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If-2x-2-11x-5-is-a-factor-of-ax-3-17x-2-bx-15-then-what-is-a-and-b-

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Question Number 157416 by cortano last updated on 23/Oct/21
If 2x^2 +11x+5 is a factor of ax^3 −17x^2 +bx−15 , then what is a and b ?
If2x2+11x+5isafactorofax3−17x2+bx−15,thenwhatisaandb?
Commented by Rasheed.Sindhi last updated on 23/Oct/21
ax^3 −17x^2 +bx−15 has afactor 2x^2 +11x+5; a=?,b=? _(−) Other factor=(a/2)x+((−15)/5)=(a/2)x−3 ▶(2x^2 +11x+5)((a/2)x−3) =ax^3 −17x^2 +bx−15 ▶ax^3 +((11a)/2)x^2 +((5a)/2)x−6x^2 −33x−15 =ax^3 −17x^2 +bx−15 ▶ax^3 +(((11a)/2)−6)x^2 +(((5a)/2)−33)x−15 =ax^3 −17x^2 +bx−15 Comparing coefficients: ((11a)/2)−6=−17⇒a=((−22)/(11))=−2 ((5a)/2)−33=b⇒b=((5(−2)−66)/2)=−38
ax3−17x2+bx−15hasafactor2x2+11x+5;a=?,b=?−Otherfactor=a2x+−155=a2x−3▸(2x2+11x+5)(a2x−3)=ax3−17x2+bx−15▸ax3+11a2x2+5a2x−6x2−33x−15=ax3−17x2+bx−15▸ax3+(11a2−6)x2+(5a2−33)x−15=ax3−17x2+bx−15Comparingcoefficients:11a2−6=−17⇒a=−2211=−25a2−33=b⇒b=5(−2)−662=−38
Commented by Tawa11 last updated on 23/Oct/21
Weldone sir
Weldonesir
Commented by Rasheed.Sindhi last updated on 23/Oct/21
Thanks miss! I like your encouraging!
Thanksmiss!Ilikeyourencouraging!
Answered by Rasheed.Sindhi last updated on 23/Oct/21
2x^2 +11x+5 is factor of ax^3 −17x^2 +bx−15; a=?,b=?_(−) Let the other factor is cx+d ▶(2x^2 +11x+5)(cx+d) =ax^3 −17x^2 +bx−15 ▶2cx^3 +(11c+2d)x^2 +(5c+11d)x+5d =ax^3 −17x^2 +bx−15 Comparing coefficients: ▶2c=a_((i)) , 11c+2d_( (ii) ) =−17, 5c+11d=b_((iii)) , 5d=−15_((iv)) ▶(iv)⇒d=((−15)/5)=−3, (ii)⇒11c+2(−3)=−17⇒c=−1 (i)⇒a=2(−1)=−2 (iii)⇒b=5(−1)+11(−3)=−38
2x2+11x+5isfactorofax3−17x2+bx−15;a=?,b=?−Lettheotherfactoriscx+d▸(2x2+11x+5)(cx+d)=ax3−17x2+bx−15▸2cx3+(11c+2d)x2+(5c+11d)x+5d=ax3−17x2+bx−15Comparingcoefficients:▸2c=a⏟(i),11c+2d⏟=−17(ii),5c+11d=b⏟(iii),5d=−15⏟(iv)▸(iv)⇒d=−155=−3,(ii)⇒11c+2(−3)=−17⇒c=−1(i)⇒a=2(−1)=−2(iii)⇒b=5(−1)+11(−3)=−38
Commented by cortano last updated on 23/Oct/21
yes sir. i got the same result
yessir.igotthesameresult
Commented by cortano last updated on 23/Oct/21
Answered by Rasheed.Sindhi last updated on 23/Oct/21
P(x)=ax^3 −17x^2 +bx−15 D(x)=2x^2 +11x+5 ; a=?,b=? _(−) ▶Two roots of P(x)=0 are also roots of D(x): 2x^2 +11x+5=0 (x+5)(2x+1)=0 x=−5 ∣ x=−(1/2) ▶Let third root is α P(x)=ax^3 −17x^2 +bx−15 =(x+5)(x+(1/2))(x−α) ▶This is an identity so it′s correct for every value of x: ⧫ x=−5 (LHS=0 for x=−5) a(−5)^3 −17(−5)^2 +b(−5)−15=0 −125a−425−5b−15=0 25a+85+b+3=0 25a+b=−88.................(i) ⧫ x=−1/2 (LHS=0 for x=−1/2) a(−(1/2))^3 −17(−(1/2))^2 +b(−(1/2))−15=0 −(1/8)a−((17)/4)−(1/2)b−15=0 a+34+4b+120=0 a+4b=−154..................(ii) ▶4(i)−(ii)⇒99a=−352+154=−198 a=−2 b=−88−25a=−88−25(−2)=−38 b=−38
P(x)=ax3−17x2+bx−15D(x)=2x2+11x+5;a=?,b=?−▸TworootsofP(x)=0arealsorootsofD(x):2x2+11x+5=0(x+5)(2x+1)=0x=−5∣x=−12▸LetthirdrootisαP(x)=ax3−17x2+bx−15=(x+5)(x+12)(x−α)▸Thisisanidentitysoit′scorrectforeveryvalueofx:⧫x=−5(LHS=0forx=−5)a(−5)3−17(−5)2+b(−5)−15=0−125a−425−5b−15=025a+85+b+3=025a+b=−88……………..(i)⧫x=−1/2(LHS=0forx=−1/2)a(−12)3−17(−12)2+b(−12)−15=0−18a−174−12b−15=0a+34+4b+120=0a+4b=−154………………(ii)▸4(i)−(ii)⇒99a=−352+154=−198a=−2b=−88−25a=−88−25(−2)=−38b=−38
Answered by ajfour last updated on 23/Oct/21
2ax^3 +11ax^2 +5ax +2px^2 +11px+5p =2ax^3 −34x^2 +2bx−30 ⇒ 11a+2p=−34 5a+11p=2b 5p=−30 so 11a=−22 ⇒ a=−2 2b=−66−10 ⇒ b=−38 (.^� .^� _(-_(∣ ∣) ) ^(⌢) ) (._( (√)) ._( (√)) )∫
2ax3+11ax2+5ax+2px2+11px+5p=2ax3−34x2+2bx−30⇒11a+2p=−345a+11p=2b5p=−30so11a=−22⇒a=−22b=−66−10⇒b=−38(.^.^⌢−∣∣)(..)∫

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