+(1/(12))}3−(1/6)}4+(1/2)}30+1 ={−(4/(15))+(1/2)}30+1 =−8+15+1 =8

anattemptwithoutsolvingeqn.andevenwithoutusingacalculator:p(1)=1⇒p(x)=a(x)(x−1)+1p(2)=a(2)(2−1)+1=1⇒a(2)=0⇒a(x)=b(x)(x−2)⇒p(x)=b(x)(x−2)(x−1)+1p(3)=b(3)(3−2)(3−1)+1=2⇒b(3)=12⇒b(x)=c(x)(x−3)+12⇒p(x)=[c(x)(x−3)+12](x−2)(x−1)+1⇒p(4)=[c(4)(4−3)+12](4−2)(4−1)+1=3⇒c(4)=−16⇒c(x)=d(x)(x−4)−16⇒p(x)={[d(x)(x−4)−16](x−3)+12}(x−2)(x−1)+1p(5)={[d(5)(5−4)−16](5−3)+12}(5−2)(5−1)+1=5⇒d(5)=112⇒d(x)=e(x)(x−5)+112⇒p(x)={[{e(x)(x−5)+112}(x−4)−16](x−3)+12}(x−2)(x−1)+1p(6)={[{e(6)(6−5)+112}(6−4)−16](6−3)+12}(6−2)(6−1)+1=8⇒e(6)=−140⇒e(x)=−140=constant,sincep(x)isquintic.⇒p(x)={{{−140(x−5)+112}(x−4)−16}(x−3)+12}(x−2)(x−1)+1⇒p(7)={{{−140(7−5)+112}(7−4)−16}(7−3)+12}(7−2)(7−1)+1={{{−120+112}3−16}4+12}30+1={−415+12}30+1=−8+15+1=8

Commented by MJS last updated on 10/Jun/20

great! method using only subtraction and addition: 1 1 2 3 5 8 8 0 1 1 2 3 0 1 0 1 1 −3 −1 1 0 −4 2 −1 −4 −3 −3

great!methodusingonlysubtractionandaddition:11235880112301011−3−110−42−1−4−3−3

Commented by mr W last updated on 10/Jun/20

nice illustration! i didn′t think of this.

niceillustration!ididn′tthinkofthis.

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