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p-x-x-1-2-Q-x-3x-8-p-x-x-2-Q-x-R-R-




Question Number 144060 by Khalmohmmad last updated on 21/Jun/21
p(x)=(x−1)^2 Q(x)+3x+8  p(x)=(x−2)Q(x)+R  R=?
p(x)=(x1)2Q(x)+3x+8p(x)=(x2)Q(x)+RR=?
Commented by Rasheed.Sindhi last updated on 21/Jun/21
Q(x) is same in both cases!!!?
Q(x)issameinbothcases!!!?
Answered by Rasheed.Sindhi last updated on 21/Jun/21
p(x)=(x−1)^2 Q_1 (x)+3x+8.....(i)  p(x)=(x−2)Q_2 (x)+R.....(ii)   ;  R=?  ^• Let Q_1 (x)=q(constant  p(x)=(x−1)^2 q+3x+8          =qx^2 −2xq+q+3x+8          =qx^2 +(3−2q)x+q+8  When divided by x−2          R=q(2)^2 +(3−2q)(2)+q+8          R=4q+6−4q+q+8              =q+14  ^(• ) Q_1 (x)=ax+b      p(x)=(x−1)^2 Q_1 (x)+3x+8  p(x)=(x−1)^2 (ax+b)+3x+8          =(x^2 −2x+1)(ax+b)+3x+8     =ax^3 +bx^2 −2ax^2 −2bx+ax+b+3x+8    =ax^3 +(b−2a)x^2 +(a+3−2b)x+b+8  When divided by x−2  R=p(2)=a(2)^3 +(b−2a)(2)^2 +(a+3−2b)(2)+b+8  =8a+4b−8a+2a+6−4b+b+8  =2a+b+14  No unique answer.  R depends upon Q_1 (x).
p(x)=(x1)2Q1(x)+3x+8..(i)p(x)=(x2)Q2(x)+R..(ii);R=?LetQ1(x)=q(constantp(x)=(x1)2q+3x+8=qx22xq+q+3x+8=qx2+(32q)x+q+8Whendividedbyx2R=q(2)2+(32q)(2)+q+8R=4q+64q+q+8=q+14Q1(x)=ax+bp(x)=(x1)2Q1(x)+3x+8p(x)=(x1)2(ax+b)+3x+8=(x22x+1)(ax+b)+3x+8=ax3+bx22ax22bx+ax+b+3x+8=ax3+(b2a)x2+(a+32b)x+b+8Whendividedbyx2R=p(2)=a(2)3+(b2a)(2)2+(a+32b)(2)+b+8=8a+4b8a+2a+64b+b+8=2a+b+14Nouniqueanswer.RdependsuponQ1(x).
Answered by Rasheed.Sindhi last updated on 21/Jun/21
p(x)=(x−1)^2 Q_1 (x)+3x+8  p(x)=(x−2)Q_2 (x)+R  ;  R=?  p(x) when is divided by x−2 giving   remainder R  (x−1)^2 Q_1 (x)+3x+8  is divided by  x−2 giving remainder R  R=p(2)=(2−1)^2 Q_1 (2)+3(2)+8  R=p(2)=Q_1 (2)+14  ^• Let  Q_1 (x)=q (constant)  R=q+14  ^• Let Q_1 (x)=ax+b  R=Q_1 (2)+14=a(2)+b+14            R=2a+b+14  ^• Let Q_1 (x)=ax^2 +bx+c      R=a(2)^2 +b(2)+c+14         R=4a+2b+c+14  R depends on Q_1 (x) (or Q_2 (x) ).  It′s not unique.
p(x)=(x1)2Q1(x)+3x+8p(x)=(x2)Q2(x)+R;R=?p(x)whenisdividedbyx2givingremainderR(x1)2Q1(x)+3x+8isdividedbyx2givingremainderRR=p(2)=(21)2Q1(2)+3(2)+8R=p(2)=Q1(2)+14LetQ1(x)=q(constant)R=q+14LetQ1(x)=ax+bR=Q1(2)+14=a(2)+b+14R=2a+b+14LetQ1(x)=ax2+bx+cR=a(2)2+b(2)+c+14R=4a+2b+c+14RdependsonQ1(x)(orQ2(x)).Itsnotunique.

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