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Question Number 33455 by Rio Mike last updated on 16/Apr/18

Please can someone help with a simplier method of solving this question Q1; Given that the expression x^3 +x^2 −4x +5 and x^3 +3x−7 leave same remainder when divided by (x−a) find the possible values of a

$$\:\:\:\:\:{Please}\:{can}\:{someone}\:{help}\:{with}\:{a} \\ $$$${simplier}\:{method}\:{of}\:{solving}\:{this}\:{question} \\ $$$${Q}\mathrm{1};\:\:\: \\ $$$$\:\:\:{Given}\:{that}\:{the}\:{expression}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{5} \\ $$$${and}\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{7}\:{leave}\:{same}\:{remainder} \\ $$$${when}\:{divided}\:{by}\:\left({x}−{a}\right)\:{find}\:{the}\:{possible} \\ $$$${values}\:{of}\:{a} \\ $$

Commented by prof Abdo imad last updated on 17/Apr/18

x^3 +x^2 −4x +5 =(x−a)Q(x) +r(x) x^3 +3x −7 =(x−a)V(x) +r(x) ⇒ r(a)= a^3 +a^2 −4a +5 =a^3 +3a −7 ⇒ a^2 −4a +5 −3a +7 =0 ⇒a^2 −7a +12 =0 Δ =49 −48 =1 ⇒ a_1 = ((7+1)/2) =4 a_2 =((7−1)/2) =3 .

$${x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\mathrm{5}\:=\left({x}−{a}\right){Q}\left({x}\right)\:+{r}\left({x}\right) \\ $$$${x}^{\mathrm{3}} \:+\mathrm{3}{x}\:−\mathrm{7}\:=\left({x}−{a}\right){V}\left({x}\right)\:+{r}\left({x}\right)\:\Rightarrow \\ $$$${r}\left({a}\right)=\:{a}^{\mathrm{3}} \:+{a}^{\mathrm{2}} \:−\mathrm{4}{a}\:+\mathrm{5}\:={a}^{\mathrm{3}} \:+\mathrm{3}{a}\:−\mathrm{7}\:\Rightarrow \\ $$$${a}^{\mathrm{2}} \:−\mathrm{4}{a}\:+\mathrm{5}\:−\mathrm{3}{a}\:+\mathrm{7}\:=\mathrm{0}\:\Rightarrow{a}^{\mathrm{2}} \:−\mathrm{7}{a}\:+\mathrm{12}\:=\mathrm{0} \\ $$$$\Delta\:=\mathrm{49}\:−\mathrm{48}\:=\mathrm{1}\:\Rightarrow\:{a}_{\mathrm{1}} =\:\frac{\mathrm{7}+\mathrm{1}}{\mathrm{2}}\:=\mathrm{4} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{7}−\mathrm{1}}{\mathrm{2}}\:=\mathrm{3}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Apr/18

a^3 +a^(2 ^ ) −4a+5=a^3 +3a−7 a^2 −7a+12=0 (a−3)(a−4)=0 a=3 or a=4

$${a}^{\mathrm{3}} \:+{a}^{\mathrm{2}\overset{} {\:}} −\mathrm{4}{a}+\mathrm{5}={a}^{\mathrm{3}} +\mathrm{3}{a}−\mathrm{7} \\ $$$${a}^{\mathrm{2}} −\mathrm{7}{a}+\mathrm{12}=\mathrm{0} \\ $$$$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{4}\right)=\mathrm{0} \\ $$$${a}=\mathrm{3}\:{or}\:{a}=\mathrm{4} \\ $$

Answered by ajfour last updated on 16/Apr/18

let remainder =R ⇒ a^3 +a^2 −4a+5=a^3 +3a−7=R ⇒ a^2 −7a+12=0 or (a−3)(a−4)=0 ⇒ a=3,4 .

$${let}\:{remainder}\:={R} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{3}} +{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{5}={a}^{\mathrm{3}} +\mathrm{3}{a}−\mathrm{7}={R} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} −\mathrm{7}{a}+\mathrm{12}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\left({a}−\mathrm{3}\right)\left({a}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{a}=\mathrm{3},\mathrm{4}\:. \\ $$

Commented by Rio Mike last updated on 16/Apr/18

thanks very simple now

$${thanks}\:{very}\:{simple}\:{now} \\ $$

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