Menu Close

Question-136693

Tinku Tara 0 Comments
Pin




Question Number 136693 by liberty last updated on 25/Mar/21
Commented by Olaf last updated on 25/Mar/21
f(x) = mx^3 +5x^2 +kx−18 = (6−x−x^2 )(−mx−3) =(x^2 +x−6)(mx+3) = mx^3 +(3+m)x^2 +(3−6m)x−18 By identification : { ((3+m = 5 (1))),((3−6m = k (2))) :} (1) : m = 2 (2) : k = 3−6m = −9 then f(x) = 2x^3 +5x^2 −9x−18 = (6−x−x^2 )(−2x−3)
f(x)=mx3+5x2+kx−18=(6−x−x2)(−mx−3)=(x2+x−6)(mx+3)=mx3+(3+m)x2+(3−6m)x−18Byidentification:{3+m=5(1)3−6m=k(2)(1):m=2(2):k=3−6m=−9thenf(x)=2x3+5x2−9x−18=(6−x−x2)(−2x−3)
Answered by Rasheed.Sindhi last updated on 25/Mar/21
6−x−x^2 =−(x+3)(x−2) ∴ x+3 & x−2 are also factors of f(x) By synthetic division determinant (((2)),m,5,k,(−18)),(,,(2m),(4m+10),(8m+2k+20)),((−3)),m,(2m+5),(4m+k+10),(8m+2k+2=0_(4m+k=−1) )),(,,(−3m),(3m−15),),(,m,(−m+5),(7m+k−5=0),)) 4m+k=−1 7m+k=5 3m=6⇒m=2 k=5−7m=5−7(2)=−9
6−x−x2=−(x+3)(x−2)∴x+3&x−2arealsofactorsoff(x)Bysyntheticdivision2)m5k−182m4m+108m+2k+20−3)m2m+54m+k+108m+2k+2=04m+k=−1−3m3m−15m−m+57m+k−5=04m+k=−17m+k=53m=6⇒m=2k=5−7m=5−7(2)=−9
Answered by bramlexs22 last updated on 25/Mar/21
method(1) by Remainder theorem consider 6−x−x^2 = 0 or x^2 +x−6=0 we get (x+3)(x−2)=0⇒x=−3, 2 since 6−x−x^2 is a factor of polynomes f(x) so f(−3) and f(2) is zeros to f(x). f(2)=8m+20+2k−18=0 8m+2k=−2→4m+k=−1...(i) f(−3)=−27m+45−3k−18=0 −27m−3k=−27→9m+k=9...(ii) eliminate (i)&(ii) gives 5m = 10 { ((m=2)),((k=−9)) :}
method(1)byRemaindertheoremconsider6−x−x2=0orx2+x−6=0weget(x+3)(x−2)=0⇒x=−3,2since6−x−x2isafactorofpolynomesf(x)sof(−3)andf(2)iszerostof(x).f(2)=8m+20+2k−18=08m+2k=−2→4m+k=−1…(i)f(−3)=−27m+45−3k−18=0−27m−3k=−27→9m+k=9…(ii)eliminate(i)&(ii)gives5m=10{m=2k=−9

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *