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If-x-2-2-is-a-factor-of-the-polynomial-f-x-mx-3-x-2-x-n-find-the-values-of-m-and-n-




Question Number 57000 by pieroo last updated on 28/Mar/19
If (x+2)^2  is a factor of the polynomial  f(x)=mx^3 +x^2 +x+n, find;   the values of m and n.
If(x+2)2isafactorofthepolynomialf(x)=mx3+x2+x+n,find;thevaluesofmandn.
Commented by maxmathsup by imad last updated on 29/Mar/19
⇒f(−2)=f^′ (−2)=0   but f(−2)=−8m+4−2+n =−8m+n+2  f^′ (x)=3mx^2  +2x+1 ⇒f^′ (−2)=12m−4+1 =12m−3 ⇒12m−3=0 and  −8m +n+2 =0 ⇒m=(1/4) and n=0 ⇒f(x)=(1/4) x^3  +x^2  +x .
f(2)=f(2)=0butf(2)=8m+42+n=8m+n+2f(x)=3mx2+2x+1f(2)=12m4+1=12m312m3=0and8m+n+2=0m=14andn=0f(x)=14x3+x2+x.
Answered by Smail last updated on 28/Mar/19
f(−2)=8m+4+2+n  f′(x)=3mx^2 +2x+1  Since (x+2)  is a double factor of f  the (x+2) is  a simple factor of ((df(x))/dx)  Which means  f′(−2)=0⇔12m+4+1=0  m=−(5/(12))  f(−2)=8×(−(5/(12)))+6+n  n=(8/3)  f(x)=((−5)/(12))x^3 +x^2 +x+(8/3)
f(2)=8m+4+2+nf(x)=3mx2+2x+1Since(x+2)isadoublefactoroffthe(x+2)isasimplefactorofdf(x)dxWhichmeansf(2)=012m+4+1=0m=512f(2)=8×(512)+6+nn=83f(x)=512x3+x2+x+83
Commented by MJS last updated on 28/Mar/19
something went wrong
somethingwentwrong
Commented by Smail last updated on 28/Mar/19
I made a mistake on the 5th line  f′(−2)=3m(−2)^2 +2×(−2)+1=0  12m−4+1=0⇔m=(1/4)  f(−2)=(1/4)×(−2)^3 +4−2+n=0  n=0
Imadeamistakeonthe5thlinef(2)=3m(2)2+2×(2)+1=012m4+1=0m=14f(2)=14×(2)3+42+n=0n=0
Answered by mr W last updated on 28/Mar/19
mx^3 +x^2 +x+n=(x+2)^2 (ax+b)  mx^3 +x^2 +x+n=(x^2 +4x+4)(ax+b)  mx^3 +x^2 +x+n=ax^3 +(4a+b)x^2 +4(a+b)x+4b  4a+b=4(a+b)=1  ⇒b=0  ⇒a=(1/4)  ⇒m=a=(1/4)  ⇒n=4b=0  (1/4)x^3 +x^2 +x=(1/4)(x+2)^2 x
mx3+x2+x+n=(x+2)2(ax+b)mx3+x2+x+n=(x2+4x+4)(ax+b)mx3+x2+x+n=ax3+(4a+b)x2+4(a+b)x+4b4a+b=4(a+b)=1b=0a=14m=a=14n=4b=014x3+x2+x=14(x+2)2x
Commented by pete last updated on 29/Mar/19
thanks prof
thanksprof
Answered by malwaan last updated on 29/Mar/19
by long division we have  n−8m+2=0 (1)  12m−3=0 (2)  ⇒m=(1/4) ;  n=0
bylongdivisionwehaven8m+2=0(1)12m3=0(2)m=14;n=0

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