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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 ?

$$\:{If}\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\:{is}\:{a}\:{factor}\:{of} \\ $$$$\:{ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15}\:,\:{then}\:{what} \\ $$$${is}\:{a}\:{and}\:{b}\:? \\ $$

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

$${ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15}\:{has}\:{afactor} \\ $$$$\underset{−} {\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5};\:\:{a}=?,{b}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} \\ $$$${Other}\:{factor}=\frac{{a}}{\mathrm{2}}{x}+\frac{−\mathrm{15}}{\mathrm{5}}=\frac{{a}}{\mathrm{2}}{x}−\mathrm{3} \\ $$$$\:\blacktriangleright\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\right)\left(\frac{{a}}{\mathrm{2}}{x}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$$\blacktriangleright{ax}^{\mathrm{3}} +\frac{\mathrm{11}{a}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{5}{a}}{\mathrm{2}}{x}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{33}{x}−\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$$\blacktriangleright{ax}^{\mathrm{3}} +\left(\frac{\mathrm{11}{a}}{\mathrm{2}}−\mathrm{6}\right){x}^{\mathrm{2}} +\left(\frac{\mathrm{5}{a}}{\mathrm{2}}−\mathrm{33}\right){x}−\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$${Comparing}\:{coefficients}: \\ $$$$\frac{\mathrm{11}{a}}{\mathrm{2}}−\mathrm{6}=−\mathrm{17}\Rightarrow{a}=\frac{−\mathrm{22}}{\mathrm{11}}=−\mathrm{2} \\ $$$$\frac{\mathrm{5}{a}}{\mathrm{2}}−\mathrm{33}={b}\Rightarrow{b}=\frac{\mathrm{5}\left(−\mathrm{2}\right)−\mathrm{66}}{\mathrm{2}}=−\mathrm{38} \\ $$

Commented by Tawa11 last updated on 23/Oct/21

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Oct/21

Thanks miss! I like your encouraging!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{miss}! \\ $$$$\mathrm{I}\:\mathrm{like}\:\mathrm{your}\:\mathrm{encouraging}! \\ $$

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

$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\:\:\mathrm{is}\:\mathrm{factor}\:\mathrm{of} \\ $$$$\underset{−} {{ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15};\:\:{a}=?,{b}=?} \\ $$$${Let}\:{the}\:{other}\:{factor}\:{is}\:{cx}+{d} \\ $$$$\blacktriangleright\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\right)\left({cx}+{d}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$$\blacktriangleright\mathrm{2}{cx}^{\mathrm{3}} +\left(\mathrm{11}{c}+\mathrm{2}{d}\right){x}^{\mathrm{2}} +\left(\mathrm{5}{c}+\mathrm{11}{d}\right){x}+\mathrm{5}{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$${Comparing}\:{coefficients}: \\ $$$$\blacktriangleright\underset{\left({i}\right)} {\underbrace{\mathrm{2}{c}={a}}}\:,\:\underset{\:\:\:\:\:\:\:\:\:\:\:\left({ii}\right)\:\:} {\underbrace{\mathrm{11}{c}+\mathrm{2}{d}}=−\mathrm{17}}, \\ $$$$\:\:\:\:\:\:\underset{\left({iii}\right)} {\underbrace{\mathrm{5}{c}+\mathrm{11}{d}={b}}}\:,\:\underset{\left({iv}\right)} {\underbrace{\mathrm{5}{d}=−\mathrm{15}}} \\ $$$$\blacktriangleright\left({iv}\right)\Rightarrow{d}=\frac{−\mathrm{15}}{\mathrm{5}}=−\mathrm{3}, \\ $$$$\:\:\:\:\left({ii}\right)\Rightarrow\mathrm{11}{c}+\mathrm{2}\left(−\mathrm{3}\right)=−\mathrm{17}\Rightarrow{c}=−\mathrm{1} \\ $$$$\:\:\:\:\:\left({i}\right)\Rightarrow{a}=\mathrm{2}\left(−\mathrm{1}\right)=−\mathrm{2} \\ $$$$\:\:\:\:\left({iii}\right)\Rightarrow{b}=\mathrm{5}\left(−\mathrm{1}\right)+\mathrm{11}\left(−\mathrm{3}\right)=−\mathrm{38} \\ $$

