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Question Number 105707 by ZiYangLee last updated on 31/Jul/20

Find a polynomial f(x) which satisfy  the following conditions:  i) f(x) of degree 4  ii) (x−1) is a factor of f(x) and f′(x)  iii) f(0)=3 and f′(0)=−5  iv) when f(x) is divided by (x−2), the         remainder is 13.

Findapolynomialf(x)whichsatisfythefollowingconditions:i)f(x)ofdegree4ii)(x1)isafactoroff(x)andf(x)iii)f(0)=3andf(0)=5iv)whenf(x)isdividedby(x2),theremainderis13.

Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

let  f(x)=(x−1)^2 (ax^2 +bx+3)          = ax^4 +x^3 (b−2a)+x^2 (a−2b+3)+x(b−6)+3  f′(x)=4ax^3 +3x^2 (b−2a)+2x(a−2b+3)+(b−6)  f^′ (0)= −5 ⇒ b−6=−5⇒ b=1     f(2)=13 ⇒ 2a+b=5 ⇒ a=2  ∴ f(x)= 2x^4 −3x^3 +3x^2 −5x+3

letf(x)=(x1)2(ax2+bx+3)=ax4+x3(b2a)+x2(a2b+3)+x(b6)+3f(x)=4ax3+3x2(b2a)+2x(a2b+3)+(b6)f(0)=5b6=5b=1f(2)=132a+b=5a=2f(x)=2x43x3+3x25x+3

Answered by bobhans last updated on 31/Jul/20

f(x)= (x−1)(2x^3 −x^2 +2x−3)∴  let f(x)=(x−1)(ax^3 +bx^2 +cx+d)  (1)f(0) = −d = 3 ; d=−3  (2)f ′(x)= (x−1)(3ax^2 +2bx+c)+(ax^3 +bx^2 +cx−3)  f ′(0)= −c −3=−5 ; c = 2   (3)f ′(1)=0 → a+b+2−3=0; b= 1−a  (4)f(x)=(x−1)(ax^3 +(1−a)x^2 +2x−3)  f(2)= 8a+4−4a+1 = 13 ; 4a = 8 ; a=2  then b = −1

f(x)=(x1)(2x3x2+2x3)letf(x)=(x1)(ax3+bx2+cx+d)(1)f(0)=d=3;d=3(2)f(x)=(x1)(3ax2+2bx+c)+(ax3+bx2+cx3)f(0)=c3=5;c=2(3)f(1)=0a+b+23=0;b=1a(4)f(x)=(x1)(ax3+(1a)x2+2x3)f(2)=8a+44a+1=13;4a=8;a=2thenb=1

Commented by bemath last updated on 31/Jul/20

cooll ....

cooll....

Answered by floor(10²Eta[1]) last updated on 31/Jul/20

f(x)=ax^4 +bx^3 +cx^2 +dx+e  f′(x)=4ax^3 +3bx^2 +2cx+d  by info 2:  f(x)=A(x)(x−1)  ⇒f′(x)=A′(x)(x−1)+A(x), but since   x−1 is a factor of f′(x), so x−1 has to  be a factor of A(x)⇒A(x)=B(x)(x−1)  ⇒f(x)=B(x)(x−1)^2   by info 3 and 4:  f(0)=e=3, f′(0)=d=−5, f(2)=13  ⇒f(x)=ax^4 +bx^3 +cx^2 −5x+3  ⇒f′(x)=4ax^3 +3bx^2 +2cx−5  ⇒f(1)=a+b+c−5+3=0  f′(1)=4a+3b+2c−5=0  and f(2)=16a+8b+4c−10+3=13  (I): a+b+c=2  (II): 4a+3b+2c=5  (III): 16a+8b+4c=20⇒4a+2b+c=5  solving this system of equations we get:  a=2, b=−3, c=3  ⇒f(x)=2x^4 −3x^3 +3x^2 −5x+3.

f(x)=ax4+bx3+cx2+dx+ef(x)=4ax3+3bx2+2cx+dbyinfo2:f(x)=A(x)(x1)f(x)=A(x)(x1)+A(x),butsincex1isafactoroff(x),sox1hastobeafactorofA(x)A(x)=B(x)(x1)f(x)=B(x)(x1)2byinfo3and4:f(0)=e=3,f(0)=d=5,f(2)=13f(x)=ax4+bx3+cx25x+3f(x)=4ax3+3bx2+2cx5f(1)=a+b+c5+3=0f(1)=4a+3b+2c5=0andf(2)=16a+8b+4c10+3=13(I):a+b+c=2(II):4a+3b+2c=5(III):16a+8b+4c=204a+2b+c=5solvingthissystemofequationsweget:a=2,b=3,c=3f(x)=2x43x3+3x25x+3.

