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Question Number 194853 by horsebrand11 last updated on 17/Jul/23

Answered by cortano12 last updated on 17/Jul/23

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Commented by horsebrand11 last updated on 17/Jul/23

Answered by Rasheed.Sindhi last updated on 17/Jul/23

Let P(x)=8x^3 +16x^2 +kx+l                          =(2x−3)(4x^2 +ax+b)  8x^3 +16x^2 +kx+l=8x^3 +(2a−12)x^2 +(2b−3a)x−3b  Comparing coefficints:  2a−12=16⇒a=14  2b−3a=k⇒k=2b−42  −3b=l⇒b=−(l/3)  k=2(−(l/3))−42  3k+2l=−126................(i)   Similarly,  let P(x)=8x^3 +16x^2 +kx+l=(2x+1)(4x^2 +cx+d)+12  8x^3 +16x^2 +kx+l  =8x^3 +(2c+4)x^2 +(2d+c)x+d+12  Comparing coefficients:  2c+4=16⇒c=6  2d+c=k⇒2d+6=k  d+12=l⇒d=l−12  2d+6=k⇒2(l−12)+6=k  k−2l=−18.................(ii)  (i)+(ii):4k=−144⇒k=−36  (i)⇒−36−2l=−18⇒l=−9

LetP(x)=8x3+16x2+kx+l=(2x3)(4x2+ax+b)8x3+16x2+kx+l=8x3+(2a12)x2+(2b3a)x3bComparingcoefficints:2a12=16a=142b3a=kk=2b423b=lb=l3k=2(l3)423k+2l=126................(i)Similarly,letP(x)=8x3+16x2+kx+l=(2x+1)(4x2+cx+d)+128x3+16x2+kx+l=8x3+(2c+4)x2+(2d+c)x+d+12Comparingcoefficients:2c+4=16c=62d+c=k2d+6=kd+12=ld=l122d+6=k2(l12)+6=kk2l=18.................(ii)(i)+(ii):4k=144k=36(i)362l=18l=9

Commented by horsebrand11 last updated on 17/Jul/23

Answered by Rasheed.Sindhi last updated on 17/Jul/23

By synthetic division   determinant (((−(1/2)),8,(16),k,l),( , ,(−4),(−6),(−(k/2)+3)),( ,8,(12),(k−6),(l−(k/2)+3=12)))   2l−k=18........(i)   determinant (((3/2),8,(16),k,l),( , ,(12),(42),(((3k)/2)+63)),( ,8,(28),(k+42),(l+((3k)/2)+63=0)))  2l+3k=−126........(ii)  (ii)−(i):4k=−144⇒k=−36  2l−(−36)=18⇒l=−9

Bysyntheticdivision12816kl46k2+3812k6lk2+3=122lk=18........(i)32816kl12423k2+63828k+42l+3k2+63=02l+3k=126........(ii)(ii)(i):4k=144k=362l(36)=18l=9

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