Question-212152

Question Number 212152 by RojaTaniya last updated on 04/Oct/24

Answered by som(math1967) last updated on 04/Oct/24

x^3 +ax+b=(x+c)(x−1)(x−2) put x=1 a+b=−1 ....case1 put x=2 2a+b=−8 ∴ a=−7,b=6 now x^3 −7x+6 =(x−1)(x^2 +x−6) =(x−1)(x−2)(x+3) ∴ c=3 log_(a+b+c) (b+c−a)^(a−b−c) log_2 (16)^(−16) [b+c−a=6+3−(−7)=16] =−16×4log_2 2=−64

$$\:{x}^{\mathrm{3}} +{ax}+{b}=\left({x}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right) \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\:{a}+{b}=−\mathrm{1}\:….{case}\mathrm{1} \\ $$$${put}\:{x}=\mathrm{2} \\ $$$$\:\mathrm{2}{a}+{b}=−\mathrm{8} \\ $$$$\therefore\:{a}=−\mathrm{7},{b}=\mathrm{6} \\ $$$${now}\:{x}^{\mathrm{3}} −\mathrm{7}{x}+\mathrm{6} \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{6}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right) \\ $$$$\therefore\:{c}=\mathrm{3} \\ $$$$\:\boldsymbol{{log}}_{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}} \left(\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}\right)^{\boldsymbol{{a}}−\boldsymbol{{b}}−\boldsymbol{{c}}} \\ $$$$\:\boldsymbol{{log}}_{\mathrm{2}} \left(\mathrm{16}\right)^{−\mathrm{16}} \:\left[{b}+{c}−{a}=\mathrm{6}+\mathrm{3}−\left(−\mathrm{7}\right)=\mathrm{16}\right] \\ $$$$=−\mathrm{16}×\mathrm{4}\boldsymbol{{log}}_{\mathrm{2}} \mathrm{2}=−\mathrm{64} \\ $$

Commented by RojaTaniya last updated on 04/Oct/24

$${Sir},\:\left({b}+{c}−{a}\right)=\mathrm{16}, \\ $$$$\:\:{answer}\:{will}\:{be}\:−\mathrm{64} \\ $$

Commented by som(math1967) last updated on 04/Oct/24

$$\:{thank}\:{you}\:,\:{I}\:{corrected} \\ $$

Answered by Rasheed.Sindhi last updated on 04/Oct/24

x^3 +ax+b=(x+c)(x^2 −3x+2) x^3 +ax+b=x^3 −3x^2 +2x+cx^2 −3cx+2c =x^3 +(c−3)x^2 +(2−3c)x+2c c−3=0⇒c=3 a=2−3c=2−3(3)=−7 b=2c=2(3)=6 log_((−7+6+3)) (6+3+7)^(−7−6−3) =log_2 16^(−16) =(−16)log_2 16=−64

$${x}^{\mathrm{3}} +{ax}+{b}=\left({x}+{c}\right)\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\right) \\ $$$${x}^{\mathrm{3}} +{ax}+{b}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+{cx}^{\mathrm{2}} −\mathrm{3}{cx}+\mathrm{2}{c} \\ $$$$\:\:\:={x}^{\mathrm{3}} +\left({c}−\mathrm{3}\right){x}^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{3}{c}\right){x}+\mathrm{2}{c} \\ $$$${c}−\mathrm{3}=\mathrm{0}\Rightarrow{c}=\mathrm{3} \\ $$$${a}=\mathrm{2}−\mathrm{3}{c}=\mathrm{2}−\mathrm{3}\left(\mathrm{3}\right)=−\mathrm{7} \\ $$$${b}=\mathrm{2}{c}=\mathrm{2}\left(\mathrm{3}\right)=\mathrm{6} \\ $$$$\mathrm{log}_{\left(−\mathrm{7}+\mathrm{6}+\mathrm{3}\right)} \left(\mathrm{6}+\mathrm{3}+\mathrm{7}\right)^{−\mathrm{7}−\mathrm{6}−\mathrm{3}} \\ $$$$=\mathrm{log}_{\mathrm{2}} \mathrm{16}^{−\mathrm{16}} =\left(−\mathrm{16}\right)\mathrm{log}_{\mathrm{2}} \mathrm{16}=−\mathrm{64}\:\:\: \\ $$

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