Commented by bemath last updated on 31/Jul/20

f(2)=16−8+8−10+3 = 9   wrong sir

f(2)=168+810+3=9wrongsir

Commented by floor(10²Eta[1]) last updated on 31/Jul/20

oh i actually missed the information 4

ohiactuallymissedtheinformation4

Answered by 1549442205PVT last updated on 31/Jul/20

General form of the a polynomyal of   degree 4 is f(x)=ax^4 +bx^3 +cx^2 +dx+e.Then  f ′(x)=4ax^3 +3bx^2 +2cx+d,  From the hypothesis iii)we get:  f(0)=3⇔e=3,f ′(0)=−5⇔d=−5.Hence  f(x)=ax^4 +bx^3 +cx^2 −5x+3  f ′(x)=4ax^3 +3bx^2 +2cx−5  From ii)we get f(1)=0⇔a+b+c−2=0(1)  f ′(1)=0⇔4a+3b+2c−5=0 (2)  From iv)we get f(2)=13⇔16a+8b+4c−7=13(3)  From(1)(2)and (3) we get   { ((a+b+c−2=0)),((4a+3b+2c−5=0)),((16a+8b+4c−7=13)) :}⇔ { ((a+b+c=2   (4))),((4a+3b+2c=5)),((4a+2b+c=5 (6))) :}(5)  Substract (5) and (6) we getb+c=0(7),  replace in (4) we obtain a=2,replace  into (6)we get 2b+c=−3(8).From  (7),(8) we obtain b=−3,c=3.  Thus,f(x)=2x^4 −3x^3 +3x^2 −5x+3

Generalformoftheapolynomyalofdegree4isf(x)=ax4+bx3+cx2+dx+e.Thenf(x)=4ax3+3bx2+2cx+d,Fromthehypothesisiii)weget:f(0)=3e=3,f(0)=5d=5.Hencef(x)=ax4+bx3+cx25x+3f(x)=4ax3+3bx2+2cx5Fromii)wegetf(1)=0a+b+c2=0(1)f(1)=04a+3b+2c5=0(2)Fromiv)wegetf(2)=1316a+8b+4c7=13(3)From(1)(2)and(3)weget{a+b+c2=04a+3b+2c5=016a+8b+4c7=13{a+b+c=2(4)4a+3b+2c=54a+2b+c=5(6)(5)Substract(5)and(6)wegetb+c=0(7),replacein(4)weobtaina=2,replaceinto(6)weget2b+c=3(8).From(7),(8)weobtainb=3,c=3.Thus,f(x)=2x43x3+3x25x+3

Answered by malwaan last updated on 31/Jul/20

f(x)=ax^4 +bx^3 +cx^2 +dx+e  f(1)=a+b+c+d+e=0...1  f′(1)=4a+3b+2c+d=0...2  f(0)=3 ⇒ e=3.....3  f′(0)=−5⇒d=−5....4  f(2)=16a+8b+4c+2d+e=13..5  1⇒a+b+c−5+3=0  ⇒  a+b+c=2....6  2⇒4a+3b+2c−5=0  ⇒ 4a+3b+2c=5....7  5⇒16a+8b+4c−10+3=13  ⇒16a+8b+4c=20  ⇒ 4a+2b+c=5...8  solve the system of equations  6  ; 7 ; 8 we get  c = −3  ; b = 3 ; a = 2  the polynomial is  f(x)= 2x^4  −3x^3 +3x^2  −5x+3

f(x)=ax4+bx3+cx2+dx+ef(1)=a+b+c+d+e=0...1f(1)=4a+3b+2c+d=0...2f(0)=3e=3.....3f(0)=5d=5....4f(2)=16a+8b+4c+2d+e=13..51a+b+c5+3=0a+b+c=2....624a+3b+2c5=04a+3b+2c=5....7516a+8b+4c10+3=1316a+8b+4c=204a+2b+c=5...8solvethesystemofequations6;7;8wegetc=3;b=3;a=2thepolynomialisf(x)=2x43x3+3x25x+3

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