Commented by cortano last updated on 23/Oct/21

yes sir. i got the same result

$${yes}\:{sir}.\:{i}\:{got}\:{the}\:{same}\:{result} \\ $$

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

$${P}\left({x}\right)={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$$\underset{−} {{D}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}\:;\:{a}=?,{b}=?\:\:\:\:\:\:\:\:\:\:} \\ $$$$\blacktriangleright\mathcal{T}{wo}\:{roots}\:{of}\:{P}\left({x}\right)=\mathrm{0}\:{are}\:{also}\:{roots} \\ $$$${of}\:{D}\left({x}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}+\mathrm{5}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\mid\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\blacktriangleright{Let}\:{third}\:{root}\:{is}\:\alpha \\ $$$${P}\left({x}\right)={ax}^{\mathrm{3}} −\mathrm{17}{x}^{\mathrm{2}} +{bx}−\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({x}+\mathrm{5}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\alpha\right) \\ $$$$\:\blacktriangleright\mathcal{T}{his}\:{is}\:{an}\:{identity}\:{so}\:{it}'{s}\:{correct} \\ $$$${for}\:{every}\:{value}\:{of}\:{x}:\: \\ $$$$\blacklozenge\:\:\:{x}=−\mathrm{5}\:\:\:\:\:\:\:\left({LHS}=\mathrm{0}\:{for}\:{x}=−\mathrm{5}\right) \\ $$$${a}\left(−\mathrm{5}\right)^{\mathrm{3}} −\mathrm{17}\left(−\mathrm{5}\right)^{\mathrm{2}} +{b}\left(−\mathrm{5}\right)−\mathrm{15}=\mathrm{0} \\ $$$$−\mathrm{125}{a}−\mathrm{425}−\mathrm{5}{b}−\mathrm{15}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{25}{a}+\mathrm{85}+{b}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{25}{a}+{b}=−\mathrm{88}.................\left({i}\right) \\ $$$$\blacklozenge\:\:\:{x}=−\mathrm{1}/\mathrm{2}\:\:\:\:\left({LHS}=\mathrm{0}\:{for}\:{x}=−\mathrm{1}/\mathrm{2}\right) \\ $$$$\:{a}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{17}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{b}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{15}=\mathrm{0} \\ $$$$\:\:−\frac{\mathrm{1}}{\mathrm{8}}{a}−\frac{\mathrm{17}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{b}−\mathrm{15}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{a}+\mathrm{34}+\mathrm{4}{b}+\mathrm{120}=\mathrm{0} \\ $$$$\:\:\:\:\:{a}+\mathrm{4}{b}=−\mathrm{154}..................\left({ii}\right) \\ $$$$\blacktriangleright\mathrm{4}\left({i}\right)−\left({ii}\right)\Rightarrow\mathrm{99}{a}=−\mathrm{352}+\mathrm{154}=−\mathrm{198} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=−\mathrm{2} \\ $$$${b}=−\mathrm{88}−\mathrm{25}{a}=−\mathrm{88}−\mathrm{25}\left(−\mathrm{2}\right)=−\mathrm{38} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=−\mathrm{38} \\ $$

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 (.^� .^� _(-_(∣ ∣) ) ^(⌢) ) (._( (√)) ._( (√)) )∫

$$\mathrm{2}{ax}^{\mathrm{3}} +\mathrm{11}{ax}^{\mathrm{2}} +\mathrm{5}{ax} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{px}^{\mathrm{2}} +\mathrm{11}{px}+\mathrm{5}{p} \\ $$$$=\mathrm{2}{ax}^{\mathrm{3}} −\mathrm{34}{x}^{\mathrm{2}} +\mathrm{2}{bx}−\mathrm{30} \\ $$$$\Rightarrow\:\:\mathrm{11}{a}+\mathrm{2}{p}=−\mathrm{34} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{5}{a}+\mathrm{11}{p}=\mathrm{2}{b} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{5}{p}=−\mathrm{30} \\ $$$${so}\:\:\:\mathrm{11}{a}=−\mathrm{22}\:\:\Rightarrow\:\:{a}=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{b}=−\mathrm{66}−\mathrm{10}\:\Rightarrow\:\:{b}=−\mathrm{38} \\ $$$$\:\:\left(\underset{\underset{\mid\:\:\mid} {-}} {\overset{\frown} {\hat {.}\:\:\hat {.}}}\right) \\ $$$$\:\:\left(\underset{\:\sqrt{}} {.}\:\:\underset{\:\sqrt{}} {.}\:\right)\int \\ $$$$ \\ $$

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